Probability Questions in English

(D) The sample space $S$ for tossing two coins is $S = \{HH, HT, TH, TT\}$.
The event $A$ (first coin shows head) is $A = \{HH, HT\}$. Thus,$P(A) = 2/4 = 1/2$.
The event $B$ (second coin shows tail) is $B = \{HT, TT\}$. Thus,$P(B) = 2/4 = 1/2$.
The intersection $A \cap B$ is the event that the first coin shows head $AND$ the second coin shows tail,which is $A \cap B = \{HT\}$.
Thus,$P(A \cap B) = 1/4$.
Since $P(A) \times P(B) = (1/2) \times (1/2) = 1/4$,we have $P(A \cap B) = P(A) \times P(B)$.
Therefore,the events $A$ and $B$ are independent.

(B) We are given the condition $P(A_1 \cup A_2) = 1 - P(A_1^c) P(A_2^c)$.
Using the complement rule,$P(A^c) = 1 - P(A)$,we can rewrite the expression as:
$P(A_1 \cup A_2) = 1 - (1 - P(A_1))(1 - P(A_2))$
Expanding the right side:
$P(A_1 \cup A_2) = 1 - (1 - P(A_1) - P(A_2) + P(A_1)P(A_2))$
$P(A_1 \cup A_2) = 1 - 1 + P(A_1) + P(A_2) - P(A_1)P(A_2)$
$P(A_1 \cup A_2) = P(A_1) + P(A_2) - P(A_1)P(A_2)$
By the definition of the union of two events,$P(A_1 \cup A_2) = P(A_1) + P(A_2) - P(A_1 \cap A_2)$.
Comparing the two expressions,we get $P(A_1 \cap A_2) = P(A_1)P(A_2)$.
This is the condition for two events to be independent.

(D) When two fair dice are tossed,the outcome of the first die does not affect the outcome of the second die.
Event $A$ is that the first die shows an even number: $A = \{2, 4, 6\}$. The probability $P(A) = 3/6 = 1/2$.
Event $B$ is that the second die shows an odd number: $B = \{1, 3, 5\}$. The probability $P(B) = 3/6 = 1/2$.
The intersection $A \cap B$ represents the event where the first die is even $AND$ the second die is odd. The number of such outcomes is $3 \times 3 = 9$ out of $36$ total outcomes.
Thus,$P(A \cap B) = 9/36 = 1/4$.
Since $P(A) \times P(B) = (1/2) \times (1/2) = 1/4$,we have $P(A \cap B) = P(A) \times P(B)$.
Therefore,the events $A$ and $B$ are independent.

(A) Total number of cards = $52$.
Number of diamond cards $n(A) = 13$,so $P(A) = 13/52 = 1/4$.
Number of ace cards $n(B) = 4$,so $P(B) = 4/52 = 1/13$.
Number of ace of diamond cards $n(A \cap B) = 1$,so $P(A \cap B) = 1/52$.
For events to be independent,$P(A \cap B)$ must be equal to $P(A) \times P(B)$.
$P(A) \times P(B) = (1/4) \times (1/13) = 1/52$.
Since $P(A \cap B) = P(A) \times P(B) = 1/52$,the events $A$ and $B$ are independent.

(B) Since $A \cap \bar B$ and $A \cap B$ are mutually exclusive events such that $A = (A \cap \bar B) \cup (A \cap B)$.
Therefore,$P(A) = P(A \cap \bar B) + P(A \cap B)$.
This implies $P(A \cap \bar B) = P(A) - P(A \cap B)$.
Since $A$ and $B$ are independent,$P(A \cap B) = P(A)P(B)$.
Substituting this,we get $P(A \cap \bar B) = P(A) - P(A)P(B) = P(A)(1 - P(B))$.
Since $P(\bar B) = 1 - P(B)$,we have $P(A \cap \bar B) = P(A)P(\bar B)$.
Thus,$A$ and $\bar B$ are also independent.

