Mix Examples - Probability Questions in English

(B) When two balanced dice are rolled simultaneously,the total number of possible outcomes is $n = 6 \times 6 = 36$.
The favorable outcomes for the sum of the numbers on the dice to be $7$ are: $(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)$.
Thus,the number of favorable outcomes is $m = 6$.
The probability of the event is given by the formula: $P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{m}{n}$.
Substituting the values,we get: $P(E) = \frac{6}{36} = \frac{1}{6}$.

(B) When two balanced dice are rolled simultaneously,the total number of possible outcomes is $6 \times 6 = 36$.
The event is defined as the sum of the numbers on the two dice being $11$.
The possible outcomes that result in a sum of $11$ are $(5, 6)$ and $(6, 5)$.
Thus,the number of favorable outcomes is $m = 2$.
The probability $P$ of the event is given by the ratio of favorable outcomes to total outcomes:
$P = \frac{m}{n} = \frac{2}{36} = \frac{1}{18}$.

(D) The maximum sum of numbers on two dice is $6 + 6 = 12$.
Since the maximum possible sum is $12$,it is impossible to obtain a sum of $13$ when rolling two dice.
Therefore,the event of getting a sum of $13$ is an impossible event.
The probability of an impossible event is always $0$.
$\therefore$ Required probability $= 0$.

(D) When a balanced die is rolled twice,the total number of possible outcomes is $6 \times 6 = 36$.
The sum of the numbers obtained in two trials is even if:
$1$. Both numbers are even (e.g.,$2, 4, 6$ for each die: $3 \times 3 = 9$ outcomes).
$2$. Both numbers are odd (e.g.,$1, 3, 5$ for each die: $3 \times 3 = 9$ outcomes).
Total number of favorable outcomes $= 9 + 9 = 18$.
Therefore,the required probability $= \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{18}{36} = \frac{1}{2}$.

(C) When a balanced die is rolled twice,the total number of possible outcomes is $6 \times 6 = 36$.
The even numbers on a die are $\{2, 4, 6\}$.
The event of getting an even number both times means the outcome must be an ordered pair $(x, y)$ where both $x$ and $y$ are from the set $\{2, 4, 6\}$.
The favourable outcomes are: $(2, 2), (2, 4), (2, 6), (4, 2), (4, 4), (4, 6), (6, 2), (6, 4), (6, 6)$.
The number of favourable outcomes is $m = 9$.
Therefore,the required probability $P = \frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}} = \frac{9}{36} = \frac{1}{4}$.

(A) When a balanced die is rolled twice,the total number of possible outcomes is $6 \times 6 = 36$.
Let $E$ be the event that the sum of the numbers obtained in two trials is more than $10$.
The outcomes where the sum is greater than $10$ are $(5, 6)$,$(6, 5)$,and $(6, 6)$.
Thus,the number of favorable outcomes is $m = 3$.
The probability $P(E)$ is given by the ratio of favorable outcomes to total outcomes:
$P(E) = \frac{m}{n} = \frac{3}{36} = \frac{1}{12}$.

(D) When a balanced die is rolled twice,the possible outcomes for the sum of the two numbers range from $1+1=2$ to $6+6=12$.
Since the maximum possible sum is $12$,any sum obtained will always be $12$ or less than $12$.
This is a certain event (or sure event).
The probability of a certain event is always $1$.
Therefore,the required probability is $1$.

(C) Total number of screws in the packet $= n = 400$.
Number of defective screws $= 120$.
Number of non-defective screws $= 400 - 120 = 280$.
Let $E$ be the event of picking up a non-defective screw.
The number of outcomes favourable to event $E$ is $m = 280$.
The probability of picking up a non-defective screw is given by $P(E) = \frac{m}{n} = \frac{280}{400}$.
Simplifying the fraction,$P(E) = \frac{28}{40} = \frac{7}{10} = 0.7$.

(C) Total number of roses in the vase $= 5 + 3 + 2 = 10$.
The number of favourable outcomes for selecting a yellow rose is $2$ (since there are $2$ yellow roses).
The probability of an event is given by the formula: $P(E) = \frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}$.
Therefore,the probability of selecting a yellow rose $= \frac{2}{10} = \frac{1}{5}$.

(B) Total number of toys in the box $= n = 10$.
Number of defective toys $= 3$.
Number of non-defective toys $= 10 - 3 = 7$.
$A$ customer purchases only a non-defective toy.
Therefore,the number of favourable outcomes for the event that the toy is purchased $= m = 7$.
The probability of an event is given by the ratio of the number of favourable outcomes to the total number of outcomes.
$\text{Probability} = \frac{m}{n} = \frac{7}{10} = 0.7$.

