A English
Mix Examples - Probability Questions in English
Class 9 Mathematics · Probability · Mix Examples - Probability
78+
Questions
English
Language
100%
With Solutions
Showing 50 of 78 questions in English
1
EasyMCQ
$A$ die is thrown $1000$ times and the outcomes were recorded as follows:
If the die is thrown once more,then the probability that it shows $5$ is:
| Outcome | $1$ | $2$ | $3$ | $4$ | $5$ | $6$ |
| Frequency | $180$ | $150$ | $160$ | $170$ | $150$ | $190$ |
If the die is thrown once more,then the probability that it shows $5$ is:
A
$\frac{3}{20}$
B
$\frac{9}{50}$
C
$\frac{4}{25}$
D
$\frac{7}{25}$
Solution
(A) The total number of trials is $1000$.
The frequency of the outcome $5$ is $150$.
The probability of an event is given by the formula: $P(E) = \frac{\text{Number of trials in which the event occurred}}{\text{Total number of trials}}$.
Therefore,the probability of getting $5$ is $P(5) = \frac{150}{1000}$.
Simplifying the fraction: $P(5) = \frac{15}{100} = \frac{3}{20}$.
Thus,the correct option is $A$.
The frequency of the outcome $5$ is $150$.
The probability of an event is given by the formula: $P(E) = \frac{\text{Number of trials in which the event occurred}}{\text{Total number of trials}}$.
Therefore,the probability of getting $5$ is $P(5) = \frac{150}{1000}$.
Simplifying the fraction: $P(5) = \frac{15}{100} = \frac{3}{20}$.
Thus,the correct option is $A$.
0 likes
View Solution2
MediumMCQ
In a sample study of $642$ people,it was found that $514$ people have a high school certificate. If a person is selected at random,the probability that the person has a high school certificate is:
A
$0.5$
B
$0.8$
C
$0.7$
D
$0.6$
Solution
(B) The empirical probability $P(E)$ of an event $E$ is defined as:
$P(E) = \frac{\text{Number of trials in which the event happened}}{\text{Total number of trials}}$
Let $E$ be the event that a person selected at random has a high school certificate.
Given:
Number of people with a high school certificate = $514$
Total number of people in the sample study = $642$
Therefore,the probability $P(E)$ is:
$P(E) = \frac{514}{642}$
Calculating the value:
$P(E) \approx 0.8006$
Rounding to one decimal place,we get $0.8$.
$P(E) = \frac{\text{Number of trials in which the event happened}}{\text{Total number of trials}}$
Let $E$ be the event that a person selected at random has a high school certificate.
Given:
Number of people with a high school certificate = $514$
Total number of people in the sample study = $642$
Therefore,the probability $P(E)$ is:
$P(E) = \frac{514}{642}$
Calculating the value:
$P(E) \approx 0.8006$
Rounding to one decimal place,we get $0.8$.
0 likes
View Solution3
MediumMCQ
In a survey of $364$ children aged $19-36$ months,it was found that $91$ liked to eat potato chips. If a child is selected at random,the probability that he/she does not like to eat potato chips is:
A
$0.25$
B
$0.50$
C
$0.75$
D
$0.80$
Solution
(C) Total number of children in the survey $= 364$.
Number of children who like to eat potato chips $= 91$.
Number of children who do not like to eat potato chips $= 364 - 91 = 273$.
Let $E$ be the event that a randomly selected child does not like to eat potato chips.
The probability $P(E)$ is given by the ratio of the number of favorable outcomes to the total number of outcomes.
$P(E) = \frac{273}{364}$.
Dividing both numerator and denominator by $91$,we get $P(E) = \frac{3}{4} = 0.75$.
Number of children who like to eat potato chips $= 91$.
Number of children who do not like to eat potato chips $= 364 - 91 = 273$.
Let $E$ be the event that a randomly selected child does not like to eat potato chips.
The probability $P(E)$ is given by the ratio of the number of favorable outcomes to the total number of outcomes.
$P(E) = \frac{273}{364}$.
Dividing both numerator and denominator by $91$,we get $P(E) = \frac{3}{4} = 0.75$.
0 likes
View Solution4
EasyMCQ
In a medical examination of students of a class,the following blood groups are recorded:
$A$ student is selected at random from the class. The probability that he/she has blood group $B$,is:
| Blood group | $A, AB, B, O$ |
| Number of students | $10, 13, 12, 5$ |
$A$ student is selected at random from the class. The probability that he/she has blood group $B$,is:
A
$\frac{1}{4}$
B
$\frac{13}{40}$
C
$\frac{1}{8}$
D
$\frac{3}{10}$
Solution
(D) Total number of students $= 10 + 13 + 12 + 5 = 40$.
Let $E$ be the event that the selected student has blood group $B$.
The number of students with blood group $B$ is $12$.
The probability $P(E)$ is given by the ratio of the number of favorable outcomes to the total number of outcomes.
$P(E) = \frac{\text{Number of students with blood group } B}{\text{Total number of students}} = \frac{12}{40}$.
Simplifying the fraction by dividing both numerator and denominator by $4$,we get $P(E) = \frac{3}{10}$.
Let $E$ be the event that the selected student has blood group $B$.
The number of students with blood group $B$ is $12$.
The probability $P(E)$ is given by the ratio of the number of favorable outcomes to the total number of outcomes.
$P(E) = \frac{\text{Number of students with blood group } B}{\text{Total number of students}} = \frac{12}{40}$.
Simplifying the fraction by dividing both numerator and denominator by $4$,we get $P(E) = \frac{3}{10}$.
0 likes
View Solution5
MediumMCQ
Two coins are tossed $1000$ times and the outcomes are recorded as below:
Based on this information,the probability for at most one head is
| Number of heads | $2$ | $1$ | $0$ |
| Frequency | $200$ | $550$ | $250$ |
Based on this information,the probability for at most one head is
A
$\frac{4}{5}$
B
$\frac{1}{4}$
C
$\frac{1}{5}$
D
$\frac{3}{4}$
Solution
(A) Total number of times the two coins are tossed $= 1000$.
Let $E$ be the event of getting at most one head,which means getting either $0$ head or $1$ head.
The number of times $E$ occurs is the sum of the frequencies of getting $0$ head and $1$ head.
Number of times $E$ happens $= 250 + 550 = 800$.
The probability $P(E)$ is given by the ratio of the number of favorable outcomes to the total number of trials.
$P(E) = \frac{800}{1000} = \frac{8}{10} = \frac{4}{5}$.
Let $E$ be the event of getting at most one head,which means getting either $0$ head or $1$ head.
The number of times $E$ occurs is the sum of the frequencies of getting $0$ head and $1$ head.
Number of times $E$ happens $= 250 + 550 = 800$.
The probability $P(E)$ is given by the ratio of the number of favorable outcomes to the total number of trials.
$P(E) = \frac{800}{1000} = \frac{8}{10} = \frac{4}{5}$.
0 likes
View Solution6
EasyMCQ
$80$ bulbs are selected at random from a lot and their lifetime (in $hrs$) is recorded in the form of a frequency table given below:
One bulb is selected at random from the lot. The probability that its lifetime is $1150$ hours is:
| Lifetime (in hours) | $300$ | $500$ | $700$ | $900$ | $1100$ |
| Frequency | $10$ | $12$ | $23$ | $25$ | $10$ |
One bulb is selected at random from the lot. The probability that its lifetime is $1150$ hours is:
A
$\frac{1}{80}$
B
$0$
C
$\frac{7}{16}$
D
$1$
Solution
(B) Total number of bulbs $= 80$.
Let $E$ be the event that the bulb selected at random from the lot has a lifetime of $1150 \text{ hours}$.
From the frequency table provided,we observe that the lifetimes recorded are $300, 500, 700, 900,$ and $1100 \text{ hours}$.
Since there is no entry for $1150 \text{ hours}$,the number of favorable outcomes for event $E$ is $0$.
Therefore,the probability $P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{0}{80} = 0$.
Let $E$ be the event that the bulb selected at random from the lot has a lifetime of $1150 \text{ hours}$.
From the frequency table provided,we observe that the lifetimes recorded are $300, 500, 700, 900,$ and $1100 \text{ hours}$.
Since there is no entry for $1150 \text{ hours}$,the number of favorable outcomes for event $E$ is $0$.
Therefore,the probability $P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{0}{80} = 0$.
0 likes
View Solution7
MediumMCQ
$80$ bulbs are selected at random from a lot and their lifetime (in $hrs$) is recorded in the form of a frequency table given below:
The probability that a bulb selected randomly from the lot has a lifetime less than $900$ hours is:
| Lifetime (in hours) | $300$ | $500$ | $700$ | $900$ | $1100$ |
| Frequency | $10$ | $12$ | $23$ | $25$ | $10$ |
The probability that a bulb selected randomly from the lot has a lifetime less than $900$ hours is:
A
$\frac{11}{40}$
B
$\frac{5}{16}$
C
$\frac{9}{16}$
D
$\frac{7}{16}$
Solution
(C) Let $E$ be the event that the bulb selected at random from the lot has a lifetime less than $900$ hours.
From the table,the number of bulbs having a lifetime less than $900$ hours is the sum of frequencies for lifetimes $300, 500,$ and $700$ hours.
