Mix Examples - Probability Questions in English

(C) In the experiment of throwing a fair die,the possible equally likely outcomes are $1, 2, 3, 4, 5,$ and $6.$
Let event $A$ be the event of 'getting $2$ on the die.' Then,the number of outcomes favourable to $A$ is only $1.$
$\therefore P(A) = \text{P(getting } 2 \text{ on the die)}$
$= \frac{\text{Number of outcomes favourable to } A}{\text{Number of all possible outcomes}}$
$= \frac{1}{6}$
Thus,the probability of getting number $2$ on the die is $\frac{1}{6}$.

(D) There are six elementary outcomes in the experiment of throwing a balanced die once: $1, 2, 3, 4, 5, 6$.
Let event $A$ be: 'The number on the die is a prime number'.
The prime numbers among the outcomes are $2, 3, 5$.
Thus,the number of outcomes favourable to event $A$ is $3$.
The total number of possible outcomes is $6$.
Therefore,the probability of event $A$ is $P(A) = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}} = \frac{3}{6} = \frac{1}{2}$.
Thus,the probability of getting a prime number on the die is $\frac{1}{2}$.

(A) There are six elementary outcomes in the experiment of throwing a balanced die once: $1, 2, 3, 4, 5, 6$.
Let event $A$ be: 'The number on the die is greater than $4$.'
The outcomes favourable to event $A$ are $5$ and $6$.
Hence,the number of outcomes favourable to event $A$ is $2$.
$\therefore P(A) = P(\text{The number on the die is greater than } 4) = \frac{\text{Number of outcomes favourable to } A}{\text{Number of all possible outcomes}}$.
$P(A) = \frac{2}{6} = \frac{1}{3}$.
Thus,the probability of getting a number greater than $4$ is $\frac{1}{3}$.

(B) There are six elementary outcomes in the experiment of throwing a balanced die once: $1, 2, 3, 4, 5, 6$.
Let event $B$ be: 'The number on the die is smaller than $4$.'
The outcomes favourable to event $B$ are $1, 2,$ and $3$.
Hence,the number of outcomes favourable to event $B$ is $3$.
Therefore,the probability $P(B)$ is given by:
$P(B) = \frac{\text{Number of outcomes favourable to } B}{\text{Total number of possible outcomes}}$
$P(B) = \frac{3}{6}$
$P(B) = \frac{1}{2}$
Thus,the probability of getting a number smaller than $4$ is $\frac{1}{2}$.

(C) In the experiment of selecting a card at random from a well-shuffled pack of $52$ cards,
the total number of elementary outcomes is $52$.
Let event $A$ be: "The selected card is a seven."
The number of outcomes favourable to event $A$ is $4$,as there are $4$ sevens in a standard pack of cards (one for each suit: Hearts,Diamonds,Clubs,and Spades).
Therefore,the probability $P(A)$ is given by the ratio of favourable outcomes to total outcomes:
$P(A) = \frac{4}{52} = \frac{1}{13}$.
Thus,the probability that the selected card is a seven is $\frac{1}{13}$.

(D) In the experiment of selecting a card at random from a well-shuffled pack of $52$ cards,the total number of elementary outcomes is $52$.
Let event $B$ be: "The selected card is a spade."
The number of outcomes favourable to event $B$ is $13$,as there are $13$ cards of spades in a standard pack of $52$ cards.
Therefore,the probability $P(B)$ is given by:
$P(B) = \frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}} = \frac{13}{52} = \frac{1}{4}$.
Thus,the probability that the selected card is a spade is $\frac{1}{4}$.

(A) In the experiment of selecting a card at random from a well-shuffled pack of $52$ cards,the total number of elementary outcomes is $52$.
Let event $C$ be the event that the selected card is of a black suit.
$A$ pack of $52$ cards contains two black suits: Spades ($13$ cards) and Clubs ($13$ cards).
Therefore,the number of outcomes favourable to event $C$ is $13 + 13 = 26$.
The probability of event $C$ is given by $P(C) = \frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}$.
$P(C) = \frac{26}{52} = \frac{1}{2}$.
Thus,the probability that the selected card is of a black suit is $1/2$.