(A) The total number of outcomes when two dice are rolled is $6 \times 6 = 36$.
The possible sums of the numbers on the two dice range from $2$ to $12$. The prime numbers in this range are $2, 3, 5, 7,$ and $11$.
Let us list the favorable outcomes for each prime sum:
- Sum $= 2$: $(1, 1)$ [$1$ outcome]
- Sum $= 3$: $(1, 2), (2, 1)$ [$2$ outcomes]
- Sum $= 5$: $(1, 4), (2, 3), (3, 2), (4, 1)$ [$4$ outcomes]
- Sum $= 7$: $(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)$ [$6$ outcomes]
- Sum $= 11$: $(5, 6), (6, 5)$ [$2$ outcomes]
Total number of favorable outcomes $= 1 + 2 + 4 + 6 + 2 = 15$.
Therefore,the required probability $= \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{15}{36} = \frac{5}{12}$.

(B) The total number of outcomes when two dice are rolled is $6 \times 6 = 36$.
The prime numbers on a single die are $2, 3,$ and $5$.
Let the event $E$ be that the numbers on both dice are prime. The favorable outcomes are:
$(2, 2), (2, 3), (2, 5), (3, 2), (3, 3), (3, 5), (5, 2), (5, 3), (5, 5)$.
The number of favorable outcomes is $m = 9$.
The probability $P(E) = \frac{m}{n} = \frac{9}{36} = \frac{1}{4}$.

(A) When two balanced dice are rolled simultaneously,the total number of possible outcomes is $6 \times 6 = 36$.
The favorable outcomes where the same number appears on both dice are $(1, 1), (2, 2), (3, 3), (4, 4), (5, 5),$ and $(6, 6)$.
Thus,the number of favorable outcomes is $m = 6$.
The probability of an event is given by the ratio of the number of favorable outcomes to the total number of possible outcomes.
Therefore,the required probability $= \frac{m}{n} = \frac{6}{36} = \frac{1}{6}$.

(B) The bag contains only orange-flavoured candies.
Since all candies in the bag are orange-flavoured,the event of taking out an orange-flavoured candy is a 'certain event',and its probability is $1$.
Since there are no lemon-flavoured candies in the bag,the event of taking out a lemon-flavoured candy is an 'impossible event',and its probability is $0$.
Therefore,the required probabilities are $1$ and $0$ respectively.

(B) Here,the total number of trousers $= n = 100$.
Manu rejects only those trousers which have major defects.
Therefore,Manu accepts the trousers that are either good or have minor defects.
The number of favorable outcomes for the event that Manu accepts the trouser $= m = 73 + 12 = 85$.
The required probability $= P = \frac{m}{n} = \frac{85}{100} = 0.85$.

(C) The total number of possible outcomes on the circular disc is $n = 12$ (since the numbers are $1$ to $12$).
The event is that the arrow points at the number $7$. There is only $1$ such outcome, so the number of favourable outcomes is $m = 1$.
The probability of an event is given by the formula: $P(E) = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}}$.
Therefore, the required probability is $P = \frac{m}{n} = \frac{1}{12}$.

(D) The number of total outcomes when an unbiased coin is tossed once is $2$ (either Head or Tail).
Since the coin is tossed thrice,the total number of outcomes is calculated by multiplying the number of outcomes for each toss.
Therefore,the total number of outcomes $= 2 \times 2 \times 2 = 2^3 = 8$.

(B) When an unbiased coin is tossed thrice,the total number of possible outcomes is $2^3 = 8$.
These outcomes are: $HHH, HHT, HTH, THH, HTT, THT, TTH, TTT$.
We are looking for the event of receiving exactly two heads.
The favorable outcomes are: $HHT, HTH, THH$.
Thus,the number of favorable outcomes is $m = 3$.
The probability of an event is given by the ratio of favorable outcomes to total outcomes.
Therefore,the required probability $= \frac{m}{n} = \frac{3}{8}$.

(A) The total number of possible outcomes when a coin is tossed thrice is $2^3 = 8$.
These outcomes are: $HHH, HHT, HTH, THH, HTT, THT, TTH, TTT$.
We are looking for the event where the number of heads is greater than the number of tails.
The favorable outcomes are:
$1$. $HHH$ ($3$ heads,$0$ tails)
$2$. $HHT$ ($2$ heads,$1$ tail)
$3$. $HTH$ ($2$ heads,$1$ tail)
$4$. $THH$ ($2$ heads,$1$ tail)
Thus,the number of favorable outcomes is $m = 4$.
The required probability is $\frac{m}{n} = \frac{4}{8} = \frac{1}{2}$.