(A) Given that $A, B,$ and $C$ are mutually independent events,we have $P(A \cap B) = P(A)P(B)$,$P(A \cap C) = P(A)P(C)$,and $P(A \cap B \cap C) = P(A)P(B)P(C)$.
For $S_1$: We check if $P(A \cap (B \cup C)) = P(A)P(B \cup C)$.
$P(A \cap (B \cup C)) = P((A \cap B) \cup (A \cap C)) = P(A \cap B) + P(A \cap C) - P(A \cap B \cap C)$
$= P(A)P(B) + P(A)P(C) - P(A)P(B)P(C) = P(A)[P(B) + P(C) - P(B)P(C)]$
$= P(A)P(B \cup C)$. Thus,$S_1$ is true.
For $S_2$: We check if $P(A \cap (B \cap C)) = P(A)P(B \cap C)$.
$P(A \cap (B \cap C)) = P(A \cap B \cap C) = P(A)P(B)P(C) = P(A)P(B \cap C)$. Thus,$S_2$ is true.
Therefore,both $S_1$ and $S_2$ are true.

(C) The formula for the union of two events is $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
Substituting the given values: $5/6 = 2/3 + 1/2 - P(A \cap B)$.
$5/6 = 4/6 + 3/6 - P(A \cap B)$.
$5/6 = 7/6 - P(A \cap B)$.
$P(A \cap B) = 7/6 - 5/6 = 2/6 = 1/3$.
Since $P(A \cap B) \neq 0$,the events are not mutually exclusive.
Now,check for independence: $P(A) \times P(B) = (2/3) \times (1/2) = 1/3$.
Since $P(A \cap B) = P(A) \times P(B) = 1/3$,the events $A$ and $B$ are independent.

(A) pack of cards contains $52$ cards in total.
The number of aces in a pack is $4$.
The probability of drawing an ace in a single draw is $P(A) = \frac{4}{52} = \frac{1}{13}$.
Since the cards are drawn with replacement,the probability of drawing an ace in the second draw remains the same,$P(A) = \frac{1}{13}$.
The probability of drawing two aces successively is $P(A \cap A) = P(A) \times P(A) = \frac{1}{13} \times \frac{1}{13} = \frac{1}{169}$.

(C) When two dice are thrown,the total number of possible outcomes is $6 \times 6 = 36$.
We need to find the probability of getting a sum greater than $7$,which means the sum can be $8, 9, 10, 11,$ or $12$.
The favorable outcomes are:
Sum $= 8$: $(2,6), (3,5), (4,4), (5,3), (6,2)$ ($5$ outcomes)
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 $= 5 + 4 + 3 + 2 + 1 = 15$.
Required probability $= \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{15}{36} = \frac{5}{12}$.

(A) Total number of balls in the bag $= 3 \text{ (black)} + 4 \text{ (white)} = 7 \text{ balls}$.
Number of favorable outcomes (drawing a white ball) $= 4$.
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 drawing a white ball $= \frac{4}{7}$.

(D) The probability of getting a head is $P(H) = 1/2$ and the probability of not getting a head is $P(T) = 1/2$.
Since $A$ starts the game,$A$ wins if he gets a head on the $1^{st}, 3^{rd}, 5^{th}, \dots$ toss.
$A$ wins on the $1^{st}$ toss with probability $1/2$.
$A$ wins on the $3^{rd}$ toss if the sequence is $T, T, H$ with probability $(1/2) \times (1/2) \times (1/2) = (1/2)^3$.
$A$ wins on the $5^{th}$ toss if the sequence is $T, T, T, T, H$ with probability $(1/2)^5$.
Thus,the total probability of $A$ winning is an infinite geometric series:
$P(A) = 1/2 + (1/2)^3 + (1/2)^5 + \dots$
This is a geometric progression with first term $a = 1/2$ and common ratio $r = (1/2)^2 = 1/4$.
The sum of an infinite geometric series is $S = a / (1 - r)$.
$P(A) = (1/2) / (1 - 1/4) = (1/2) / (3/4) = (1/2) \times (4/3) = 2/3$.

(C) When two balanced dice are tossed,the total number of possible outcomes is $6 \times 6 = 36$.
We are looking for the event where the sum of the numbers on the upper faces is $9$.
The possible pairs $(d_1, d_2)$ that result in a sum of $9$ are: $(3, 6), (4, 5), (5, 4), (6, 3)$.
There are $4$ such favorable outcomes.
The probability of the event is given by the ratio of favorable outcomes to total outcomes:
$P = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{4}{36} = \frac{1}{9}$.

(A) standard pack of cards contains $52$ cards in total.
There are $4$ aces in a deck of cards (one for each suit: hearts,diamonds,clubs,and spades).
The probability of an event is given by the ratio of the number of favorable outcomes to the total number of possible outcomes.
Probability $P(\text{Ace}) = \frac{\text{Number of Aces}}{\text{Total number of cards}} = \frac{4}{52}$.
Simplifying the fraction,we get $\frac{4}{52} = \frac{1}{13}$.