Number of favorable outcomes $= 10 + 12 + 23 = 45$.
Total number of bulbs $= 80$.
The probability $P(E)$ is given by the ratio of the number of favorable outcomes to the total number of outcomes.
$P(E) = \frac{45}{80} = \frac{9}{16}$.
From the table,the number of bulbs having a lifetime less than $900$ hours is the sum of frequencies for lifetimes $300, 500,$ and $700$ hours.
Number of favorable outcomes $= 10 + 12 + 23 = 45$.
Total number of bulbs $= 80$.
The probability $P(E)$ is given by the ratio of the number of favorable outcomes to the total number of outcomes.
$P(E) = \frac{45}{80} = \frac{9}{16}$.
0 likes
View Solution8
Easy
Can the experimental probability of an event be a negative number? If not,why?
Solution
(B) No,the experimental probability of an event cannot be a negative number.
Probability is defined as the ratio of the number of trials in which the event occurred to the total number of trials.
Since both the number of trials in which the event occurred and the total number of trials are non-negative integers (and the total number of trials is always greater than $0$),the resulting ratio must be greater than or equal to $0$.
Probability is defined as the ratio of the number of trials in which the event occurred to the total number of trials.
Since both the number of trials in which the event occurred and the total number of trials are non-negative integers (and the total number of trials is always greater than $0$),the resulting ratio must be greater than or equal to $0$.
0 likes
View Solution9
EasyMCQ
Can the experimental probability of an event be greater than $1$? Justify your answer.
A
Yes,it can be greater than $1$.
B
No,it cannot be greater than $1$.
C
It is always equal to $1$.
D
It is always equal to $0$.
Solution
(B) The experimental probability of an event is defined as the ratio of the number of trials in which the event occurred to the total number of trials.
Mathematically,$P(E) = \frac{\text{Number of trials in which the event occurred}}{\text{Total number of trials}}$.
Since the number of trials in which the event occurred can never exceed the total number of trials,the numerator is always less than or equal to the denominator.
Therefore,the experimental probability of an event can never be greater than $1$.
Mathematically,$P(E) = \frac{\text{Number of trials in which the event occurred}}{\text{Total number of trials}}$.
Since the number of trials in which the event occurred can never exceed the total number of trials,the numerator is always less than or equal to the denominator.
Therefore,the experimental probability of an event can never be greater than $1$.
0 likes
View Solution10
Easy
As the number of tosses of a coin increases,the ratio of the number of heads to the total number of tosses will be $\frac{1}{2}$. Is it correct? If not,write the correct one.
Solution
(B) No,it is not correct. As the number of trials increases,the experimental probability (the ratio of the number of heads to the total number of tosses) tends to get closer to $\frac{1}{2}$,but it is not necessarily exactly $\frac{1}{2}$ in every case.
0 likes
View Solution11
MediumMCQ
Here is an extract from a mortality table.
Based on this information,what is the probability of a person aged $60$ dying within a year?
| Age (in years) | Number of persons surviving out of a sample of one million |
| $60$ | $16090$ |
| $61$ | $11490$ |
| $62$ | $8012$ |
| $63$ | $5448$ |
| $64$ | $3607$ |
| $65$ | $2320$ |
Based on this information,what is the probability of a person aged $60$ dying within a year?
A
$\frac{46}{100}$
B
$\frac{46}{1600}$
C
$\frac{460}{1609}$
D
$\frac{460}{1600}$
Solution
(C) We observe that out of $16090$ persons aged $60$,the number of persons who died before reaching their $61^{\text{st}}$ birthday is $(16090 - 11490) = 4600$.
Therefore,the probability of a person aged $60$ dying within a year is given by the ratio of the number of deaths to the total number of persons at that age.
$P(\text{a person aged } 60 \text{ dies within a year}) = \frac{4600}{16090}$
$= \frac{460}{1609}$
Therefore,the probability of a person aged $60$ dying within a year is given by the ratio of the number of deaths to the total number of persons at that age.
$P(\text{a person aged } 60 \text{ dies within a year}) = \frac{4600}{16090}$
$= \frac{460}{1609}$
0 likes
View Solution12
MediumMCQ
Here is an extract from a mortality table.
What is the probability that a person aged $61$ will live for $4$ years?
| Age (in years) | Number of persons surviving out of a sample of one million |
| $60$ | $16090$ |
| $61$ | $11490$ |
| $62$ | $8012$ |
| $63$ | $5448$ |
| $64$ | $3607$ |
| $65$ | $2320$ |
What is the probability that a person aged $61$ will live for $4$ years?
A
$\frac{232}{1100}$
B
$\frac{2320}{1149}$
C
$\frac{232}{11490}$
D
$\frac{232}{1149}$
Solution
(D) The number of persons aged $61$ years is $11490$.
$A$ person aged $61$ living for $4$ years means they will reach the age of $61 + 4 = 65$ years.
The number of persons surviving at age $65$ is $2320$.
The probability that a person aged $61$ will live for $4$ years is given by the ratio of the number of survivors at age $65$ to the number of persons at age $61$.
$P = \frac{2320}{11490} = \frac{232}{1149}$.
$A$ person aged $61$ living for $4$ years means they will reach the age of $61 + 4 = 65$ years.
The number of persons surviving at age $65$ is $2320$.
The probability that a person aged $61$ will live for $4$ years is given by the ratio of the number of survivors at age $65$ to the number of persons at age $61$.
$P = \frac{2320}{11490} = \frac{232}{1149}$.
0 likes
View Solution13
MediumMCQ
$A$ company selected $4000$ households at random and surveyed them to find out a relationship between income level and the number of television sets in a home. The information obtained is listed in the following table:
Find the probability of a household earning $Rs. 10000 - 14999$ per month and having exactly one television.
| Monthly income (in Rs) | $0$ Television | $1$ Television | $2$ Televisions | Above $2$ Televisions |
| $< 10000$ | $20$ | $80$ | $10$ | $0$ |
| $10000-14999$ | $10$ | $240$ | $60$ | $0$ |
| $15000-19999$ | $0$ | $380$ | $120$ | $30$ |
| $20000-24999$ | $0$ | $520$ | $370$ | $80$ |
| $25000$ and above | $0$ | $1100$ | $760$ | $220$ |
Find the probability of a household earning $Rs. 10000 - 14999$ per month and having exactly one television.
A
$0.06$
B
$0.08$
C
$0.6$
D
$0.02$
Solution
(A) Total number of households selected at random $= 4000$.
Number of households earning $Rs. 10000 - 14999$ per month and having exactly one television $= 240$.
The probability $P$ of an event is given by the formula:
$P = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}$
Required probability $= \frac{240}{4000} = \frac{24}{400} = \frac{6}{100} = 0.06$.
Number of households earning $Rs. 10000 - 14999$ per month and having exactly one television $= 240$.
The probability $P$ of an event is given by the formula:
$P = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}$
Required probability $= \frac{240}{4000} = \frac{24}{400} = \frac{6}{100} = 0.06$.
0 likes
View Solution14
MediumMCQ
$A$ company selected $4000$ households at random and surveyed them to find out a relationship between income level and the number of television sets in a home. The information so obtained is listed in the following table:
Find the probability of a household earning $Rs. 25000$ and more per month and owning $2$ televisions.
| Monthly income (in Rs) | $0$ Televisions | $1$ Television | $2$ Televisions | Above $2$ Televisions |
| $< 10000$ | $20$ | $80$ | $10$ | $0$ |
| $10000-14999$ | $10$ | $240$ | $60$ | $0$ |
| $15000-19999$ | $0$ | $380$ | $120$ | $30$ |
| $20000-24999$ | $0$ | $520$ | $370$ | $80$ |
| $25000$ and above | $0$ | $1100$ | $760$ | $220$ |
Find the probability of a household earning $Rs. 25000$ and more per month and owning $2$ televisions.
A
$0.109$
B
$0.19$
C
$0.99$
D
$0.59$
Solution
(B) Total number of households surveyed $= 4000$.
Number of households earning $Rs. 25000$ and more per month and owning $2$ televisions $= 760$.
The probability of an event is given by the formula:
$P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}$
Therefore,the required probability $= \frac{760}{4000} = \frac{76}{400} = \frac{19}{100} = 0.19$.
Number of households earning $Rs. 25000$ and more per month and owning $2$ televisions $= 760$.
The probability of an event is given by the formula:
$P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}$
Therefore,the required probability $= \frac{760}{4000} = \frac{76}{400} = \frac{19}{100} = 0.19$.
0 likes
View Solution15
EasyMCQ
$A$ company selected $4000$ households at random and surveyed them to find out a relationship between income level and the number of television sets in a home. The information obtained is listed in the following table:
Find the probability of a household not having any television.
| Monthly income (in Rs) | Number of Televisions per household |
| $0, 1, 2, \text{Above } 2$ | |
| $< 10000$ | $20, 80, 10, 0$ |
| $10000-14999$ | $10, 240, 60, 0$ |
| $15000-19999$ | $0, 380, 120, 30$ |
| $20000-24999$ | $0, 520, 370, 80$ |
| $25000$ and above | $0, 1100, 760, 220$ |
Find the probability of a household not having any television.
