Textbook - Probability Questions in English

(A) Since the coin is tossed $1000$ times,the total number of trials is $1000$.
Let $E$ be the event of getting a head and $F$ be the event of getting a tail.
The number of times $E$ occurs is $455$.
The probability of event $E$ is given by $P(E) = \frac{\text{Number of heads}}{\text{Total number of trials}} = \frac{455}{1000} = 0.455$.
The number of times $F$ occurs is $545$.
The probability of event $F$ is given by $P(F) = \frac{\text{Number of tails}}{\text{Total number of trials}} = \frac{545}{1000} = 0.545$.
Thus,the probabilities are $0.455$ and $0.545$ respectively.

(N/A) Let us denote the events of getting two heads,one head,and no head by $E_1, E_2,$ and $E_3$ respectively.
The total number of trials is $500$.
For $E_1$ (two heads): $P(E_1) = \frac{105}{500} = 0.21$
For $E_2$ (one head): $P(E_2) = \frac{275}{500} = 0.55$
For $E_3$ (no head): $P(E_3) = \frac{120}{500} = 0.24$
Observe that $P(E_1) + P(E_2) + P(E_3) = 0.21 + 0.55 + 0.24 = 1.0$. These events cover all possible outcomes of the trial.

(N/A) Let $E_i$ denote the event of getting the outcome $i$,where $i = 1, 2, 3, 4, 5, 6$.
The probability of an event is given by the formula: $P(E) = \frac{\text{Number of trials in which the event happened}}{\text{Total number of trials}}$.
Here,the total number of trials is $1000$.
$P(E_1) = \frac{179}{1000} = 0.179$
$P(E_2) = \frac{150}{1000} = 0.150$
$P(E_3) = \frac{157}{1000} = 0.157$
$P(E_4) = \frac{149}{1000} = 0.149$
$P(E_5) = \frac{175}{1000} = 0.175$
$P(E_6) = \frac{190}{1000} = 0.190$
Note that the sum of all probabilities is $0.179 + 0.150 + 0.157 + 0.149 + 0.175 + 0.190 = 1$.

(D) The probability of an event is given by the formula: $P(E) = \frac{\text{Number of trials in which the event happened}}{\text{Total number of trials}}$.
Here,the total number of telephone numbers is $200$.
The frequency of the digit $6$ in the unit place is $14$.
Therefore,the probability that the digit in the unit place is $6$ is:
$P(\text{digit } 6) = \frac{14}{200} = \frac{7}{100} = 0.07$.

(A) The total number of days for which the record is available $= 250$.
$(i)$ The number of days when the forecast was correct $= 175$.
$P$ (the forecast was correct on a given day) $= \frac{\text{Number of days when the forecast was correct}}{\text{Total number of days}} = \frac{175}{250} = 0.7$.
$(ii)$ The number of days when the forecast was not correct $= 250 - 175 = 75$.
$P$ (the forecast was not correct on a given day) $= \frac{\text{Number of days when the forecast was not correct}}{\text{Total number of days}} = \frac{75}{250} = 0.3$.

(N/A) $(i)$ The total number of trials $= 1000$.
The frequency of a tyre that needs to be replaced before it covers $4000 \, km$ is $20$.
So,$P(\text{tyre to be replaced before it covers } 4000 \, km) = \frac{20}{1000} = 0.02$.
$(ii)$ The frequency of a tyre that will last more than $9000 \, km$ is the sum of frequencies for the ranges $9001$ to $14000$ and more than $14000$,which is $325 + 445 = 770$.
So,$P(\text{tyre will last more than } 9000 \, km) = \frac{770}{1000} = 0.77$.
$(iii)$ The frequency of a tyre that requires replacement between $4000 \, km$ and $14000 \, km$ is the sum of frequencies for the ranges $4000$ to $9000$ and $9001$ to $14000$,which is $210 + 325 = 535$.
So,$P(\text{tyre requiring replacement between } 4000 \, km \text{ and } 14000 \, km) = \frac{535}{1000} = 0.535$.

(C) The total number of unit tests held is $5$.
The percentage of marks obtained in the unit tests are $69, 71, 73, 68, 74$.
The unit tests in which the student obtained more than $70\%$ marks are $II$ $(71\%)$,$III$ $(73\%)$,and $V$ $(74\%)$.
Thus,the number of unit tests in which the student obtained more than $70\%$ marks is $3$.
The probability $P$ of an event is given by the formula: $P = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}$.
Therefore,$P$ (scoring more than $70\%$ marks) $= \frac{3}{5} = 0.6$.