(A) When three unbiased coins are tossed simultaneously,the total number of possible outcomes is $2^3 = 8$.
The sample space is $S = \{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT\}$.
We need to find the probability of getting at most two tails.
'At most two tails' means getting $0, 1,$ or $2$ tails.
The only outcome that does not satisfy this condition is $TTT$ (three tails).
Thus,the favorable outcomes are ${HHH, HHT, HTH, THH, HTT, THT, TTH}$,which gives $m = 7$ outcomes.
Therefore,the required probability $P = \frac{m}{n} = \frac{7}{8}$.

(C) When three unbiased coins are tossed simultaneously,the total number of possible outcomes is $2^3 = 8$.
These outcomes are: $\{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT\}$.
We are looking for the probability of receiving at least two heads.
The favorable outcomes are those that contain two or three heads: $\{HHT, HTH, THH, HHH\}$.
The number of favorable outcomes is $m = 4$.
Therefore,the required probability $P = \frac{m}{n} = \frac{4}{8} = \frac{1}{2}$.

(D) The total number of possible outcomes when two dice are rolled is $6 \times 6 = 36$.
The product of the numbers on the two dice is odd if and only if both numbers are odd. The odd numbers on a die are ${1, 3, 5}$.
The number of outcomes where both dice show an odd number is $3 \times 3 = 9$. These outcomes are $(1,1), (1,3), (1,5), (3,1), (3,3), (3,5), (5,1), (5,3), (5,5)$.
The product is even in all other cases. Therefore,the number of favorable outcomes is $36 - 9 = 27$.
The probability that the product is even is $\frac{27}{36} = \frac{3}{4}$.

(A) The event of the sun rising in the east is a certain event.
$\therefore$ The probability of a certain event is always $1$.
Hence,the required probability is $1$.

(C) In a question paper of $100$ marks,the possible marks one can score are $0, 1, 2, \dots, 100$.
$\therefore$ The total number of possible outcomes $n = 101$.
The number of outcomes favourable to the event of scoring exactly $100$ marks is $m = 1$.
$\therefore$ The required probability $P = \frac{m}{n} = \frac{1}{101}$.

(B) Total number of balls in the bag $= 6 + 5 + 4 = 15$.
$\therefore$ Number of total outcomes $= n = 15$.
The number of outcomes favouring the event that the ball drawn is not red $= m = 6 \text{ (green)} + 4 \text{ (blue)} = 10$.
$\therefore$ Required probability $= \frac{m}{n} = \frac{10}{15} = \frac{2}{3}$.

(D) The number of total outcomes in the experiment of rolling one balanced die is $6$.
Therefore,the number of total outcomes in the experiment of rolling three balanced dice simultaneously is $6 \times 6 \times 6 = 6^3 = 216$.

(C) non-leap year consists of $365$ days.
$365$ days $= 52$ weeks and $1$ extra day.
The $1$ extra day can be any one of the following: {Sunday,Monday,Tuesday,Wednesday,Thursday,Friday,Saturday}.
There are $7$ total possible outcomes.
For the year to have $53$ Saturdays,the extra day must be a Saturday.
There is only $1$ favorable outcome (Saturday) out of $7$ total outcomes.
Therefore,the required probability $= \frac{1}{7}$.

(D) leap year consists of $366$ days.
$366$ days = $52$ weeks and $2$ extra days.
The $2$ extra days can be any of the following pairs: (Sunday,Monday),(Monday,Tuesday),(Tuesday,Wednesday),(Wednesday,Thursday),(Thursday,Friday),(Friday,Saturday),(Saturday,Sunday).
There are $7$ possible outcomes for these $2$ extra days.
For a leap year to have $53$ Wednesdays,one of the $2$ extra days must be a Wednesday.
From the list above,the pairs containing Wednesday are (Tuesday,Wednesday) and (Wednesday,Thursday).
Thus,there are $2$ favorable outcomes.
Therefore,the required probability = $\frac{2}{7}$.

(D) non-leap year has $28$ days in the month of February.
$28$ days consist of exactly $4$ weeks ($4 \times 7 = 28$ days).
Since there are exactly $4$ weeks,every day of the week occurs exactly $4$ times.
Therefore,it is impossible for the month of February in a non-leap year to have $5$ Thursdays.
Since this is an impossible event,the probability is $0$.

(B) The month of February in a leap year has $29$ days.
$29$ days consist of $4$ complete weeks and $1$ extra day.
This extra day can be any one of the $7$ days of the week (Monday,Tuesday,Wednesday,Thursday,Friday,Saturday,Sunday).
For the month to have $5$ Fridays,the extra day must be a Friday.
Since there is only $1$ favorable outcome out of $7$ possible outcomes,
$\text{Required probability} = \frac{1}{7}$.