A
$\frac{35}{1600}$
B
$\frac{30}{400}$
C
$\frac{3}{400}$
D
$\frac{3}{4000}$
Solution
(C) Total number of households surveyed $= 4000$.
Number of households having $0$ televisions is the sum of households in the first column of the table:
$20 + 10 + 0 + 0 + 0 = 30$.
Probability of a household not having any television $= \frac{\text{Number of households with } 0 \text{ televisions}}{\text{Total number of households}}$.
Probability $= \frac{30}{4000} = \frac{3}{400}$.
Number of households having $0$ televisions is the sum of households in the first column of the table:
$20 + 10 + 0 + 0 + 0 = 30$.
Probability of a household not having any television $= \frac{\text{Number of households with } 0 \text{ televisions}}{\text{Total number of households}}$.
Probability $= \frac{30}{4000} = \frac{3}{400}$.
0 likes
View Solution16
MediumMCQ
Two dice are thrown simultaneously $500$ times. Each time the sum of two numbers appearing on their tops is noted and recorded as given in the following table:
If the dice are thrown once more,what is the probability of getting a sum $3\,?$
| Sum | Frequency |
| $2$ | $14$ |
| $3$ | $30$ |
| $4$ | $42$ |
| $5$ | $55$ |
| $6$ | $72$ |
| $7$ | $75$ |
| $8$ | $70$ |
| $9$ | $53$ |
| $10$ | $46$ |
| $11$ | $28$ |
| $12$ | $15$ |
If the dice are thrown once more,what is the probability of getting a sum $3\,?$
A
$0.6$
B
$0.006$
C
$0.36$
D
$0.06$
Solution
(D) The total number of trials is $500$.
Let $E$ be the event of getting a sum of $3$ on the uppermost faces of the two dice.
From the given table,the frequency of the sum $3$ is $30$.
The probability of an event is given by the formula:
$P(E) = \frac{\text{Number of trials in which the event occurred}}{\text{Total number of trials}}$
Substituting the values:
$P(E) = \frac{30}{500} = \frac{3}{50} = 0.06$
Therefore,the probability of getting a sum of $3$ is $0.06$.
Let $E$ be the event of getting a sum of $3$ on the uppermost faces of the two dice.
From the given table,the frequency of the sum $3$ is $30$.
The probability of an event is given by the formula:
$P(E) = \frac{\text{Number of trials in which the event occurred}}{\text{Total number of trials}}$
Substituting the values:
$P(E) = \frac{30}{500} = \frac{3}{50} = 0.06$
Therefore,the probability of getting a sum of $3$ is $0.06$.
0 likes
View Solution17
MediumMCQ
Two dice are thrown simultaneously $500$ times. Each time the sum of two numbers appearing on their tops is noted and recorded as given in the following table:
If the dice are thrown once more,what is the probability of getting a sum more than $10\,?$
| Sum | Frequency |
| $2$ | $14$ |
| $3$ | $30$ |
| $4$ | $42$ |
| $5$ | $55$ |
| $6$ | $72$ |
| $7$ | $75$ |
| $8$ | $70$ |
| $9$ | $53$ |
| $10$ | $46$ |
| $11$ | $28$ |
| $12$ | $15$ |
If the dice are thrown once more,what is the probability of getting a sum more than $10\,?$
A
$0.086$
B
$0.86$
C
$0.09$
D
$0.5$
Solution
(A) The total number of trials is $500$.
Let $E$ be the event that the sum of the two numbers appearing on the uppermost faces of the two dice is more than $10$. This means the sum must be $11$ or $12$.
From the table,the frequency of getting a sum of $11$ is $28$ and the frequency of getting a sum of $12$ is $15$.
Number of trials in which the event $E$ occurred $= 28 + 15 = 43$.
The probability $P(E)$ is given by the formula:
$P(E) = \frac{\text{Number of trials in which the event occurred}}{\text{Total number of trials}}$
$P(E) = \frac{43}{500} = 0.086$.
Let $E$ be the event that the sum of the two numbers appearing on the uppermost faces of the two dice is more than $10$. This means the sum must be $11$ or $12$.
From the table,the frequency of getting a sum of $11$ is $28$ and the frequency of getting a sum of $12$ is $15$.
Number of trials in which the event $E$ occurred $= 28 + 15 = 43$.
The probability $P(E)$ is given by the formula:
$P(E) = \frac{\text{Number of trials in which the event occurred}}{\text{Total number of trials}}$
$P(E) = \frac{43}{500} = 0.086$.
0 likes
View Solution18
MediumMCQ
Two dice are thrown simultaneously $500$ times. Each time the sum of two numbers appearing on their tops is noted and recorded as given in the following table:
If the dice are thrown once more,what is the probability of getting a sum less than or equal to $5$?
| Sum | Frequency |
| $2$ | $14$ |
| $3$ | $30$ |
| $4$ | $42$ |
| $5$ | $55$ |
| $6$ | $72$ |
| $7$ | $75$ |
| $8$ | $70$ |
| $9$ | $53$ |
| $10$ | $46$ |
| $11$ | $28$ |
| $12$ | $15$ |
If the dice are thrown once more,what is the probability of getting a sum less than or equal to $5$?
A
$0.582$
B
$0.282$
C
$0.141$
D
$0.550$
Solution
(B) The total number of trials is $500$.
Let $E$ be the event that the sum of the two numbers appearing on the uppermost faces of the two dice is less than or equal to $5$.
This includes the sums $2, 3, 4,$ and $5$.
The number of times the event $E$ occurred is the sum of the frequencies of these outcomes:
Number of trials for $E = 14 + 30 + 42 + 55 = 141$.
The probability $P(E)$ is given by the formula:
$P(E) = \frac{\text{Number of trials in which the event happened}}{\text{Total number of trials}}$
$P(E) = \frac{141}{500} = 0.282$.
Let $E$ be the event that the sum of the two numbers appearing on the uppermost faces of the two dice is less than or equal to $5$.
This includes the sums $2, 3, 4,$ and $5$.
The number of times the event $E$ occurred is the sum of the frequencies of these outcomes:
Number of trials for $E = 14 + 30 + 42 + 55 = 141$.
The probability $P(E)$ is given by the formula:
$P(E) = \frac{\text{Number of trials in which the event happened}}{\text{Total number of trials}}$
$P(E) = \frac{141}{500} = 0.282$.
0 likes
View Solution19
MediumMCQ
Two dice are thrown simultaneously $500$ times. Each time the sum of two numbers appearing on their tops is noted and recorded as given in the following table:
If the dice are thrown once more,what is the probability of getting a sum between $8$ and $12\,?$
| Sum | Frequency |
| $2$ | $14$ |
| $3$ | $30$ |
| $4$ | $42$ |
| $5$ | $55$ |
| $6$ | $72$ |
| $7$ | $75$ |
| $8$ | $70$ |
| $9$ | $53$ |
| $10$ | $46$ |
| $11$ | $28$ |
| $12$ | $15$ |
If the dice are thrown once more,what is the probability of getting a sum between $8$ and $12\,?$
A
$0.954$
B
$0.127$
C
$0.254$
D
$0.554$
Solution
(C) The total number of trials is $500$.
Let $E$ be the event of getting a sum between $8$ and $12$. This means the sum can be $9, 10,$ or $11$.
From the table,the frequencies for these sums are:
Frequency of sum $9 = 53$
Frequency of sum $10 = 46$
Frequency of sum $11 = 28$
Total number of trials in which the event $E$ occurred $= 53 + 46 + 28 = 127$.
The probability $P(E)$ is given by:
$P(E) = \frac{\text{Number of trials in which the event occurred}}{\text{Total number of trials}}$
$P(E) = \frac{127}{500} = 0.254$.
Let $E$ be the event of getting a sum between $8$ and $12$. This means the sum can be $9, 10,$ or $11$.
From the table,the frequencies for these sums are:
Frequency of sum $9 = 53$
Frequency of sum $10 = 46$
Frequency of sum $11 = 28$
Total number of trials in which the event $E$ occurred $= 53 + 46 + 28 = 127$.
The probability $P(E)$ is given by:
$P(E) = \frac{\text{Number of trials in which the event occurred}}{\text{Total number of trials}}$
$P(E) = \frac{127}{500} = 0.254$.
0 likes
View Solution20
MediumMCQ
Bulbs are packed in cartons,each containing $40$ bulbs. Seven hundred cartons were examined for defective bulbs,and the results are given in the following table:
One carton was selected at random. What is the probability that it has no defective bulb?
| Number of defective bulbs | $0$ | $1$ | $2$ | $3$ | $4$ | $5$ | $6$ | more than $6$ |
| Frequency | $400$ | $180$ | $48$ | $41$ | $18$ | $8$ | $3$ | $2$ |
One carton was selected at random. What is the probability that it has no defective bulb?
A
$\frac{7}{9}$
B
$\frac{5}{7}$
C
$\frac{3}{7}$
D
$\frac{4}{7}$
Solution
(D) Total number of cartons examined $= 700$.
Let $E$ be the event that the selected carton has no defective bulb.
From the table,the number of cartons with $0$ defective bulbs is $400$.
The probability of an event is given by the formula:
$P(E) = \frac{\text{Number of trials in which the event happened}}{\text{Total number of trials}}$
Substituting the values:
$P(E) = \frac{400}{700} = \frac{4}{7}$.