(N/A) Total number of drivers $= 2000$.
$(i)$ The number of drivers who are $18-29$ years old and have exactly $3$ accidents in one year is $61$.
So,$P$ (driver is $18-29$ years old with exactly $3$ accidents) $= \frac{61}{2000} = 0.0305$.
$(ii)$ The number of drivers $30-50$ years of age and having one or more accidents in one year $= 125 + 60 + 22 + 18 = 225$.
So,$P$ (driver is $30-50$ years of age and having one or more accidents) $= \frac{225}{2000} = 0.1125$.
$(iii)$ The number of drivers having no accidents in one year $= 440 + 505 + 360 = 1305$.
Therefore,$P$ (drivers with no accident) $= \frac{1305}{2000} = 0.6525$.

(A) $(i)$ The total number of students is $38$. The number of students with weight in the interval $46-50 \, kg$ is $3$.
Therefore,the probability $P$ is given by: $P = \frac{3}{38} \approx 0.079$.
$(ii)$ An event with probability $0$ is an impossible event. For example,the event that a student weighs $30 \, kg$ (since no student in the table has this weight),the probability is $0$.
An event with probability $1$ is a sure event. For example,the event that a student weighs more than $30 \, kg$ (since all $38$ students weigh more than $30 \, kg$),the probability is $\frac{38}{38} = 1$.

(A-D) The total number of bags is $5$.
$(i)$ The number of bags in which more than $40$ seeds germinated is $3$ (Bags $2, 3,$ and $5$).
Probability $= \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{3}{5} = 0.6$.
$(ii)$ The number of bags in which $49$ seeds germinated is $0$.
Probability $= \frac{0}{5} = 0$.
$(iii)$ The number of bags in which more than $35$ seeds germinated is $5$ (All bags).
Probability $= \frac{5}{5} = 1$.

(C) Number of times the batswoman hits a boundary $= 6$.
Total number of balls played $= 30$.
Number of times that the batswoman does not hit a boundary $= 30 - 6 = 24$.
The probability $P$ that she does not hit a boundary is given by the ratio of the number of times she does not hit a boundary to the total number of balls played.
$P(\text{not hitting a boundary}) = \frac{24}{30}$.
Simplifying the fraction by dividing both numerator and denominator by $6$,we get:
$P(\text{not hitting a boundary}) = \frac{4}{5}$.

(N/A) Total number of families $= 475 + 814 + 211 = 1500$.
$(i)$ Number of families having $2$ girls $= 475$.
Probability $(P_1)$ of a randomly chosen family having $2$ girls $= \frac{475}{1500} = \frac{19}{60}$.
$(ii)$ Number of families having $1$ girl $= 814$.
Probability $(P_2)$ of a randomly chosen family having $1$ girl $= \frac{814}{1500} = \frac{407}{750}$.
$(iii)$ Number of families having no girl $= 211$.
Probability $(P_3)$ of a randomly chosen family having no girl $= \frac{211}{1500}$.
Sum of all these probabilities $= P_1 + P_2 + P_3 = \frac{475}{1500} + \frac{814}{1500} + \frac{211}{1500} = \frac{475 + 814 + 211}{1500} = \frac{1500}{1500} = 1$.
Therefore,the sum of all these probabilities is $1$.

(A) From the given bar graph,the number of students born in the month of August is $6$.
The total number of students is the sum of students born in all months:
$3 + 4 + 2 + 2 + 5 + 1 + 2 + 6 + 3 + 4 + 4 + 4 = 40$.
The probability $P$ that a student was born in August is given by:
$P(\text{August}) = \frac{\text{Number of students born in August}}{\text{Total number of students}}$
$P(\text{August}) = \frac{6}{40} = \frac{3}{20}$.

(B) The number of times $2$ heads appear is $72$.
The total number of times the coins were tossed is $200$.
The probability of an event is given by the ratio of the number of favorable outcomes to the total number of trials.
$P(2 \text{ heads}) = \frac{\text{Number of times } 2 \text{ heads come up}}{\text{Total number of trials}}$
$P(2 \text{ heads}) = \frac{72}{200}$
Dividing both the numerator and the denominator by $8$,we get:
$P(2 \text{ heads}) = \frac{9}{25}$

(N/A) Total number of families surveyed $= 2400$.
$(i)$ Number of families earning Rs. $10000 - 13000$ per month and owning exactly $2$ vehicles $= 29$.
Probability $= \frac{29}{2400}$.
$(ii)$ Number of families earning Rs. $16000$ or more per month and owning exactly $1$ vehicle $= 579$.
Probability $= \frac{579}{2400}$.
$(iii)$ Number of families earning less than Rs. $7000$ per month and owning no vehicle $= 10$.
Probability $= \frac{10}{2400} = \frac{1}{240}$.
$(iv)$ Number of families earning Rs. $13000 - 16000$ per month and owning more than $2$ vehicles $= 25$.
Probability $= \frac{25}{2400} = \frac{1}{96}$.
$(v)$ Number of families owning not more than $1$ vehicle (i.e.,$0$ or $1$ vehicle) $= (10 + 160) + (0 + 305) + (1 + 535) + (2 + 469) + (1 + 579) = 170 + 305 + 536 + 471 + 580 = 2062$.
Probability $= \frac{2062}{2400} = \frac{1031}{1200}$.