(B) In the experiment of selecting a card at random from a well-shuffled pack of $52$ cards,the total number of elementary outcomes is $52$.
Let event $D$ be the event that the selected card is not a king.
Let event $\bar{D}$ be the event that the selected card is a king.
The number of outcomes favourable to event $\bar{D}$ is $4$,as there are $4$ kings in a standard pack of cards.
Therefore,$P(\bar{D}) = \frac{4}{52} = \frac{1}{13}$.
Using the complement rule,$P(D) = 1 - P(\bar{D})$.
Therefore,$P(D) = 1 - \frac{1}{13} = \frac{12}{13}$.
Thus,the probability that the selected card is not a king is $\frac{12}{13}$.

(C) In the experiment of selecting a card at random from a well-shuffled pack of $52$ cards,the total number of elementary outcomes is $52$.
Let event $E$ be: 'The selected card is the queen of hearts.'
The number of outcomes favourable to event $E$ is $1$,as there is only one queen of hearts in a standard pack of $52$ cards.
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,$P(E) = \frac{1}{52}$.
Thus,the probability that the selected card is the queen of hearts is $\frac{1}{52}$.

(D) When two unbiased dice are thrown,the total number of possible outcomes is $6 \times 6 = 36$. The sample space $S$ is given by:
$S = \{(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)\}$
Let $A$ be the event that the sum of the numbers on the two dice is $8$.
The outcomes favourable to event $A$ are $(2,6), (3,5), (4,4), (5,3), (6,2)$.
The number of favourable outcomes is $n(A) = 5$.
The probability of event $A$ is given by $P(A) = \frac{n(A)}{n(S)} = \frac{5}{36}$.

(A) When two unbiased dice are thrown,the sample space $S$ consists of all possible ordered pairs $(x, y)$ where $x, y \in \{1, 2, 3, 4, 5, 6\}$.
The total number of elementary outcomes is $6 \times 6 = 36$.
The sample space is:
$S = \{(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)\}$
Let $B$ be the event that the sum of the numbers on the two dice is $4$.
The outcomes favourable to event $B$ are $(1,3), (2,2), (3,1)$.
The number of favourable outcomes is $n(B) = 3$.
The probability of event $B$ is given by $P(B) = \frac{n(B)}{n(S)} = \frac{3}{36} = \frac{1}{12}$.

(B) When two unbiased dice are thrown,the total number of possible outcomes is $6 \times 6 = 36$. The sample space $S$ is given by:
$S = \{(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)\}$
Let event $C$ be the event that the sum of the numbers on the two dice is greater than $9$. This means the sum can be $10, 11,$ or $12$.
The outcomes favourable to event $C$ are:
Sum $= 10: (4,6), (5,5), (6,4)$
Sum $= 11: (5,6), (6,5)$
Sum $= 12: (6,6)$
Thus,the number of favourable outcomes is $n(C) = 3 + 2 + 1 = 6$.
The probability of event $C$ is given by $P(C) = \frac{n(C)}{n(S)} = \frac{6}{36} = \frac{1}{6}$.

(C) When two unbiased dice are thrown,the sample space $S$ consists of $6 \times 6 = 36$ possible outcomes:
$S = \{(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)\}$
Total number of outcomes $n(S) = 36$.
Let $E$ be the event that the sum of the numbers on the two dice is smaller than $5$.
The possible sums smaller than $5$ are $2, 3,$ and $4$.
The outcomes favourable to event $E$ are:
Sum $= 2: (1,1)$
Sum $= 3: (1,2), (2,1)$
Sum $= 4: (1,3), (2,2), (3,1)$
Thus,the number of favourable outcomes $n(E) = 1 + 2 + 3 = 6$.
The probability $P(E) = \frac{n(E)}{n(S)} = \frac{6}{36} = \frac{1}{6}$.