Therefore,the probability that the selected carton has no defective bulb is $\frac{4}{7}$.
Let $E$ be the event that the selected carton has no defective bulb.
From the table,the number of cartons with $0$ defective bulbs is $400$.
The probability of an event is given by the formula:
$P(E) = \frac{\text{Number of trials in which the event happened}}{\text{Total number of trials}}$
Substituting the values:
$P(E) = \frac{400}{700} = \frac{4}{7}$.
Therefore,the probability that the selected carton has no defective bulb is $\frac{4}{7}$.
0 likes
View Solution21
MediumMCQ
Bulbs are packed in cartons,each containing $40$ bulbs. Seven hundred cartons were examined for defective bulbs,and the results are given in the following table:
One carton was selected at random. What is the probability that it has defective bulbs from $2$ to $6$?
| Number of defective bulbs | $0$ | $1$ | $2$ | $3$ | $4$ | $5$ | $6$ | More than $6$ |
| Frequency | $400$ | $180$ | $48$ | $41$ | $18$ | $8$ | $3$ | $2$ |
One carton was selected at random. What is the probability that it has defective bulbs from $2$ to $6$?
A
$\frac{59}{350}$
B
$\frac{59}{750}$
C
$\frac{118}{350}$
D
$\frac{18}{35}$
Solution
(A) Total number of cartons examined $= 700$.
Let $E$ be the event that the selected carton has defective bulbs from $2$ to $6$.
From the given table,the number of cartons having $2, 3, 4, 5,$ or $6$ defective bulbs is the sum of their frequencies:
Number of favorable outcomes $= 48 + 41 + 18 + 8 + 3 = 118$.
The probability $P(E)$ is given by the formula:
$P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of trials}}$
$P(E) = \frac{118}{700} = \frac{59}{350}$.
Let $E$ be the event that the selected carton has defective bulbs from $2$ to $6$.
From the given table,the number of cartons having $2, 3, 4, 5,$ or $6$ defective bulbs is the sum of their frequencies:
Number of favorable outcomes $= 48 + 41 + 18 + 8 + 3 = 118$.
The probability $P(E)$ is given by the formula:
$P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of trials}}$
$P(E) = \frac{118}{700} = \frac{59}{350}$.
0 likes
View Solution22
MediumMCQ
Bulbs are packed in cartons,each containing $40$ bulbs. Seven hundred cartons were examined for defective bulbs,and the results are given in the following table:
One carton was selected at random. What is the probability that it has less than $4$ defective bulbs?
| Number of defective bulbs | $0$ | $1$ | $2$ | $3$ | $4$ | $5$ | $6$ | more than $6$ |
| Frequency | $400$ | $180$ | $48$ | $41$ | $18$ | $8$ | $3$ | $2$ |
One carton was selected at random. What is the probability that it has less than $4$ defective bulbs?
A
$\frac{569}{700}$
B
$\frac{669}{700}$
C
$\frac{269}{700}$
D
$\frac{66}{700}$
Solution
(B) Total number of cartons examined $= 700$.
Let $E$ be the event that the selected carton has less than $4$ defective bulbs.
This means the carton has $0, 1, 2,$ or $3$ defective bulbs.
From the table,the number of cartons with $0, 1, 2,$ or $3$ defective bulbs is:
$400 + 180 + 48 + 41 = 669$.
Therefore,the probability $P(E)$ is given by:
$P(E) = \frac{\text{Number of cartons with less than } 4 \text{ defective bulbs}}{\text{Total number of cartons}}$
$P(E) = \frac{669}{700}$.
Let $E$ be the event that the selected carton has less than $4$ defective bulbs.
This means the carton has $0, 1, 2,$ or $3$ defective bulbs.
From the table,the number of cartons with $0, 1, 2,$ or $3$ defective bulbs is:
$400 + 180 + 48 + 41 = 669$.
Therefore,the probability $P(E)$ is given by:
$P(E) = \frac{\text{Number of cartons with less than } 4 \text{ defective bulbs}}{\text{Total number of cartons}}$
$P(E) = \frac{669}{700}$.
0 likes
View Solution23
MediumMCQ
Over the past $200$ working days,the number of defective parts produced by a machine is given in the following table:
Determine the probability that tomorrow's output will have no defective part.
| Number of defective parts | $0$ | $1$ | $2$ | $3$ | $4$ | $5$ | $6$ | $7$ | $8$ | $9$ | $10$ | $11$ | $12$ | $13$ |
| Days | $50$ | $32$ | $22$ | $18$ | $12$ | $12$ | $10$ | $10$ | $10$ | $8$ | $6$ | $6$ | $2$ | $2$ |
Determine the probability that tomorrow's output will have no defective part.
A
$0.85$
B
$0.75$
C
$0.25$
D
$0.55$
Solution
(C) Total number of past working days $= 200$.
Let $E$ be the event that tomorrow's output will have no defective part.
From the table,the number of days with $0$ defective parts is $50$.
The probability of an event $E$ is given by:
$P(E) = \frac{\text{Number of trials in which the event happened}}{\text{Total number of trials}}$
$P(E) = \frac{50}{200} = \frac{1}{4} = 0.25$.
Thus,the probability that tomorrow's output will have no defective part is $0.25$.
Let $E$ be the event that tomorrow's output will have no defective part.
From the table,the number of days with $0$ defective parts is $50$.
The probability of an event $E$ is given by:
$P(E) = \frac{\text{Number of trials in which the event happened}}{\text{Total number of trials}}$
$P(E) = \frac{50}{200} = \frac{1}{4} = 0.25$.
Thus,the probability that tomorrow's output will have no defective part is $0.25$.
0 likes
View Solution24
MediumMCQ
Over the past $200$ working days,the number of defective parts produced by a machine is given in the following table:
Determine the probability that tomorrow's output will have at least one defective part.
| Number of defective parts | $0$ | $1$ | $2$ | $3$ | $4$ | $5$ | $6$ | $7$ | $8$ | $9$ | $10$ | $11$ | $12$ | $13$ |
| Days | $50$ | $32$ | $22$ | $18$ | $12$ | $12$ | $10$ | $10$ | $10$ | $8$ | $6$ | $6$ | $2$ | $2$ |
Determine the probability that tomorrow's output will have at least one defective part.
A
$0.65$
B
$0.25$
C
$0.50$
D
$0.75$
Solution
(D) The total number of past working days is $200$.
Let $E$ be the event that tomorrow's output will have at least one defective part.
From the table,the number of days with $0$ defective parts is $50$.
The number of days with at least one defective part is the total number of days minus the number of days with $0$ defective parts.
Number of days with at least one defective part $= 200 - 50 = 150$.
The probability $P(E)$ is given by the ratio of the number of days with at least one defective part to the total number of days.
$P(E) = \frac{\text{Number of days with at least one defective part}}{\text{Total number of days}}$
$P(E) = \frac{150}{200} = \frac{3}{4} = 0.75$.
Let $E$ be the event that tomorrow's output will have at least one defective part.
From the table,the number of days with $0$ defective parts is $50$.
The number of days with at least one defective part is the total number of days minus the number of days with $0$ defective parts.
Number of days with at least one defective part $= 200 - 50 = 150$.
The probability $P(E)$ is given by the ratio of the number of days with at least one defective part to the total number of days.
$P(E) = \frac{\text{Number of days with at least one defective part}}{\text{Total number of days}}$
$P(E) = \frac{150}{200} = \frac{3}{4} = 0.75$.
0 likes
View Solution25
MediumMCQ
Over the past $200$ working days,the number of defective parts produced by a machine is given in the following table:
Determine the probability that tomorrow's output will have not more than $5$ defective parts.
| Number of defective parts | $0$ | $1$ | $2$ | $3$ | $4$ | $5$ | $6$ | $7$ | $8$ | $9$ | $10$ | $11$ | $12$ | $13$ |
| Days | $50$ | $32$ | $22$ | $18$ | $12$ | $12$ | $10$ | $10$ | $10$ | $8$ | $6$ | $6$ | $2$ | $2$ |
Determine the probability that tomorrow's output will have not more than $5$ defective parts.
A
$0.73$
B
$0.63$
C
$0.53$
D
$0.43$
Solution
(A) The total number of past working days is $200$.
Let $E$ be the event that tomorrow's output will have not more than $5$ defective parts.
'Not more than $5$' means the number of defective parts can be $0, 1, 2, 3, 4,$ or $5$.
From the given table,the number of days corresponding to these values are:
Days with $0$ parts: $50$
Days with $1$ part: $32$
Days with $2$ parts: $22$
Days with $3$ parts: $18$
Days with $4$ parts: $12$
Days with $5$ parts: $12$
Total number of favorable outcomes $= 50 + 32 + 22 + 18 + 12 + 12 = 146$.
The probability $P(E)$ is given by:
$P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of trials}}$
$P(E) = \frac{146}{200} = 0.73$.
Let $E$ be the event that tomorrow's output will have not more than $5$ defective parts.
'Not more than $5$' means the number of defective parts can be $0, 1, 2, 3, 4,$ or $5$.