(D) Total number of chits in the box is $n = 120$.
$A$ $3$-digit number ranges from $100$ to $120$.
The number of favourable outcomes $m$ is the count of integers from $100$ to $120$ inclusive,which is calculated as $120 - 100 + 1 = 21$.
The probability $P$ of drawing a $3$-digit number is given by the ratio of favourable outcomes to total outcomes:
$P = \frac{m}{n} = \frac{21}{120}$.
Simplifying the fraction by dividing both numerator and denominator by $3$:
$P = \frac{7}{40}$.

(A) The total number of outcomes is $n = 100$.
The numbers from $1$ to $100$ that are $2$-digit numbers are $10, 11, 12, \ldots, 99$.
The number of favorable outcomes $m$ is calculated as $99 - 10 + 1 = 90$.
Therefore,the required probability $P = \frac{m}{n} = \frac{90}{100} = 0.9$.

(C) The sum of the probabilities of an event and its complement is always $1$.
Given that $P(A) = 0.6$.
We know that $P(A) + P(\bar{A}) = 1$.
Therefore,$P(\bar{A}) = 1 - P(A)$.
$P(\bar{A}) = 1 - 0.6 = 0.4$.

(A) The probability of an event $E$,denoted by $P(E)$,always lies in the range $0 \le P(E) \le 1$. Since the probability value is restricted to this interval,it can never be negative.

(D) When an unbiased coin is tossed once,the sample space is $S = \{H, T\}$.
The probability of getting a tail in one toss is $P(T) = \frac{1}{2}$.
Since the coin is tossed thrice,the events are independent.
The probability of getting a tail all three times is $P(T \cap T \cap T) = P(T) \times P(T) \times P(T)$.
Substituting the values: $P = \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{8}$.

(C) The sum of the probability of an event and its complement is always $1$.
Given that $P(\bar{B}) = 0.55$.
We know that $P(B) + P(\bar{B}) = 1$.
Therefore,$P(B) = 1 - P(\bar{B})$.
Substituting the value: $P(B) = 1 - 0.55 = 0.45$.
Thus,the correct option is $C$.

(B) Given the ratio $P(A) : P(\bar{A}) = 3 : 4$.
Let $P(A) = 3x$ and $P(\bar{A}) = 4x$,where $x$ is a constant.
We know that the sum of the probability of an event and its complement is always $1$,i.e.,$P(A) + P(\bar{A}) = 1$.
Substituting the values,we get $3x + 4x = 1$.
$7x = 1$,which implies $x = \frac{1}{7}$.
Therefore,$P(A) = 3x = 3 \times \frac{1}{7} = \frac{3}{7}$.

(D) We know that for any event $A$,the sum of the probability of the event and its complement is $P(A) + P(\bar{A}) = 1$.
Given the ratio $P(A) : P(\bar{A}) = 5 : 3$,we can assume $P(A) = 5k$ and $P(\bar{A}) = 3k$ for some constant $k$.
Substituting these into the identity: $5k + 3k = 1$.
$8k = 1$,which implies $k = \frac{1}{8}$.
Therefore,$P(\bar{A}) = 3k = 3 \times \frac{1}{8} = \frac{3}{8}$.

(C) The month of August has $31$ days.
$31$ days consist of $4$ full weeks and $3$ extra days.
$4 \times 7 = 28$ days.
$31 - 28 = 3$ extra days.
The possible combinations for these $3$ consecutive days are:
$1$. (Monday,Tuesday,Wednesday)
$2$. (Tuesday,Wednesday,Thursday)
$3$. (Wednesday,Thursday,Friday)
$4$. (Thursday,Friday,Saturday)
$5$. (Friday,Saturday,Sunday)
$6$. (Saturday,Sunday,Monday)
$7$. (Sunday,Monday,Tuesday)
Out of these $7$ possibilities,the cases where Monday occurs are:
(Monday,Tuesday,Wednesday),(Saturday,Sunday,Monday),and (Sunday,Monday,Tuesday).
There are $3$ favorable outcomes.
Therefore,the probability is $\frac{3}{7}$.

(B) The month of April has $30$ days.
$30$ days consist of $4$ complete weeks and $2$ extra days.
$4$ weeks contain $4$ Sundays.
The $2$ extra days can be any of the following pairs: (Sunday,Monday),(Monday,Tuesday),(Tuesday,Wednesday),(Wednesday,Thursday),(Thursday,Friday),(Friday,Saturday),or (Saturday,Sunday).
There are $7$ possible outcomes for these $2$ extra days.
For the month to have $5$ Sundays,one of the extra days must be a Sunday.
This happens in $2$ cases: (Sunday,Monday) and (Saturday,Sunday).
Therefore,the probability is $\frac{2}{7}$.