(D) When two unbiased dice are thrown,the total number of possible outcomes is $6 \times 6 = 36$. The sample space $S$ is given by:
$S = \{(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)\}$
Let $E$ be the event that the sum of the numbers on the two dice is a multiple of $3$ or $4$. The possible sums range from $2$ to $12$. Multiples of $3$ are $3, 6, 9, 12$. Multiples of $4$ are $4, 8, 12$.
Thus,we need the sum to be in the set $\{3, 4, 6, 8, 9, 12\}$.
Outcomes with sum $3$: $(1,2), (2,1)$
Outcomes with sum $4$: $(1,3), (2,2), (3,1)$
Outcomes with sum $6$: $(1,5), (2,4), (3,3), (4,2), (5,1)$
Outcomes with sum $8$: $(2,6), (3,5), (4,4), (5,3), (6,2)$
Outcomes with sum $9$: $(3,6), (4,5), (5,4), (6,3)$
Outcomes with sum $12$: $(6,6)$
Total favorable outcomes = $2 + 3 + 5 + 5 + 4 + 1 = 20$.
Probability $P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{20}{36} = \frac{5}{9}$.

(A) When two unbiased dice are thrown,the total number of possible outcomes is $6 \times 6 = 36$. The sample space $S$ is given by:
$S = \{(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)\}$
We need the sum of the numbers on the two dice to be a multiple of both $3$ and $4$. The least common multiple of $3$ and $4$ is $12$. Thus,we are looking for outcomes where the sum is $12$.
The only outcome where the sum is $12$ is $(6,6)$.
Number of favorable outcomes $= 1$.
Probability $P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{1}{36}$.

(B) When two unbiased dice are thrown,the total number of possible outcomes is $6 \times 6 = 36$. The sample space $S$ is given by:
$S = \{(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)\}$
Let $E$ be the event that the sum of the numbers on the two dice is a prime number. The possible prime sums are $2, 3, 5, 7,$ and $11$.
The outcomes favorable to event $E$ are:
Sum $= 2: (1, 1)$
Sum $= 3: (1, 2), (2, 1)$
Sum $= 5: (1, 4), (2, 3), (3, 2), (4, 1)$
Sum $= 7: (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)$
Sum $= 11: (5, 6), (6, 5)$
Total number of favorable outcomes $= 1 + 2 + 4 + 6 + 2 = 15$.
Therefore,the probability $P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{15}{36} = \frac{5}{12}$.

(C) When two unbiased dice are thrown,the total number of possible outcomes is $6 \times 6 = 36$.
The possible outcomes are:
$(1,1), (1,2), (1,3), (1,4), (1,5), (1,6)$
$(2,1), (2,2), (2,3), (2,4), (2,5), (2,6)$
$(3,1), (3,2), (3,3), (3,4), (3,5), (3,6)$
$(4,1), (4,2), (4,3), (4,4), (4,5), (4,6)$
$(5,1), (5,2), (5,3), (5,4), (5,5), (5,6)$
$(6,1), (6,2), (6,3), (6,4), (6,5), (6,6)$
Let $H$ be the event that the sum of the numbers on the two dice is a factor of $18$. The factors of $18$ are $1, 2, 3, 6, 9, 18$. Since the minimum sum is $2$ and the maximum sum is $12$,the possible sums that are factors of $18$ are $2, 3, 6, 9$.
Favorable outcomes for each sum:
Sum $= 2: (1,1)$
Sum $= 3: (1,2), (2,1)$
Sum $= 6: (1,5), (2,4), (3,3), (4,2), (5,1)$
Sum $= 9: (3,6), (4,5), (5,4), (6,3)$
Total number of favorable outcomes $= 1 + 2 + 5 + 4 = 12$.
Probability $P(H) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{12}{36} = \frac{1}{3}$.