From the given table,the number of days corresponding to these values are:
Days with $0$ parts: $50$
Days with $1$ part: $32$
Days with $2$ parts: $22$
Days with $3$ parts: $18$
Days with $4$ parts: $12$
Days with $5$ parts: $12$
Total number of favorable outcomes $= 50 + 32 + 22 + 18 + 12 + 12 = 146$.
The probability $P(E)$ is given by:
$P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of trials}}$
$P(E) = \frac{146}{200} = 0.73$.
0 likes
View Solution26
MediumMCQ
Over the past $200$ working days,the number of defective parts produced by a machine is given in the following table:
Determine the probability that tomorrow's output will have more than $13$ defective parts.
| Number of defective parts | $0$ | $1$ | $2$ | $3$ | $4$ | $5$ | $6$ | $7$ | $8$ | $9$ | $10$ | $11$ | $12$ | $13$ |
| Days | $50$ | $32$ | $22$ | $18$ | $12$ | $12$ | $10$ | $10$ | $10$ | $8$ | $6$ | $6$ | $2$ | $2$ |
Determine the probability that tomorrow's output will have more than $13$ defective parts.
A
$1$
B
$0$
C
$\frac{1}{200}$
D
$\frac{13}{200}$
Solution
(B) The total number of past working days is $200$.
Let $E$ be the event that tomorrow's output will have more than $13$ defective parts.
From the given table,we observe that the maximum number of defective parts recorded is $13$.
Therefore,the number of days on which more than $13$ defective parts were produced is $0$.
The probability $P(E)$ is calculated as:
$P(E) = \frac{\text{Number of trials in which the event happened}}{\text{Total number of trials}}$
$P(E) = \frac{0}{200} = 0$.
Let $E$ be the event that tomorrow's output will have more than $13$ defective parts.
From the given table,we observe that the maximum number of defective parts recorded is $13$.
Therefore,the number of days on which more than $13$ defective parts were produced is $0$.
The probability $P(E)$ is calculated as:
$P(E) = \frac{\text{Number of trials in which the event happened}}{\text{Total number of trials}}$
$P(E) = \frac{0}{200} = 0$.
0 likes
View Solution27
MediumMCQ
$A$ recent survey found that the ages of workers in a factory is distributed as follows:
If a person is selected at random,find the probability that the person is $40$ years or more.
| Age (in years) | $20-29$ | $30-39$ | $40-49$ | $50-59$ | $60$ and above |
| Number of workers | $38$ | $27$ | $86$ | $46$ | $3$ |
If a person is selected at random,find the probability that the person is $40$ years or more.
A
$0.75$
B
$0.375$
C
$0.675$
D
$0.775$
Solution
(C) Total number of workers in the factory $= 38 + 27 + 86 + 46 + 3 = 200$.
Let $E$ be the event that the person selected at random is $40$ years or more.
The number of workers who are $40$ years or more is the sum of workers in the age groups $40-49$,$50-59$,and $60$ and above.
Number of favorable outcomes $= 86 + 46 + 3 = 135$.
The probability $P(E)$ is given by the formula:
$P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}$
$P(E) = \frac{135}{200} = 0.675$.
Let $E$ be the event that the person selected at random is $40$ years or more.
The number of workers who are $40$ years or more is the sum of workers in the age groups $40-49$,$50-59$,and $60$ and above.
Number of favorable outcomes $= 86 + 46 + 3 = 135$.
The probability $P(E)$ is given by the formula:
$P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}$
$P(E) = \frac{135}{200} = 0.675$.
0 likes
View Solution28
MediumMCQ
$A$ recent survey found that the ages of workers in a factory is distributed as follows:
If a person is selected at random,find the probability that the person is under $40$ years.
| Age (in years) | $20-29$ | $30-39$ | $40-49$ | $50-59$ | $60$ and above |
| Number of workers | $38$ | $27$ | $86$ | $46$ | $3$ |
If a person is selected at random,find the probability that the person is under $40$ years.
A
$0.925$
B
$0.125$
C
$0.525$
D
$0.325$
Solution
(D) Total number of workers in the factory $= 38 + 27 + 86 + 46 + 3 = 200$.
Let $E$ be the event that the person selected at random is under $40$ years of age.
The workers under $40$ years are those in the age groups $20-29$ and $30-39$.
Number of workers under $40$ years $= 38 + 27 = 65$.
The probability $P(E)$ is given by the formula:
$P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}$
$P(E) = \frac{65}{200} = \frac{32.5}{100} = 0.325$.
Let $E$ be the event that the person selected at random is under $40$ years of age.
The workers under $40$ years are those in the age groups $20-29$ and $30-39$.
Number of workers under $40$ years $= 38 + 27 = 65$.
The probability $P(E)$ is given by the formula:
$P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}$
$P(E) = \frac{65}{200} = \frac{32.5}{100} = 0.325$.
0 likes
View Solution29
MediumMCQ
$A$ recent survey found that the ages of workers in a factory is distributed as follows:
If a person is selected at random,find the probability that the person is having age from $30$ to $39$ years.
| Age (in years) | $20-29$ | $30-39$ | $40-49$ | $50-59$ | $60$ and above |
| Number of workers | $38$ | $27$ | $86$ | $46$ | $3$ |
If a person is selected at random,find the probability that the person is having age from $30$ to $39$ years.
A
$0.135$
B
$0.145$
C
$0.185$
D
$0.155$
Solution
(A) Total number of workers in a factory $= 38 + 27 + 86 + 46 + 3 = 200$.
Let $E$ be the event that the person selected at random is having age from $30$ to $39$ years.
From the table given above,the number of workers in the age group $30-39$ years is $27$.
The probability $P(E)$ is given by the formula:
$P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}$
$P(E) = \frac{27}{200} = 0.135$.
Let $E$ be the event that the person selected at random is having age from $30$ to $39$ years.
From the table given above,the number of workers in the age group $30-39$ years is $27$.
The probability $P(E)$ is given by the formula:
$P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}$
$P(E) = \frac{27}{200} = 0.135$.
0 likes
View Solution30
MediumMCQ
$A$ recent survey found that the ages of workers in a factory is distributed as follows:
If a person is selected at random,find the probability that the person is under $60$ but over $39$ years.
| Age (in years) | $20-29$ | $30-39$ | $40-49$ | $50-59$ | $60$ and above |
| Number of workers | $38$ | $27$ | $86$ | $46$ | $3$ |
If a person is selected at random,find the probability that the person is under $60$ but over $39$ years.
A
$0.16$
B
$0.66$
C
$0.76$
D
$0.96$
Solution
(B) Total number of workers in the factory $= 38 + 27 + 86 + 46 + 3 = 200$.
Let $E$ be the event that the person selected at random is under $60$ but over $39$ years of age.
From the table,the age groups satisfying this condition are $40-49$ and $50-59$.
The number of workers in these age groups $= 86 + 46 = 132$.
The probability $P(E)$ is given by the ratio of the number of favorable outcomes to the total number of outcomes:
$P(E) = \frac{\text{Number of workers aged between } 40 \text{ and } 59}{\text{Total number of workers}}$
$P(E) = \frac{132}{200} = 0.66$.
Let $E$ be the event that the person selected at random is under $60$ but over $39$ years of age.
From the table,the age groups satisfying this condition are $40-49$ and $50-59$.
The number of workers in these age groups $= 86 + 46 = 132$.
The probability $P(E)$ is given by the ratio of the number of favorable outcomes to the total number of outcomes:
$P(E) = \frac{\text{Number of workers aged between } 40 \text{ and } 59}{\text{Total number of workers}}$
$P(E) = \frac{132}{200} = 0.66$.
0 likes
View Solution31
Medium
Three coins are tossed simultaneously $500$ times with the following frequencies of different outcomes:
Find the probability of getting $0, 1, 2$ or $3$ heads.
Verify that the sum of all these probabilities is $1$.
| Number of heads | $0$ | $1$ | $2$ | $3$ |
| Frequency | $70$ | $190$ | $175$ | $65$ |
Find the probability of getting $0, 1, 2$ or $3$ heads.
Verify that the sum of all these probabilities is $1$.
Solution
(A-D) Since the coins are tossed $500$ times,the total number of trials is $500$.
Let us denote the events of getting $0$ head,$1$ head,$2$ heads,and $3$ heads by $A_1, A_2, A_3$,and $A_4$ respectively.
The probability of an event is given by the formula: $P(E) = \frac{\text{Number of trials in which the event happened}}{\text{Total number of trials}}$.
$P(A_1) = \frac{70}{500} = 0.14$
$P(A_2) = \frac{190}{500} = 0.38$
$P(A_3) = \frac{175}{500} = 0.35$
$P(A_4) = \frac{65}{500} = 0.13$
Now,verify the sum of all these probabilities:
$P(A_1) + P(A_2) + P(A_3) + P(A_4) = 0.14 + 0.38 + 0.35 + 0.13 = 1$.
Thus,the sum of all probabilities is $1$.
Let us denote the events of getting $0$ head,$1$ head,$2$ heads,and $3$ heads by $A_1, A_2, A_3$,and $A_4$ respectively.
The probability of an event is given by the formula: $P(E) = \frac{\text{Number of trials in which the event happened}}{\text{Total number of trials}}$.