(B) Let event $A$ be the event that the numbers appearing on both the dice are different.
Then,the complementary event $\overline{A}$ is the event that the numbers appearing on both the dice are the same.
The total number of possible outcomes when two dice are rolled is $6 \times 6 = 36$.
The outcomes favorable to event $\overline{A}$ are $(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)$.
Thus,the number of favorable outcomes for $\overline{A}$ is $m = 6$.
The probability of event $\overline{A}$ is $P(\overline{A}) = \frac{m}{n} = \frac{6}{36} = \frac{1}{6}$.
Since $P(A) + P(\overline{A}) = 1$,the probability that the numbers are different is $P(A) = 1 - P(\overline{A}) = 1 - \frac{1}{6} = \frac{5}{6}$.

(C) The total number of possible outcomes when a balanced die is rolled is $S = \{1, 2, 3, 4, 5, 6\}$,so the total number of outcomes $n(S) = 6$.
Let $E$ be the event that the number on the die is greater than $3$.
The favorable outcomes are $E = \{4, 5, 6\}$,so the number of favorable outcomes $n(E) = 3$.
The probability $P(E)$ is given by the formula $P(E) = \frac{n(E)}{n(S)}$.
Substituting the values,we get $P(E) = \frac{3}{6} = \frac{1}{2}$.

(A) When two balanced dice are rolled,the total number of possible outcomes is $6 \times 6 = 36$.
The sample space $S$ is given by: $S = \{(1,1), (1,2), \dots, (6,6)\}$.
We are looking for the event $E$ where the sum of the numbers on the two dice is $12$.
The only outcome that satisfies this condition is $(6, 6)$.
Thus,the number of favorable outcomes is $n(E) = 1$.
The probability $P(E)$ is calculated as: $P(E) = \frac{n(E)}{n(S)} = \frac{1}{36}$.

(B) When two balanced dice are rolled,the total number of possible outcomes is $6 \times 6 = 36$.
Let $E$ be the event that the sum of the numbers on the two dice is $8$.
The possible outcomes for the sum to be $8$ are: $(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)$.
The number of favorable outcomes is $5$.
Therefore,the probability $P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}} = \frac{5}{36}$.

(D) We know that the sum of the probability of an event and its complement is $1$,i.e.,$P(A) + P(\bar{A}) = 1$.
We are given the equation: $P(A) - P(\bar{A}) = 0.5$.
Adding these two equations:
$(P(A) + P(\bar{A})) + (P(A) - P(\bar{A})) = 1 + 0.5$
$2P(A) = 1.5$
Dividing by $2$:
$P(A) = \frac{1.5}{2} = 0.75$.

(B) Total number of cards in a well-shuffled pack = $52$.
There are two black suits in a deck of cards: Spades and Clubs.
Each suit contains exactly one jack.
Therefore,the total number of black jacks (Jack of Spades and Jack of Clubs) = $2$.
The probability of drawing a jack of a black suit is given by the ratio of the number of favorable outcomes to the total number of possible outcomes.
Probability = $\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{2}{52} = \frac{1}{26}$.

(B) The probability of any event $K$,denoted by $P(K)$,is a measure of the likelihood of that event occurring.
By definition,the probability of an event always lies between $0$ and $1$,inclusive.
This means that the minimum possible probability is $0$ (for an impossible event) and the maximum possible probability is $1$ (for a sure event).
Therefore,the condition satisfied by the probability of an event $K$ is $0 \leqslant P(K) \leqslant 1$.

(C) The sample space $S$ for rolling a die is $\{1, 2, 3, 4, 5, 6\}$,so the total number of outcomes is $6$.
The prime numbers between $1$ and $6$ are $2, 3,$ and $5$.
Thus,the number of favorable outcomes is $3$.
The probability of getting a prime number is given by the ratio of the number of favorable outcomes to the total number of outcomes.
Probability $= \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{3}{6} = \frac{1}{2}$.

(A) The total outcomes when a die is rolled are ${1, 2, 3, 4, 5, 6}$.
Total number of outcomes $= 6$.
$A$ composite number is a positive integer greater than $1$ that has at least one divisor other than $1$ and itself.
In the set ${1, 2, 3, 4, 5, 6}$,the composite numbers are $4$ and $6$.
Number of favorable outcomes $= 2$.
Probability $= \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{2}{6} = \frac{1}{3}$.