(D) The sample space $S$ for throwing two unbiased dice is given by the set of all possible ordered pairs $(x, y)$ where $x, y \in \{1, 2, 3, 4, 5, 6\}$.
There are $6 \times 6 = 36$ possible outcomes:
$S = \{(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)\}$
Let $E$ be the event that the sum of the numbers on the two dice is smaller than $13$.
The minimum sum is $1 + 1 = 2$ and the maximum sum is $6 + 6 = 12$.
Since the maximum possible sum is $12$,which is always less than $13$,every outcome in the sample space satisfies the condition.
Thus,$E = S$ and the number of favorable outcomes is $36$.
The probability $P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{36}{36} = 1$.
Therefore,the probability is $1$.

(A) All the possible outcomes in the experiment of throwing two unbiased dice are as follows:
$(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)$
$(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6)$
$(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)$
$(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)$
$(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)$
$(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)$
Hence,the total number of elementary outcomes is $36$.
Let event $J$ be: 'The sum of numbers on two dice is smaller than $2$.'
The minimum sum of numbers on two dice is $1 + 1 = 2$.
Since the sum of numbers on two dice is always from $2$ to $12$,it is never smaller than $2$.
Therefore,event $J$ is an impossible event.
The probability of an impossible event is $0$.
Thus,$P(J) = 0$.

(B) The total number of elementary outcomes in the experiment of selecting a bulb at random from a carton carrying $20$ bulbs is $20$.
Total number of bulbs in the carton $= 20$.
Number of defective bulbs in the carton $= 4$.
Therefore,the number of non-defective bulbs in the carton $= 20 - 4 = 16$.
Let event $A$ be the event that the bulb selected at random is non-defective.
The number of outcomes favourable to event $A = 16$.
The probability of event $A$ is given by $P(A) = \frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}$.
$P(A) = \frac{16}{20} = \frac{4}{5}$.
Thus,the probability that the bulb selected at random is non-defective is $\frac{4}{5}$.

(C) Total number of balls in the box $= 5 + 8 + 4 = 17$.
Therefore,the total number of possible outcomes in the experiment of drawing a ball from the box $= 17$.
Let event $A$ be the event that the ball drawn is red.
There are $5$ red balls in the box.
Therefore,the number of outcomes favourable to event $A = 5$.
The probability of an event is given by the formula: $P(A) = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}}$.
Thus,$P(A) = \frac{5}{17}$.

(D) Total number of balls in the box $= 5 + 8 + 4 = 17$.
Therefore,the total number of elementary outcomes in the experiment of drawing a ball from the box $= 17$.
Let event $B$ be the event that the ball drawn is white.
There are $8$ white balls in the box.
Therefore,the number of outcomes favourable to event $B = 8$.
The probability of an event is given by the formula: $P(E) = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}}$.
Thus,$P(B) = \frac{8}{17}$.

(A) Total number of balls in the box $= 5 + 8 + 4 = 17$.
Therefore,the total number of elementary outcomes in the experiment of drawing a ball from the box $= 17$.
Let event $C$ be the event that the ball drawn is not green.
Let event $\overline{C}$ be the event that the ball drawn is green.
There are $4$ green balls in the box.
Therefore,the number of outcomes favourable to event $\overline{C} = 4$.
Thus,$P(\overline{C}) = \frac{4}{17}$.
Using the property of complementary events,$P(C) = 1 - P(\overline{C})$.
Therefore,$P(C) = 1 - \frac{4}{17} = \frac{17 - 4}{17} = \frac{13}{17}$.

(B) The month of July in any year has $31$ days.
These $31$ days consist of $4$ complete weeks and $3$ additional days.
Therefore,every day of the week will occur at least $4$ times.
The remaining $3$ days can be any of the following $7$ possible triplets:
$1$. (Sunday,Monday,Tuesday)
$2$. (Monday,Tuesday,Wednesday)
$3$. (Tuesday,Wednesday,Thursday)
$4$. (Wednesday,Thursday,Friday)
$5$. (Thursday,Friday,Saturday)
$6$. (Friday,Saturday,Sunday)
$7$. (Saturday,Sunday,Monday)
Total number of elementary outcomes $= 7$.
The outcomes that contain a Sunday are:
$1$. (Sunday,Monday,Tuesday)
$6$. (Friday,Saturday,Sunday)
$7$. (Saturday,Sunday,Monday)
Number of favourable outcomes $= 3$.
Therefore,the required probability $= \frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}} = \frac{3}{7}$.