$P(A_1) = \frac{70}{500} = 0.14$
$P(A_2) = \frac{190}{500} = 0.38$
$P(A_3) = \frac{175}{500} = 0.35$
$P(A_4) = \frac{65}{500} = 0.13$
Now,verify the sum of all these probabilities:
$P(A_1) + P(A_2) + P(A_3) + P(A_4) = 0.14 + 0.38 + 0.35 + 0.13 = 1$.
Thus,the sum of all probabilities is $1$.
0 likes
View Solution32
Medium
$A$ die is thrown $400$ times with the frequencies for the outcomes $1, 2, 3, 4, 5$ and $6$ as given in the following table:
Find the probability of getting each outcome.
| Outcome | $1$ | $2$ | $3$ | $4$ | $5$ | $6$ |
| Frequency | $68$ | $70$ | $74$ | $54$ | $70$ | $64$ |
Find the probability of getting each outcome.
Solution
(N/A) The total number of trials is $400$.
Let $A_i$ be the event of getting the outcome $i$,where $i \in \{1, 2, 3, 4, 5, 6\}$.
The probability of an event is given by the formula: $P(A_i) = \frac{\text{Number of times the outcome occurred}}{\text{Total number of trials}}$.
$P(A_1) = \frac{68}{400} = 0.17$
$P(A_2) = \frac{70}{400} = 0.175$
$P(A_3) = \frac{74}{400} = 0.185$
$P(A_4) = \frac{54}{400} = 0.135$
$P(A_5) = \frac{70}{400} = 0.175$
$P(A_6) = \frac{64}{400} = 0.16$
Let $A_i$ be the event of getting the outcome $i$,where $i \in \{1, 2, 3, 4, 5, 6\}$.
The probability of an event is given by the formula: $P(A_i) = \frac{\text{Number of times the outcome occurred}}{\text{Total number of trials}}$.
$P(A_1) = \frac{68}{400} = 0.17$
$P(A_2) = \frac{70}{400} = 0.175$
$P(A_3) = \frac{74}{400} = 0.185$
$P(A_4) = \frac{54}{400} = 0.135$
$P(A_5) = \frac{70}{400} = 0.175$
$P(A_6) = \frac{64}{400} = 0.16$
0 likes
View Solution33
MediumMCQ
From the list of telephone numbers of $100$ employees of a factory, the following frequency distribution of their unit place digit (for example, in the number $25828573$, the unit place digit is $3$) is given in the table below:
Find the probability that a randomly selected employee has a telephone number with unit digit $3$.
| Unit digit | $0$ | $1$ | $2$ | $3$ | $4$ | $5$ | $6$ | $7$ | $8$ | $9$ |
| Frequency | $8$ | $9$ | $12$ | $15$ | $10$ | $7$ | $12$ | $10$ | $9$ | $8$ |
Find the probability that a randomly selected employee has a telephone number with unit digit $3$.
A
$0.17$
B
$0.15$
C
$0.13$
D
$0.10$
Solution
(B) The probability of an event is given by the ratio of the number of favorable outcomes to the total number of possible outcomes.
Here, the total number of employees (total outcomes) = $100$.
The frequency of the unit digit $3$ (favorable outcomes) = $15$.
Therefore, the probability that a randomly selected employee has a telephone number with unit digit $3$ is:
$P(\text{unit digit } 3) = \frac{\text{Frequency of } 3}{\text{Total number of employees}}$
$P(\text{unit digit } 3) = \frac{15}{100} = 0.15$.
Here, the total number of employees (total outcomes) = $100$.
The frequency of the unit digit $3$ (favorable outcomes) = $15$.
Therefore, the probability that a randomly selected employee has a telephone number with unit digit $3$ is:
$P(\text{unit digit } 3) = \frac{\text{Frequency of } 3}{\text{Total number of employees}}$
$P(\text{unit digit } 3) = \frac{15}{100} = 0.15$.
0 likes
View Solution34
MediumMCQ
Following is the information about vehicles passing in front of a city temple in one hour:
Find the probability that the vehicle passing through the temple is a two-wheeler.
| Vehicle type | Two-wheelers | Three-wheelers | Four-wheelers |
| Frequency | $90$ | $35$ | $25$ |
Find the probability that the vehicle passing through the temple is a two-wheeler.
A
$3/2$
B
$1/5$
C
$3/5$
D
$2/3$
Solution
(C) Total number of vehicles passing in front of the temple $= 90 + 35 + 25 = 150$.
The probability that a passing vehicle is a two-wheeler is given by the formula:
$P(\text{Two-wheeler}) = \frac{\text{Number of two-wheelers}}{\text{Total number of vehicles}}$
Substituting the values:
$P(\text{Two-wheeler}) = \frac{90}{150}$
Simplifying the fraction by dividing both numerator and denominator by $30$:
$P(\text{Two-wheeler}) = \frac{3}{5}$.
The probability that a passing vehicle is a two-wheeler is given by the formula:
$P(\text{Two-wheeler}) = \frac{\text{Number of two-wheelers}}{\text{Total number of vehicles}}$
Substituting the values:
$P(\text{Two-wheeler}) = \frac{90}{150}$
Simplifying the fraction by dividing both numerator and denominator by $30$:
$P(\text{Two-wheeler}) = \frac{3}{5}$.
0 likes
View Solution35
Medium
$1000$ families with $2$ children were selected randomly,and the following data were recorded:
Compute the probability of a family,chosen at random,having $(1)$ two girls,$(2)$ one girl.
| Number of girls in a family | $0$ | $1$ | $2$ |
| Number of families | $128$ | $672$ | $200$ |
Compute the probability of a family,chosen at random,having $(1)$ two girls,$(2)$ one girl.
Solution
(N/A) The total number of families is $1000$.
$(1)$ Let $E_1$ be the event that the chosen family has $2$ girls.
The number of families having $2$ girls is $200$.
Probability $P(E_1) = \frac{\text{Number of families with } 2 \text{ girls}}{\text{Total number of families}} = \frac{200}{1000} = 0.2$.
$(2)$ Let $E_2$ be the event that the chosen family has $1$ girl.
The number of families having $1$ girl is $672$.
Probability $P(E_2) = \frac{\text{Number of families with } 1 \text{ girl}}{\text{Total number of families}} = \frac{672}{1000} = 0.672$.
$(1)$ Let $E_1$ be the event that the chosen family has $2$ girls.
The number of families having $2$ girls is $200$.
Probability $P(E_1) = \frac{\text{Number of families with } 2 \text{ girls}}{\text{Total number of families}} = \frac{200}{1000} = 0.2$.
$(2)$ Let $E_2$ be the event that the chosen family has $1$ girl.
The number of families having $1$ girl is $672$.
Probability $P(E_2) = \frac{\text{Number of families with } 1 \text{ girl}}{\text{Total number of families}} = \frac{672}{1000} = 0.672$.
0 likes
View Solution36
MediumMCQ
An insurance company selected $1600$ drivers at random in a particular city to find a relationship between age and accidents. The data is as below:
Find the probability of the following event for a driver selected at random from the city: being $25-40$ years of age and having exactly $2$ accidents in a year.
| Age group of drivers | Number of accidents in one year $(0, 1, 2, 3, >3)$ |
| $18-25$ | $320, 125, 75, 45, 30$ |
| $25-40$ | $400, 45, 50, 15, 10$ |
| $40-55$ | $150, 85, 13, 8, 10$ |
| Above $55$ | $150, 25, 17, 20, 7$ |
Find the probability of the following event for a driver selected at random from the city: being $25-40$ years of age and having exactly $2$ accidents in a year.
A
$\frac{1}{32}$
B
$\frac{1}{23}$
C
$\frac{1}{50}$
D
$\frac{1}{25}$
Solution
(A) Total number of drivers $= 1600$.
The number of drivers who are $25-40$ years old and have exactly $2$ accidents in one year is $50$.
Therefore,the probability $P$ of the event is given by:
$P = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}$
$P = \frac{50}{1600} = \frac{1}{32}$.
The number of drivers who are $25-40$ years old and have exactly $2$ accidents in one year is $50$.
Therefore,the probability $P$ of the event is given by:
$P = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}$
$P = \frac{50}{1600} = \frac{1}{32}$.
0 likes
View Solution37
MediumMCQ
In cricket,Rohit hits a century in $4$ innings out of $10$ innings. Find the probability that he did not hit a century.
A
$0.2$
B
$0.6$
C
$0.4$
D
$0.8$
Solution
(B) The total number of innings is $10$.
The number of innings in which Rohit hits a century is $4$.
The number of innings in which he did not hit a century is $10 - 4 = 6$.
The probability of not hitting a century is given by the ratio of the number of innings without a century to the total number of innings.
Probability $= \frac{6}{10} = 0.6$.
The number of innings in which Rohit hits a century is $4$.
The number of innings in which he did not hit a century is $10 - 4 = 6$.
The probability of not hitting a century is given by the ratio of the number of innings without a century to the total number of innings.
Probability $= \frac{6}{10} = 0.6$.
0 likes
View Solution38
Easy
$A$ coin is tossed $50$ times,in which $23$ times head comes up and $27$ times tail comes up. Calculate the probability for each event.
Solution
(N/A) The total number of trials is $50$.