(C) The word $PROBABILITY$ contains $11$ letters in total: $P, R, O, B, A, B, I, L, I, T, Y$.
The vowels in the word are $O, A, I, I$.
There are $4$ vowels in total.
The total number of possible outcomes is $11$.
The probability of selecting a vowel is given by the ratio of the number of favorable outcomes to the total number of outcomes.
Therefore,the required probability $= 4/11$.

(B) The total number of ways to place $n$ letters into $n$ envelopes is $n!$.
There is only $1$ way in which all letters are placed in their correct corresponding envelopes.
Therefore,the probability that all letters are placed in the correct envelopes is $P(\text{correct}) = \frac{1}{n!}$.
The probability that all the letters are not kept in the right envelope is the complement of the probability that all are in the right envelope.
Required probability $= 1 - P(\text{correct}) = 1 - \frac{1}{n!}$.

(A) The total number of pages is $100$. The page numbers are from $1$ to $100$.
We need to find the numbers between $1$ and $100$ such that the sum of their digits is $11$.
For a two-digit number $xy$,the sum of digits is $x + y = 11$,where $1 \le x \le 9$ and $0 \le y \le 9$.
The possible pairs $(x, y)$ are $(2, 9), (3, 8), (4, 7), (5, 6), (6, 5), (7, 4), (8, 3), (9, 2)$.
These correspond to the page numbers: $29, 38, 47, 56, 65, 74, 83, 92$.
There are $8$ such numbers.
For the number $100$,the sum of digits is $1 + 0 + 0 = 1 \neq 11$.
Thus,the number of favourable outcomes is $8$.
The total number of outcomes is $100$.
The probability is $\frac{8}{100} = \frac{2}{25}$.

(C) The sample space $S$ for two children in a family is given by $S = \{BB, BG, GB, GG\}$,where $B$ denotes a boy and $G$ denotes a girl.
Total number of possible outcomes $n(S) = 4$.
The event $E$ that both children are boys is $E = \{BB\}$.
Number of favorable outcomes $n(E) = 1$.
The probability $P(E) = \frac{n(E)}{n(S)} = \frac{1}{4}$.

(C) When a dice is thrown,the total number of possible outcomes is $6$. The probability of getting $1$ in a single throw is $\frac{1}{6}$.
The probability of $NOT$ getting $1$ in a single throw is $1 - \frac{1}{6} = \frac{5}{6}$.
Since the two throws are independent events,the probability of getting $1$ in the first throw $AND$ not getting $1$ in the second throw is calculated as follows:
Required Probability $= P(\text{getting } 1 \text{ in first throw}) \times P(\text{not getting } 1 \text{ in second throw})$
Required Probability $= \frac{1}{6} \times \frac{5}{6} = \frac{5}{36}$.

(C) standard pack of cards contains $52$ cards,out of which $4$ are kings.
When the first card is drawn,the probability of it being a king is $P(K_1) = \frac{4}{52} = \frac{1}{13}$.
After drawing one king,there are $51$ cards remaining in the pack,and only $3$ kings left.
Therefore,the probability of the second card being a king,given that the first was a king,is $P(K_2 | K_1) = \frac{3}{51} = \frac{1}{17}$.
The probability that both cards are kings is $P(K_1 \cap K_2) = P(K_1) \times P(K_2 | K_1) = \frac{4}{52} \times \frac{3}{51} = \frac{1}{13} \times \frac{1}{17} = \frac{1}{221}$.

(B) The sample space for tossing a coin is ${H, T}$,so the probability of getting a head is $P(H) = \frac{1}{2}$.
The sample space for rolling a dice is ${1, 2, 3, 4, 5, 6}$,so the probability of getting a $6$ is $P(6) = \frac{1}{6}$.
Since these are independent events,the joint probability is $P(H \cap 6) = P(H) \times P(6) = \frac{1}{2} \times \frac{1}{6} = \frac{1}{12}$.

(B) When a coin is tossed twice,the sample space $S$ is given by: $S = \{(H, H), (H, T), (T, H), (T, T)\}$.
The total number of possible outcomes is $n(S) = 4$.
Let $E$ be the event of getting a head both times. The favorable outcome is $E = \{(H, H)\}$.
The number of favorable outcomes is $n(E) = 1$.
The probability $P(E)$ is given by the formula: $P(E) = \frac{n(E)}{n(S)} = \frac{1}{4}$.