(C) leap year has $366$ days,which is equivalent to $52$ complete weeks and $2$ extra days.
Since $52$ weeks contain $52$ Thursdays,the occurrence of a $53$rd Thursday depends on the remaining $2$ days.
The possible pairs for these $2$ days are:
(Sunday,Monday),(Monday,Tuesday),(Tuesday,Wednesday),(Wednesday,Thursday),(Thursday,Friday),(Friday,Saturday),(Saturday,Sunday).
There are $7$ total possible outcomes.
The outcomes that include a Thursday are (Wednesday,Thursday) and (Thursday,Friday).
Thus,there are $2$ favorable outcomes.
The probability is calculated as $\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{2}{7}$.

(D) Area of the rectangle $=$ Length $\times$ Breadth
$= 28 \, cm \times 22 \, cm = 616 \, cm^2$
For the circle,radius $r = 7 \, cm$
Area of the circle $= \pi r^2 = \frac{22}{7} \times 7 \times 7 = 154 \, cm^2$
The probability that a ball hitting the rectangle hits the circle is given by the ratio of the area of the circle to the area of the rectangle.
Required Probability $= \frac{\text{Area of the circle}}{\text{Area of the rectangle}}$
$= \frac{154}{616} = \frac{1}{4}$
Thus,the probability is $\frac{1}{4}$.

(A) In the experiment of tossing three balanced coins simultaneously,the sample space $S$ consists of the following elementary outcomes:
$S = \{(HHH), (HHT), (HTH), (THH), (HTT), (THT), (TTH), (TTT)\}$
where $H$ represents head and $T$ represents tail.
Total number of elementary outcomes $= 8$.
Let event $A$ be the event of receiving at least two heads (i.e.,$2$ or $3$ heads).
The outcomes favourable to event $A$ are: $(HHH), (HHT), (HTH), (THH)$.
Number of favourable outcomes $= 4$.
The probability of event $A$ is given by $P(A) = \frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}$.
$P(A) = \frac{4}{8} = \frac{1}{2}$.
Thus,the probability of receiving at least two heads is $\frac{1}{2}$.

(B) The total number of possible outcomes when a balanced die is thrown is $S = \{1, 2, 3, 4, 5, 6\}$,so the total number of outcomes $n(S) = 6$.
Let $E$ be the event of getting an even number on the die.
The even numbers on a die are $\{2, 4, 6\}$,so the number of favorable outcomes $n(E) = 3$.
The probability of the event $E$ is given by $P(E) = \frac{n(E)}{n(S)} = \frac{3}{6} = \frac{1}{2}$.

(C) When a balanced die is thrown,the sample space $S$ is given by $S = \{1, 2, 3, 4, 5, 6\}$.
The total number of possible outcomes is $n(S) = 6$.
Let $E$ be the event that the number on the die is a multiple of $3$.
The multiples of $3$ in the sample space are $3$ and $6$.
Thus,the favorable outcomes are $E = \{3, 6\}$.
The number of favorable outcomes is $n(E) = 2$.
The probability $P(E)$ is given by the formula $P(E) = \frac{n(E)}{n(S)}$.
Substituting the values,$P(E) = \frac{2}{6} = \frac{1}{3}$.

(D) When a balanced die is thrown,the total number of possible outcomes is $S = \{1, 2, 3, 4, 5, 6\}$.
Thus,the total number of outcomes $n(S) = 6$.
Let $E$ be the event that the number on the die is smaller than $3$.
The favorable outcomes are $E = \{1, 2\}$.
Thus,the number of favorable outcomes $n(E) = 2$.
The probability $P(E)$ is given by the formula $P(E) = \frac{n(E)}{n(S)}$.
$P(E) = \frac{2}{6} = \frac{1}{3}$.