$1$. Probability of getting a head:
$P(\text{Head}) = \frac{\text{Number of heads}}{\text{Total number of trials}} = \frac{23}{50} = 0.46$.
$2$. Probability of getting a tail:
$P(\text{Tail}) = \frac{\text{Number of tails}}{\text{Total number of trials}} = \frac{27}{50} = 0.54$.
$1$. Probability of getting a head:
$P(\text{Head}) = \frac{\text{Number of heads}}{\text{Total number of trials}} = \frac{23}{50} = 0.46$.
$2$. Probability of getting a tail:
$P(\text{Tail}) = \frac{\text{Number of tails}}{\text{Total number of trials}} = \frac{27}{50} = 0.54$.
0 likes
View Solution39
Medium
$50$ families with $2$ children were selected randomly and the following data were recorded:
If a family is chosen at random,compute the probability that it has: $(1)$ two girls,$(2)$ one girl,$(3)$ no girl.
| Number of girls in a family | $0$ | $1$ | $2$ |
| Number of families | $10$ | $32$ | $8$ |
If a family is chosen at random,compute the probability that it has: $(1)$ two girls,$(2)$ one girl,$(3)$ no girl.
Solution
The total number of families is $50$.
$(1)$ Probability of having $2$ girls:
Number of families with $2$ girls = $8$.
Probability = $\frac{8}{50} = 0.16$.
$(2)$ Probability of having $1$ girl:
Number of families with $1$ girl = $32$.
Probability = $\frac{32}{50} = 0.64$.
$(3)$ Probability of having no girl:
Number of families with $0$ girls = $10$.
Probability = $\frac{10}{50} = 0.20$.
$(1)$ Probability of having $2$ girls:
Number of families with $2$ girls = $8$.
Probability = $\frac{8}{50} = 0.16$.
$(2)$ Probability of having $1$ girl:
Number of families with $1$ girl = $32$.
Probability = $\frac{32}{50} = 0.64$.
$(3)$ Probability of having no girl:
Number of families with $0$ girls = $10$.
Probability = $\frac{10}{50} = 0.20$.
0 likes
View Solution40
Medium
Two coins are tossed simultaneously $400$ times with the following frequencies of different outcomes:
Two heads: $100$ times
One head: $220$ times
No head: $80$ times
Find the probability of occurrence of each of these events.
Two heads: $100$ times
One head: $220$ times
No head: $80$ times
Find the probability of occurrence of each of these events.
Solution
(N/A) The total number of trials is $400$.
$1$. Probability of getting two heads = (Number of times two heads occurred) / (Total number of trials) = $100 / 400 = 0.25$.
$2$. Probability of getting one head = (Number of times one head occurred) / (Total number of trials) = $220 / 400 = 0.55$.
$3$. Probability of getting no head = (Number of times no head occurred) / (Total number of trials) = $80 / 400 = 0.20$.
$1$. Probability of getting two heads = (Number of times two heads occurred) / (Total number of trials) = $100 / 400 = 0.25$.
$2$. Probability of getting one head = (Number of times one head occurred) / (Total number of trials) = $220 / 400 = 0.55$.
$3$. Probability of getting no head = (Number of times no head occurred) / (Total number of trials) = $80 / 400 = 0.20$.
0 likes
View Solution41
Medium
The marks obtained by a student in five tests are as follows:
If one of the tests is chosen at random,find the probability that the student scored $(1)$ more than $90$ marks and $(2)$ between $70$ and $80$ marks in that test.
| < strong>Test Number | < strong>Marks obtained |
| $1$ | $62$ |
| $2$ | $72$ |
| $3$ | $78$ |
| $4$ | $91$ |
| $5$ | $85$ |
If one of the tests is chosen at random,find the probability that the student scored $(1)$ more than $90$ marks and $(2)$ between $70$ and $80$ marks in that test.
Solution
(N/A) Total number of tests = $5$.
$(1)$ Number of tests where the student scored more than $90$ marks is $1$ (Test $4$ with $91$ marks).
Therefore,the probability = $\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{1}{5}$.
$(2)$ Number of tests where the student scored between $70$ and $80$ marks is $2$ (Test $2$ with $72$ marks and Test $3$ with $78$ marks).
Therefore,the probability = $\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{2}{5}$.
$(1)$ Number of tests where the student scored more than $90$ marks is $1$ (Test $4$ with $91$ marks).
Therefore,the probability = $\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{1}{5}$.
$(2)$ Number of tests where the student scored between $70$ and $80$ marks is $2$ (Test $2$ with $72$ marks and Test $3$ with $78$ marks).
Therefore,the probability = $\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{2}{5}$.
0 likes
View Solution42
Medium
$100$ seeds were selected at random from each of $5$ bags of seeds,and were kept under standardised conditions equally favourable to germination. After some days,the number of seeds which had germinated in each collection were counted and recorded as follows:
What is the probability of germination of:
$(1)$ more than $60$ seeds in a bag?
$(2)$ less than $60$ seeds in a bag?
$(3)$ more than $90$ seeds in a bag?
| Bag | $1, 2, 3, 4, 5$ |
| Number of seeds germinated | $76, 89, 65, 58, 85$ |
What is the probability of germination of:
$(1)$ more than $60$ seeds in a bag?
$(2)$ less than $60$ seeds in a bag?
$(3)$ more than $90$ seeds in a bag?
Solution
(A) Total number of bags = $5$.
$(1)$ Number of bags in which more than $60$ seeds germinated = $4$ (Bags $1, 2, 3, 5$ have $76, 89, 65, 85$ seeds respectively).
Probability = $\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}} = \frac{4}{5} = 0.8$.
$(2)$ Number of bags in which less than $60$ seeds germinated = $1$ (Bag $4$ has $58$ seeds).
Probability = $\frac{1}{5} = 0.2$.
$(3)$ Number of bags in which more than $90$ seeds germinated = $0$.
Probability = $\frac{0}{5} = 0$.
$(1)$ Number of bags in which more than $60$ seeds germinated = $4$ (Bags $1, 2, 3, 5$ have $76, 89, 65, 85$ seeds respectively).
Probability = $\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}} = \frac{4}{5} = 0.8$.
$(2)$ Number of bags in which less than $60$ seeds germinated = $1$ (Bag $4$ has $58$ seeds).
Probability = $\frac{1}{5} = 0.2$.
$(3)$ Number of bags in which more than $90$ seeds germinated = $0$.
Probability = $\frac{0}{5} = 0$.
0 likes
View Solution43
MediumMCQ
At a fair, $70$ boys, $80$ girls, $20$ men, and $30$ women have come. Find the probability that a person selected at random to be a prize winner is a woman.
A
$0.17$
B
$0.12$
C
$0.15$
D
$0.10$
Solution
(C) To find the probability, we first calculate the total number of people at the fair.
Total number of people = (Number of boys) + (Number of girls) + (Number of men) + (Number of women)
Total = $70 + 80 + 20 + 30 = 200$.
The number of women at the fair is $30$.
The probability $P$ that a person selected at random is a woman is given by the formula:
$P(\text{woman}) = \frac{\text{Number of women}}{\text{Total number of people}}$
$P(\text{woman}) = \frac{30}{200} = \frac{3}{20} = 0.15$.
Thus, the probability is $0.15$.
Total number of people = (Number of boys) + (Number of girls) + (Number of men) + (Number of women)
Total = $70 + 80 + 20 + 30 = 200$.
The number of women at the fair is $30$.
The probability $P$ that a person selected at random is a woman is given by the formula:
$P(\text{woman}) = \frac{\text{Number of women}}{\text{Total number of people}}$
$P(\text{woman}) = \frac{30}{200} = \frac{3}{20} = 0.15$.
Thus, the probability is $0.15$.
0 likes
View Solution44
MediumMCQ
The distribution of the weights of $50$ students in a class is as follows:
Find the probability that a student selected at random weighs $56 \ kg$ or more.
| Weight (in $kg$) | Number of students |
| $50-52$ | $10$ |
| $52-54$ | $12$ |
| $54-56$ | $13$ |
| $56-58$ | $9$ |
| $58-60$ | $6$ |
Find the probability that a student selected at random weighs $56 \ kg$ or more.
A
$0.7$
B
$0.4$
C
$0.1$
D
$0.3$
Solution
(D) The total number of students is $50$.
We need to find the probability that a student weighs $56 \ kg$ or more.
Students weighing $56 \ kg$ or more fall into the intervals $56-58$ and $58-60$.
The number of students in the interval $56-58$ is $9$.
The number of students in the interval $58-60$ is $6$.
Total number of students weighing $56 \ kg$ or more = $9 + 6 = 15$.
The probability $P$ is given by the formula: $P = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}$.
$P = \frac{15}{50} = \frac{3}{10} = 0.3$.
Therefore,the probability is $0.3$.
We need to find the probability that a student weighs $56 \ kg$ or more.
Students weighing $56 \ kg$ or more fall into the intervals $56-58$ and $58-60$.
The number of students in the interval $56-58$ is $9$.
The number of students in the interval $58-60$ is $6$.
Total number of students weighing $56 \ kg$ or more = $9 + 6 = 15$.
The probability $P$ is given by the formula: $P = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}$.
$P = \frac{15}{50} = \frac{3}{10} = 0.3$.