(C) Total number of cards in a pack $= 52$.
Since the cards are drawn with replacement,the two events are independent.
The probability of drawing a diamond card in the first draw is $P(\text{Diamond}) = \frac{13}{52} = \frac{1}{4}$.
The probability of drawing a king in the second draw is $P(\text{King}) = \frac{4}{52} = \frac{1}{13}$.
Since the events are independent,the required probability is $P(\text{Diamond}) \times P(\text{King}) = \frac{1}{4} \times \frac{1}{13} = \frac{1}{52}$.

(B) When two dice are thrown simultaneously,the total number of possible outcomes is $6 \times 6 = 36$.
Let $E$ be the event of getting a sum of $2$,$8$,or $12$.
$1$. Outcomes for sum $2$: $(1, 1)$ - Total $1$ outcome.
$2$. Outcomes for sum $8$: $(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)$ - Total $5$ outcomes.
$3$. Outcomes for sum $12$: $(6, 6)$ - Total $1$ outcome.
Total favorable outcomes $= 1 + 5 + 1 = 7$.
Therefore,the required probability $P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{7}{36}$.

(B) Let $A$ be the event of getting $4, 5$ or $6$ in the first throw. The total outcomes for a single throw are $6$. The favorable outcomes are ${4, 5, 6}$,so the number of favorable outcomes is $3$.
$P(A) = \frac{3}{6} = \frac{1}{2}$.
Let $B$ be the event of getting $1, 2, 3$ or $4$ in the second throw. The favorable outcomes are ${1, 2, 3, 4}$,so the number of favorable outcomes is $4$.
$P(B) = \frac{4}{6} = \frac{2}{3}$.
Since the two throws are independent events,the probability of both events occurring is $P(A \cap B) = P(A) \times P(B)$.
$P(A \cap B) = \frac{1}{2} \times \frac{2}{3} = \frac{2}{6} = \frac{1}{3}$.

(A) The total number of ways to draw $2$ cards from $52$ is $^{52}C_2 = \frac{52 \times 51}{2} = 1326$.
The number of ways to draw $2$ cards such that none of them is an ace (i.e.,choosing from the $48$ non-ace cards) is $^{48}C_2 = \frac{48 \times 47}{2} = 1128$.
The probability of drawing no ace is $P(\text{no ace}) = \frac{1128}{1326} = \frac{48}{52} \times \frac{47}{51} = \frac{12}{13} \times \frac{47}{51} = \frac{4 \times 47}{13 \times 17} = \frac{188}{221}$.
The probability that at least one card is an ace is $1 - P(\text{no ace}) = 1 - \frac{188}{221} = \frac{221 - 188}{221} = \frac{33}{221}$.

(A) Let $A$ be the event that the card drawn from the first pack is an ace of heart,and $B$ be the event that the card drawn from the second pack is an ace of heart.
The probability of drawing an ace of heart from a pack of $52$ cards is $P(A) = P(B) = \frac{1}{52}$.
The probability of $NOT$ drawing an ace of heart from a pack is $P(A') = P(B') = 1 - \frac{1}{52} = \frac{51}{52}$.
The probability that at least one of them is an ace of heart is given by $1 - P(\text{none is an ace of heart})$.
Since the draws are independent,$P(\text{none is an ace of heart}) = P(A') \times P(B') = \frac{51}{52} \times \frac{51}{52} = \frac{2601}{2704}$.
Therefore,the required probability is $1 - \frac{2601}{2704} = \frac{2704 - 2601}{2704} = \frac{103}{2704}$.

(C) Total number of items in the box $= 6 + 10 = 16$.
Number of rusted nails $= \frac{1}{2} \times 6 = 3$.
Number of rusted nuts $= \frac{1}{2} \times 10 = 5$.
Total number of rusted items $= 3 + 5 = 8$.
Number of non-rusted nails $= 6 - 3 = 3$.
We need to find the probability that the chosen item is rusted or is a nail.
Using the formula $P(A \cup B) = P(A) + P(B) - P(A \cap B)$:
Let $A$ be the event that the item is rusted,and $B$ be the event that the item is a nail.
$P(A) = \frac{8}{16}$,$P(B) = \frac{6}{16}$,and $P(A \cap B) = \frac{3}{16}$ (rusted nails).
Probability $= \frac{8}{16} + \frac{6}{16} - \frac{3}{16} = \frac{11}{16}$.