(A) When two dice are rolled,the total number of possible outcomes is $6 \times 6 = 36$.
Let the outcomes be $(x, y)$ where $x$ and $y$ are the numbers on the first and second die respectively.
The sum $(x + y)$ is even if:
$1.$ Both $x$ and $y$ are odd: There are $3$ choices for $x$ $(1, 3, 5)$ and $3$ choices for $y$ $(1, 3, 5)$. Total outcomes = $3 \times 3 = 9$.
$2.$ Both $x$ and $y$ are even: There are $3$ choices for $x$ $(2, 4, 6)$ and $3$ choices for $y$ $(2, 4, 6)$. Total outcomes = $3 \times 3 = 9$.
Total favorable outcomes = $9 + 9 = 18$.
Probability = $\frac{\text{Favorable outcomes}}{\text{Total outcomes}} = \frac{18}{36} = \frac{1}{2}$.

(B) When two balanced dice are rolled,the total number of possible outcomes is $6 \times 6 = 36$.
The sum of the numbers on the two dice is odd if one die shows an even number and the other shows an odd number.
Let $E$ be the event that the sum is odd.
Possible outcomes for (Odd,Even) are: $(1,2), (1,4), (1,6), (3,2), (3,4), (3,6), (5,2), (5,4), (5,6)$ (Total $9$ outcomes).
Possible outcomes for (Even,Odd) are: $(2,1), (2,3), (2,5), (4,1), (4,3), (4,5), (6,1), (6,3), (6,5)$ (Total $9$ outcomes).
Total favorable outcomes = $9 + 9 = 18$.
Probability $P(E) = \frac{\text{Favorable outcomes}}{\text{Total outcomes}} = \frac{18}{36} = \frac{1}{2}$.

(C) When two balanced dice are rolled,the total number of possible outcomes is $6 \times 6 = 36$.
Let $S$ be the sum of the numbers on the two dice. The possible values for $S$ range from $2$ to $12$.
The multiples of $5$ in this range are $5$ and $10$.
We identify the pairs $(d_1, d_2)$ that result in these sums:
For $S = 5$: $(1, 4), (2, 3), (3, 2), (4, 1)$ (Total $4$ outcomes).
For $S = 10$: $(4, 6), (5, 5), (6, 4)$ (Total $3$ outcomes).
The total number of favorable outcomes is $4 + 3 = 7$.
Therefore,the probability $P = \frac{\text{Favorable outcomes}}{\text{Total outcomes}} = \frac{7}{36}$.

(D) When two balanced dice are rolled,the total number of possible outcomes is $6 \times 6 = 36$.
The possible sums of the numbers on the two dice range from $2$ to $12$.
The sums that are multiples of $6$ are $6$ and $12$.
For a sum of $6$,the possible outcomes are $(1, 5), (2, 4), (3, 3), (4, 2), (5, 1)$. There are $5$ such outcomes.
For a sum of $12$,the only possible outcome is $(6, 6)$. There is $1$ such outcome.
Total favorable outcomes = $5 + 1 = 6$.
Probability = $\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{6}{36} = \frac{1}{6}$.

(A) When two balanced dice are rolled,the total number of possible outcomes is $6 \times 6 = 36$.
We need to find the probability that the sum of the numbers on the two dice is greater than $8$.
The possible sums greater than $8$ are $9, 10, 11,$ and $12$.
The outcomes resulting in these sums are:
Sum $= 9: (3, 6), (4, 5), (5, 4), (6, 3)$ ($4$ outcomes)
Sum $= 10: (4, 6), (5, 5), (6, 4)$ ($3$ outcomes)
Sum $= 11: (5, 6), (6, 5)$ ($2$ outcomes)
Sum $= 12: (6, 6)$ ($1$ outcome)
Total favorable outcomes $= 4 + 3 + 2 + 1 = 10$.
The probability is $\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{10}{36} = \frac{5}{18}$.