Therefore,the probability is $0.3$.
0 likes
View Solution45
MediumMCQ
$A$ bag contains $40$ balls of the same shape and weight. Among the balls, $18$ are red, $12$ are yellow, and the rest are blue. What is the probability that a ball drawn at random from the bag is blue?
A
$0.25$
B
$0.20$
C
$0.33$
D
$0.9$
Solution
(A) Total number of balls = $40$.
Number of red balls = $18$.
Number of yellow balls = $12$.
Number of blue balls = $\text{Total balls} - (\text{Red balls} + \text{Yellow balls}) = 40 - (18 + 12) = 40 - 30 = 10$.
The probability of drawing a blue ball is given by the ratio of the number of blue balls to the total number of balls.
$P(\text{Blue}) = \frac{\text{Number of blue balls}}{\text{Total number of balls}} = \frac{10}{40} = \frac{1}{4} = 0.25$.
Number of red balls = $18$.
Number of yellow balls = $12$.
Number of blue balls = $\text{Total balls} - (\text{Red balls} + \text{Yellow balls}) = 40 - (18 + 12) = 40 - 30 = 10$.
The probability of drawing a blue ball is given by the ratio of the number of blue balls to the total number of balls.
$P(\text{Blue}) = \frac{\text{Number of blue balls}}{\text{Total number of balls}} = \frac{10}{40} = \frac{1}{4} = 0.25$.
0 likes
View Solution46
Difficult
$A$ box contains $100$ cards marked with numbers $1$ to $100$. If one card is drawn at random from the box,find the probability that the number on the card is: $(1)$ a multiple of $7$,$(2)$ a number smaller than $30$,$(3)$ a prime number.
Solution
(N/A) Total number of cards $n(S) = 100$.
$(1)$ Multiples of $7$ between $1$ and $100$ are ${7, 14, 21, 28, 35, 42, 49, 56, 63, 70, 77, 84, 91, 98}$. There are $14$ such numbers. Probability $P = 14/100 = 0.14$.
$(2)$ Numbers smaller than $30$ are ${1, 2, 3, dots, 29}$. There are $29$ such numbers. Probability $P = 29/100 = 0.29$.
$(3)$ Prime numbers between $1$ and $100$ are ${2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97}$. There are $25$ such numbers. Probability $P = 25/100 = 0.25$.
$(1)$ Multiples of $7$ between $1$ and $100$ are ${7, 14, 21, 28, 35, 42, 49, 56, 63, 70, 77, 84, 91, 98}$. There are $14$ such numbers. Probability $P = 14/100 = 0.14$.
$(2)$ Numbers smaller than $30$ are ${1, 2, 3, dots, 29}$. There are $29$ such numbers. Probability $P = 29/100 = 0.29$.
$(3)$ Prime numbers between $1$ and $100$ are ${2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97}$. There are $25$ such numbers. Probability $P = 25/100 = 0.25$.
0 likes
View Solution47
Medium
State whether each of the following statements is true or false:
$(1)$ For the probability $P(A)$ of any event $A$,$0 < P(A) < 1$.
$(2)$ The probability of the month of March having $5$ Sundays is $\frac{1}{7}$.
$(3)$ If the probability of an event $A$ occurring is $\frac{3}{7}$,then the probability of event $A$ not occurring is $\frac{4}{7}$.
$(1)$ For the probability $P(A)$ of any event $A$,$0 < P(A) < 1$.
$(2)$ The probability of the month of March having $5$ Sundays is $\frac{1}{7}$.
$(3)$ If the probability of an event $A$ occurring is $\frac{3}{7}$,then the probability of event $A$ not occurring is $\frac{4}{7}$.
Solution
(C) $(1)$ False. The probability $P(A)$ of any event $A$ satisfies $0 \le P(A) \le 1$. It can be $0$ (impossible event) or $1$ (sure event).
$(2)$ False. March has $31$ days,which means $4$ weeks and $3$ extra days. The extra days can be (Sun,Mon,Tue),(Mon,Tue,Wed),(Tue,Wed,Thu),(Wed,Thu,Fri),(Thu,Fri,Sat),(Fri,Sat,Sun),or (Sat,Sun,Mon). Out of $7$ possibilities,$3$ contain a Sunday. Thus,the probability is $\frac{3}{7}$.
$(3)$ True. The probability of an event not occurring is $P(\text{not } A) = 1 - P(A)$. Given $P(A) = \frac{3}{7}$,then $P(\text{not } A) = 1 - \frac{3}{7} = \frac{4}{7}$.
$(2)$ False. March has $31$ days,which means $4$ weeks and $3$ extra days. The extra days can be (Sun,Mon,Tue),(Mon,Tue,Wed),(Tue,Wed,Thu),(Wed,Thu,Fri),(Thu,Fri,Sat),(Fri,Sat,Sun),or (Sat,Sun,Mon). Out of $7$ possibilities,$3$ contain a Sunday. Thus,the probability is $\frac{3}{7}$.
$(3)$ True. The probability of an event not occurring is $P(\text{not } A) = 1 - P(A)$. Given $P(A) = \frac{3}{7}$,then $P(\text{not } A) = 1 - \frac{3}{7} = \frac{4}{7}$.
0 likes
View Solution48
Medium
State whether each of the following statements is true or false:
$(1)$ The probability that the month of February in a leap year has $5$ Saturdays is $0$.
$(2)$ The probability of getting $51$ marks in a $100$ marks examination is $\frac{1}{101}$.
$(1)$ The probability that the month of February in a leap year has $5$ Saturdays is $0$.
$(2)$ The probability of getting $51$ marks in a $100$ marks examination is $\frac{1}{101}$.
Solution
(B) $(1)$ False. In a leap year,February has $29$ days. $29$ days consist of $4$ weeks and $1$ extra day. The extra day can be any of the $7$ days of the week. The probability of this extra day being a Saturday is $\frac{1}{7}$,which is not $0$.
$(2)$ False. In an examination of $100$ marks,the possible marks are integers from $0$ to $100$. There are $101$ possible outcomes $(0, 1, 2, ..., 100)$. Assuming each mark is equally likely,the probability of getting exactly $51$ marks is $\frac{1}{101}$.
$(2)$ False. In an examination of $100$ marks,the possible marks are integers from $0$ to $100$. There are $101$ possible outcomes $(0, 1, 2, ..., 100)$. Assuming each mark is equally likely,the probability of getting exactly $51$ marks is $\frac{1}{101}$.
0 likes
View Solution49
EasyMCQ
$A$ balanced die is tossed once. The probability of getting number $6$ on the die is ..............
A
$1/6$
B
$1/2$
C
$1/4$
D
$1/3$
Solution
$(A)$ standard die has $6$ faces,numbered $1, 2, 3, 4, 5,$ and $6$.
When a die is tossed once,the total number of possible outcomes is $6$ (i.e.,${1, 2, 3, 4, 5, 6}$).
The favorable outcome of getting the number $6$ is ${6}$,so the number of favorable outcomes is $1$.
The probability of an event is given by the formula: $P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}}$.
Therefore,the probability of getting $6$ is $P(6) = \frac{1}{6}$.
When a die is tossed once,the total number of possible outcomes is $6$ (i.e.,${1, 2, 3, 4, 5, 6}$).
The favorable outcome of getting the number $6$ is ${6}$,so the number of favorable outcomes is $1$.
The probability of an event is given by the formula: $P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}}$.
Therefore,the probability of getting $6$ is $P(6) = \frac{1}{6}$.
0 likes
View Solution50
EasyMCQ
The probability of getting two heads when three balanced coins are tossed is ............
A
$1/8$
B
$3/8$
C
$1/4$
D
$3/4$
Solution
(B) When three balanced coins are tossed,the total number of possible outcomes is $2^3 = 8$.
The sample space $S$ is given by: $S = \{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT\}$.
We are looking for the outcomes with exactly two heads. These are: $HHT, HTH, THH$.
The number of favorable outcomes is $3$.
Therefore,the probability $P$ of getting two heads is $\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{3}{8}$.
The sample space $S$ is given by: $S = \{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT\}$.
We are looking for the outcomes with exactly two heads. These are: $HHT, HTH, THH$.
The number of favorable outcomes is $3$.
Therefore,the probability $P$ of getting two heads is $\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{3}{8}$.
0 likes
View SolutionProbability — Mix Examples - Probability · Frequently Asked Questions
1Are these Probability questions useful for JEE and NEET?
Yes. All questions in this section are mapped to JEE Main and NEET exam patterns. Previous year questions from JEE Main, NEET, GUJCET and state-level exams are included with full solutions.
2Can I switch to Hindi or Gujarati for these questions?
Yes. Use the language tabs in the hero section or the sidebar to view the same questions and solutions in English, Hindi or Gujarati.
3How do I generate a question paper from this subtopic?
Use the Vedclass Exam Paper Generator — select the chapter and subtopic, set difficulty, and generate Sets A, B, C, D automatically. First 3 chapters of every subject are free.
Vedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D papers from this chapter in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See DemoFor Teachers & Institutes
Generate a Probability Exam Paper in 2 Minutes
Select subtopic & difficulty — Sets A, B, C, D auto-generated with No Repeat logic.
First 3 chapters of every subject are free — no payment required.