(D) Total number of students $= 90$.
$(i)$ Number of students getting less than $20$ marks (which is $20\%$ of $100$) $= 7$.
Probability $P = \frac{\text{Number of students with } < 20 \text{ marks}}{\text{Total number of students}} = \frac{7}{90}$.
$(ii)$ Number of students obtaining marks $60$ or above $= (\text{Students in } 60-70) + (\text{Students in } 70 \text{ and above}) = 15 + 8 = 23$.
Probability $P = \frac{\text{Number of students with } \ge 60 \text{ marks}}{\text{Total number of students}} = \frac{23}{90}$.

(A) Total number of students $= 135 + 65 = 200$.
$(i)$ Number of students who like statistics $= 135$.
Probability (student likes statistics) $= \frac{\text{Number of students who like statistics}}{\text{Total number of students}} = \frac{135}{200} = \frac{27}{40}$.
$(ii)$ Number of students who do not like statistics $= 65$.
Probability (student does not like statistics) $= \frac{\text{Number of students who do not like statistics}}{\text{Total number of students}} = \frac{65}{200} = \frac{13}{40}$.

(A) $(i)$ Total number of engineers $= 40$.
Number of engineers living less than $7 \, km$ from their place of work are: $5, 3, 3, 2, 7, 9, 7, 3, 2, 6, 6$ (Wait,let us re-count: $5, 3, 3, 2, 7$ is not less than $7$,so $5, 3, 3, 2, 3, 5, 2, 6, 6$ are less than $7$).
Counting values less than $7$: $5, 3, 2, 3, 5, 2, 6, 6$ (Total $8$ values). Let's re-verify: $5, 3, 2, 3, 5, 2, 6, 6$ are $8$ values. Actually,looking at the table: $5, 3, 3, 2, 7, 9, 7, 3, 2, 6, 6$. The values less than $7$ are $5, 3, 2, 3, 5, 2, 6, 6$. Total count is $8$.
Wait,let's list all values $< 7$: $5, 3, 2, 3, 5, 2, 6, 6$. Total $= 8$.
Probability $P = \frac{8}{40} = \frac{1}{5} = 0.2$.
$(ii)$ Number of engineers living more than or equal to $7 \, km = 40 - 8 = 32$.
Probability $P = \frac{32}{40} = \frac{4}{5} = 0.8$.
$(iii)$ Number of engineers living within $\frac{1}{2} \, km = 0$.
Probability $P = 0$.

(N/A) To find the probability, follow these steps:
$1$. Let $n_1$ be the number of two-wheelers, $n_2$ be the number of three-wheelers, and $n_3$ be the number of four-wheelers observed.
$2$. Calculate the total number of vehicles observed: $N = n_1 + n_2 + n_3$.
$3$. The probability $P$ of selecting a two-wheeler is given by the ratio of the number of two-wheelers to the total number of vehicles.
$4$. Formula: $P(\text{two-wheeler}) = \frac{n_1}{N} = \frac{n_1}{n_1 + n_2 + n_3}$.

(A) The total number of $3-$digit numbers ranges from $100$ to $999$.
The total count of $3-$digit numbers is $999 - 100 + 1 = 900$.
$A$ number is divisible by $3$ if the sum of its digits is divisible by $3$.
In any sequence of $3$ consecutive integers,exactly one is divisible by $3$.
Since $900$ is a multiple of $3$,exactly one-third of these numbers are divisible by $3$.
Number of favorable outcomes $= \frac{900}{3} = 300$.
The probability $P$ is given by the ratio of favorable outcomes to total outcomes.
$P = \frac{300}{900} = \frac{1}{3}$.

(A) Total number of bags $= 11$.
Bags containing more than $5 \,kg$ of flour are: $5.05, 5.08, 5.03, 5.06, 5.08, 5.04, 5.07$.
Number of bags containing more than $5 \,kg$ of flour $= 7$.
Probability of an event $= \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}$.
Therefore,the required probability $P = \frac{7}{11}$.

(B) To find the probability,we use the formula: $P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}$.
From the given frequency distribution table:
$1$. The number of days for which the concentration of sulphur dioxide was in the interval $0.12 - 0.16$ is $2$.
$2$. The total number of days is $30$.
Therefore,the required probability $P$ is:
$P = \frac{2}{30} = \frac{1}{15}$.

(C) The total number of students in the class is $30$.
The number of students having blood group $AB$ is $3$.
The probability $P$ of an event is given by the formula:
$P = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}}$
Here,the number of favorable outcomes (students with blood group $AB$) is $3$,and the total number of possible outcomes (total students) is $30$.
Therefore,the probability $P = \frac{3}{30} = \frac{1}{10}$.