(B) When two balanced dice are rolled,the total number of possible outcomes is $6 \times 6 = 36$.
We need to find the probability that the sum of the numbers on the two dice is less than $7$.
The possible outcomes where the sum is less than $7$ are:
Sum $= 2$: $(1, 1)$ ($1$ outcome)
Sum $= 3$: $(1, 2), (2, 1)$ ($2$ outcomes)
Sum $= 4$: $(1, 3), (2, 2), (3, 1)$ ($3$ outcomes)
Sum $= 5$: $(1, 4), (2, 3), (3, 2), (4, 1)$ ($4$ outcomes)
Sum $= 6$: $(1, 5), (2, 4), (3, 3), (4, 2), (5, 1)$ ($5$ outcomes)
Total favorable outcomes $= 1 + 2 + 3 + 4 + 5 = 15$.
The probability is given by $\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{15}{36}$.
Simplifying the fraction by dividing both numerator and denominator by $3$,we get $\frac{15 \div 3}{36 \div 3} = \frac{5}{12}$.

(C) The total number of cards in a well-shuffled pack is $52$.
Therefore,the total number of possible outcomes is $n(S) = 52$.
In a standard deck of cards,there are $4$ queens (one for each suit: hearts,diamonds,clubs,and spades).
Therefore,the number of favorable outcomes for drawing a queen is $n(E) = 4$.
The probability $P(E)$ of an event $E$ is given by the formula:
$P(E) = \frac{n(E)}{n(S)}$
Substituting the values:
$P(E) = \frac{4}{52} = \frac{1}{13}$
Thus,the probability of drawing a queen is $\frac{1}{13}$.

(D) The total number of cards in a well-shuffled pack is $52$.
Face cards are the cards that have faces on them,which are Kings,Queens,and Jacks.
In each suit,there are $3$ face cards (one King,one Queen,and one Jack).
Since there are $4$ suits in a deck (Hearts,Diamonds,Clubs,and Spades),the total number of face cards is $4 \times 3 = 12$.
The probability of drawing a face card is given by the ratio of the number of favorable outcomes to the total number of possible outcomes.
Probability $P = \frac{\text{Number of face cards}}{\text{Total number of cards}} = \frac{12}{52}$.
Simplifying the fraction by dividing both the numerator and the denominator by $4$,we get $P = \frac{3}{13}$.

(A) The total number of cards in a well-shuffled pack is $52$.
There are $4$ suits in a deck of cards: hearts,diamonds,clubs,and spades.
Each suit contains $13$ cards.
Therefore,the number of diamond cards is $13$.
The probability of drawing a diamond card is given by the ratio of the number of favorable outcomes to the total number of possible outcomes.
Probability $P = \frac{\text{Number of diamond cards}}{\text{Total number of cards}} = \frac{13}{52} = \frac{1}{4}$.

(B) The total number of cards in a well-shuffled pack is $n(S) = 52$.
There are two black suits in a deck of cards: Spades and Clubs.
Each suit contains exactly one king.
Therefore,the number of kings of a black suit is $n(E) = 2$ (one King of Spades and one King of Clubs).
The probability $P(E)$ of drawing a king of a black suit is given by the formula $P(E) = \frac{n(E)}{n(S)}$.
Substituting the values,we get $P(E) = \frac{2}{52} = \frac{1}{26}$.

(C) In a test of $50$ marks,the possible scores a student can obtain range from $0$ to $50$ inclusive.
Total number of possible outcomes = $50 - 0 + 1 = 51$.
The student can score any integer value from $0, 1, 2, \dots, 50$.
The favorable outcomes are scoring $30$ marks or $35$ marks.
Number of favorable outcomes = $2$ (specifically the scores $30$ and $35$).
Probability = $\frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}} = \frac{2}{51}$.