(A) The total number of outcomes when a coin is tossed $4$ times is $2^4 = 16$.
Let $E$ be the event of getting at least one tail.
The complement event $E'$ is the event of getting no tails,which means getting all heads.
The only outcome for $E'$ is $(H, H, H, H)$,so the number of outcomes for $E'$ is $1$.
The probability of $E'$ is $P(E') = \frac{1}{16}$.
The probability of getting at least one tail is $P(E) = 1 - P(E') = 1 - \frac{1}{16} = \frac{15}{16}$.

(D) The total number of ways to place $3$ distinct letters into $3$ distinct envelopes is given by the permutation of $3$ items,which is $3! = 3 \times 2 \times 1 = 6$.
Out of these $6$ possible arrangements,only $1$ arrangement corresponds to the case where every letter is placed in its correct envelope.
Therefore,the required probability is the ratio of the number of favorable outcomes to the total number of possible outcomes:
Probability $= \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{1}{3!} = \frac{1}{6}$.

(B) When two dice are thrown,the total number of possible outcomes is $6 \times 6 = 36$.
We are looking for the probability that the sum of the numbers on the two dice is more than $10$.
The possible sums greater than $10$ are $11$ and $12$.
The outcomes that result in a sum of $11$ are $(5, 6)$ and $(6, 5)$.
The outcome that results in a sum of $12$ is $(6, 6)$.
Thus,the favorable outcomes are ${(5, 6), (6, 5), (6, 6)}$.
The number of favorable outcomes is $3$.
The probability is $\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{3}{36} = \frac{1}{12}$.

(B) When $2$ dice are thrown,the total number of possible outcomes is $6 \times 6 = 36$.
We need to find the probability of getting a sum of $5$ or $6$.
Favorable outcomes for a sum of $5$ are: $(1, 4), (2, 3), (3, 2), (4, 1)$,which are $4$ outcomes.
Favorable outcomes for a sum of $6$ are: $(1, 5), (2, 4), (3, 3), (4, 2), (5, 1)$,which are $5$ outcomes.
Total favorable outcomes = $4 + 5 = 9$.
The required probability = $\frac{\text{Total favorable outcomes}}{\text{Total possible outcomes}} = \frac{9}{36} = \frac{1}{4}$.

(B) sure event is an event that is certain to occur.
By definition,the probability of the entire sample space $S$ is $P(S) = 1$.
Therefore,the probability of a sure event is $1$.

(B) Given,the probability of event $A$ occurring in one trial is $P(A) = 0.4$.
Therefore,the probability of event $A$ not occurring in one trial is $P(\bar{A}) = 1 - 0.4 = 0.6$.
For three independent trials,the probability that event $A$ does not occur at all is $P(\text{none}) = P(\bar{A})^3 = (0.6)^3 = 0.216$.
The probability that event $A$ happens at least once is given by $1 - P(\text{none})$.
Thus,the required probability is $1 - 0.216 = 0.784$.

(B) When two dice are thrown,the total number of possible outcomes is $6 \times 6 = 36$.
An odd sum is obtained if one die shows an odd number and the other shows an even number,or vice versa.
The possible outcomes for an odd sum are: $(1,2), (1,4), (1,6), (2,1), (2,3), (2,5), (3,2), (3,4), (3,6), (4,1), (4,3), (4,5), (5,2), (5,4), (5,6), (6,1), (6,3), (6,5)$.
Counting these,we find there are $18$ such outcomes.
The probability $P$ of obtaining an odd sum is $\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{18}{36} = \frac{1}{2}$.

(C) The total number of lottery tickets is $10,000$.
To find the number of tickets divisible by $20$,we divide the total number of tickets by $20$: $\frac{10,000}{20} = 500$.
These $500$ tickets represent the favorable outcomes.
The probability of an event is given by the ratio of favorable outcomes to the total number of possible outcomes.
Therefore,the required probability $= \frac{500}{10,000} = \frac{1}{20}$.