(D) Total number of students in the class = $35 \text{ (boys)} + 25 \text{ (girls)} = 60$.
Number of favorable outcomes (selecting a girl) = $25$.
The probability $P$ of an event is given by the formula: $P(\text{Girl}) = \frac{\text{Number of girls}}{\text{Total number of students}}$.
$P(\text{Girl}) = \frac{25}{60}$.
Simplifying the fraction by dividing both numerator and denominator by $5$, we get: $P(\text{Girl}) = \frac{5}{12}$.

(A) Total number of defective ballpens = $12$.
Total number of non-defective ballpens = $132$.
Total number of ballpens in the lot = $12 + 132 = 144$.
The probability of selecting a defective ballpen is given by the ratio of the number of defective ballpens to the total number of ballpens.
$P(\text{defective}) = \frac{\text{Number of defective ballpens}}{\text{Total number of ballpens}} = \frac{12}{144} = \frac{1}{12}$.
Thus,the probability is $\frac{1}{12}$.

(A) Total number of tea packets = $50$.
Number of under-weight packets = $5$.
The probability $P(E)$ of an event $E$ is given by the formula: $P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}}$.
Here, the favorable outcome is selecting an under-weight packet.
Therefore, $P(\text{under-weight}) = \frac{5}{50} = \frac{1}{10} = 0.1$.
Thus, the probability of selecting an under-weight packet is $0.1$.

(B) Total number of balls = $5 + 7 + 4 + 2 = 18$.
Let $E$ be the event of selecting a white or blue ball.
Number of white balls = $5$.
Number of blue balls = $2$.
Number of favorable outcomes = $5 + 2 = 7$.
Probability $P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{7}{18}$.

(A) Total number of balls $= 5 + 7 + 4 + 2 = 18$.
Number of red balls $= 7$.
Number of black balls $= 4$.
Number of favorable outcomes (red or black) $= 7 + 4 = 11$.
Probability $= \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{11}{18}$.

(D) Total number of balls = $5 + 7 + 4 + 2 = 18$.
Number of white balls = $5$.
Number of balls that are not white = Total balls - White balls = $18 - 5 = 13$.
Probability of selecting a ball that is not white = $\frac{\text{Number of non-white balls}}{\text{Total number of balls}} = \frac{13}{18}$.

(A) Total number of balls = $5 + 7 + 4 + 2 = 18$.
Number of white balls = $5$.
Number of black balls = $4$.
Number of balls that are neither white nor black = Total balls - (White balls + Black balls) = $18 - (5 + 4) = 18 - 9 = 9$.
The balls that are neither white nor black are the red and blue balls,which are $7 + 2 = 9$.
Probability = $\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{9}{18} = \frac{1}{2}$.

(B) When two dice are rolled,the total number of possible outcomes is $6 \times 6 = 36$.
We need to find the probability that the sum $S$ of the numbers on the two dice lies between $5$ and $9$,which means $5 < S < 9$.
This implies the sum $S$ can be $6, 7,$ or $8$.
- For $S = 6$,the possible outcomes are $(1, 5), (2, 4), (3, 3), (4, 2), (5, 1)$,which are $5$ outcomes.
- For $S = 7$,the possible outcomes are $(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)$,which are $6$ outcomes.
- For $S = 8$,the possible outcomes are $(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)$,which are $5$ outcomes.
Total favorable outcomes = $5 + 6 + 5 = 16$.
Probability = $\frac{\text{Favorable outcomes}}{\text{Total outcomes}} = \frac{16}{36} = \frac{4}{9}$.

(C) When two balanced dice are rolled,the total number of possible outcomes is $6 \times 6 = 36$.
For the product of the numbers on the two dice to be odd,both numbers must be odd.
The odd numbers on a die are $\{1, 3, 5\}$.
Thus,the number of favorable outcomes is $3 \times 3 = 9$.
The favorable outcomes are: $(1,1), (1,3), (1,5), (3,1), (3,3), (3,5), (5,1), (5,3), (5,5)$.
The probability $P$ is given by the ratio of favorable outcomes to total outcomes:
$P = \frac{9}{36} = \frac{1}{4}$.