(B) When two dice are thrown,the total number of outcomes is $6 \times 6 = 36$.
Let $A$ be the event of getting a multiple of $2$ on the first die: $A = \{2, 4, 6\}$,so $n(A) = 3$.
Let $B$ be the event of getting a multiple of $3$ on the second die: $B = \{3, 6\}$,so $n(B) = 2$.
The number of outcomes where the first die is a multiple of $2$ and the second is a multiple of $3$ is $3 \times 2 = 6$. These are $(2,3), (2,6), (4,3), (4,6), (6,3), (6,6)$.
Similarly,the number of outcomes where the first die is a multiple of $3$ and the second is a multiple of $2$ is $2 \times 3 = 6$. These are $(3,2), (3,4), (3,6), (6,2), (6,4), (6,6)$.
The intersection of these two sets is the outcome $(6,6)$,which is counted in both cases.
Using the Principle of Inclusion-Exclusion,the total number of favourable outcomes is $6 + 6 - 1 = 11$.
Therefore,the required probability is $\frac{11}{36}$.

(D) Let $A, B,$ and $C$ be the events that the three students solve the problem respectively.
Given probabilities are $P(A) = \frac{1}{3}$,$P(B) = \frac{1}{4}$,and $P(C) = \frac{1}{5}$.
The probability that the students do not solve the problem are $P(A') = 1 - \frac{1}{3} = \frac{2}{3}$,$P(B') = 1 - \frac{1}{4} = \frac{3}{4}$,and $P(C') = 1 - \frac{1}{5} = \frac{4}{5}$.
The probability that the problem is not solved by any of them is $P(\text{none}) = P(A') \times P(B') \times P(C') = \frac{2}{3} \times \frac{3}{4} \times \frac{4}{5} = \frac{2}{5}$.
Therefore,the probability that the problem is solved is $P(\text{solved}) = 1 - P(\text{none}) = 1 - \frac{2}{5} = \frac{3}{5}$.

(C) standard six-sided die has $6$ faces,numbered $1, 2, 3, 4, 5,$ and $6$.
When throwing the die,the total number of possible outcomes is $6$.
The event of getting the number $5$ is a single favorable outcome.
Therefore,the probability $P$ of getting the number $5$ is given by the formula: $P = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}}$.
Substituting the values,we get $P = \frac{1}{6}$.

(B) standard pack of cards contains $52$ cards.
There is only $1$ queen of clubs in the deck,so the probability of drawing it is $P(A) = \frac{1}{52}$.
There is only $1$ king of hearts in the deck,so the probability of drawing it is $P(B) = \frac{1}{52}$.
Since these two events are mutually exclusive (a card cannot be both a queen of clubs and a king of hearts simultaneously),the probability of either event occurring is the sum of their individual probabilities.
Required probability $P(A \cup B) = P(A) + P(B) = \frac{1}{52} + \frac{1}{52} = \frac{2}{52} = \frac{1}{26}$.

(C) When three coins are tossed simultaneously,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 probability of getting at least $2$ tails. This includes outcomes with $2$ tails or $3$ tails.
The favorable outcomes are: $\{HTT, THT, TTH, TTT\}$.
The number of favorable outcomes is $4$.
Therefore,the probability $P = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{4}{8} = \frac{1}{2}$.

(B) The sample space $S$ for a single throw of a fair die is ${1, 2, 3, 4, 5, 6}$.
The total number of possible outcomes is $n(S) = 6$.
Let $E$ be the event of getting a number less than $7$.
Since all numbers ${1, 2, 3, 4, 5, 6}$ are less than $7$,the favorable outcomes are ${1, 2, 3, 4, 5, 6}$.
The number of favorable outcomes is $n(E) = 6$.
The probability $P(E)$ is given by the formula $P(E) = \frac{n(E)}{n(S)}$.
$P(E) = \frac{6}{6} = 1$.
Therefore,the probability is $1$.

(C) When two dice are thrown,the total number of possible outcomes is $6 \times 6 = 36$.
Let $E$ be the event of getting a sum less than $11$. It is easier to find the probability of the complement event $E'$,which is getting a sum of $11$ or more.
The outcomes for a sum of $11$ or more are: $(5, 6), (6, 5), (6, 6)$.
Number of favourable outcomes for $E'$ is $3$.
Therefore,the number of favourable outcomes for $E$ (sum less than $11$) is $36 - 3 = 33$.
The required probability is $P(E) = \frac{33}{36} = \frac{11}{12}$.

(B) An ordinary year (non-leap year) consists of $365$ days.
Dividing $365$ by $7$,we get $365 = 52 \times 7 + 1$.
This means an ordinary year has $52$ full weeks and $1$ extra day.
The extra day can be any of the $7$ days of the week: {Sunday,Monday,Tuesday,Wednesday,Thursday,Friday,Saturday}.
For the year to have $53$ Sundays,the extra day must be a Sunday.
Since there is only $1$ favorable outcome (Sunday) out of $7$ possible outcomes,the probability is $1/7$.

(B) Total number of cards in a pack $= 52$.
Court cards (face cards) include jacks,queens,and kings.
There are $4$ jacks,$4$ queens,and $4$ kings in a deck.
Total number of court cards $= 4 + 4 + 4 = 12$.
The probability of drawing a court card is given by the ratio of the number of court cards to the total number of cards.
Probability $= \frac{12}{52}$.
Simplifying the fraction by dividing both numerator and denominator by $4$,we get $\frac{3}{13}$.
Thus,the correct option is $B$.

(C) When two dice are thrown,the total number of possible outcomes is $6 \times 6 = 36$.
We are looking for pairs $(x, y)$ such that $x + y$ is a multiple of $4$. The possible sums are $4, 8, 12$.
The favorable outcomes are:
Sum $= 4$: $(1, 3), (2, 2), (3, 1)$
Sum $= 8$: $(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)$
Sum $= 12$: $(6, 6)$
The total number of favorable outcomes is $3 + 5 + 1 = 9$.
The probability is $\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{9}{36} = \frac{1}{4}$.

(A) Total number of outcomes = (Number of prizes) + (Number of blanks) = $5 + 20 = 25$.
Number of favorable outcomes (getting a prize) = $5$.
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 probability of getting a prize = $\frac{5}{25} = \frac{1}{5}$.

(B) When a fair die is thrown,the total number of possible outcomes is $S = \{1, 2, 3, 4, 5, 6\}$,so the total number of outcomes $n(S) = 6$.
We want to find the probability of getting a number greater than $2$.
The favorable outcomes are $E = \{3, 4, 5, 6\}$.
The number of favorable outcomes is $n(E) = 4$.
The probability $P(E)$ is given by the formula $P(E) = \frac{n(E)}{n(S)}$.
Therefore,$P(E) = \frac{4}{6} = \frac{2}{3}$.

(D) The total number of ways to draw $2$ cards from a pack of $52$ cards is given by $^52C_2 = \frac{52 \times 51}{2 \times 1} = 1326$.
There are $4$ queens and $4$ aces in a pack of $52$ cards.
The number of ways to select $1$ queen and $1$ ace is $^4C_1 \times ^4C_1 = 4 \times 4 = 16$.
The probability is given by $\frac{\text{Favorable outcomes}}{\text{Total outcomes}} = \frac{16}{1326}$.
Simplifying the fraction by dividing both numerator and denominator by $2$,we get $\frac{8}{663}$.
Since $\frac{8}{663}$ is not among the options $A, B,$ or $C$,the correct answer is $D$ (None of these).

(C) When two dice are thrown,the total number of possible outcomes is $6 \times 6 = 36$.
The condition given is that the numbers appearing on the two dice are different. The cases where the numbers are the same are $(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)$,which are $6$ outcomes.
Therefore,the number of outcomes where the numbers are different is $36 - 6 = 30$.
Now,we need the probability that the sum is $6$,given that the numbers are different. The pairs that result in a sum of $6$ are $(1, 5), (2, 4), (3, 3), (4, 2), (5, 1)$.
Since the numbers must be different,we exclude $(3, 3)$. The remaining favorable outcomes are $(1, 5), (2, 4), (4, 2), (5, 1)$,which are $4$ outcomes.
The required probability is $\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes with different numbers}} = \frac{4}{30} = \frac{2}{15}$.

(A) Let $E_1$ be the event that the man is selected and $E_2$ be the event that the woman is selected.
Given,$P(E_1) = 1/4$ and $P(E_2) = 1/3$.
The probability that the man is not selected is $P(\bar{E}_1) = 1 - P(E_1) = 1 - 1/4 = 3/4$.
The probability that the woman is not selected is $P(\bar{E}_2) = 1 - P(E_2) = 1 - 1/3 = 2/3$.
Since the selection of the man and the woman are independent events,the probability that none of them will be selected is $P(\bar{E}_1 \cap \bar{E}_2) = P(\bar{E}_1) \times P(\bar{E}_2)$.
Therefore,$P(\bar{E}_1 \cap \bar{E}_2) = 3/4 \times 2/3 = 6/12 = 1/2$.