Mix Examples-Probability Questions in English

(B) For the quadratic expression $lx^2 + mx + 1 > 0$ to hold for all $x \in R$,the conditions are:
$1$. The coefficient of $x^2$ must be positive: $l > 0$. Since $l \in \{1, 2, 3, 4, 5, 6, 7\}$,this is always satisfied.
$2$. The discriminant $D < 0$: $m^2 - 4(l)(1) < 0$,which implies $m^2 < 4l$.
Given $l, m \in \{1, 2, 3, 4, 5, 6, 7\}$,the total number of possible pairs $(l, m)$ is $7 \times 7 = 49$.
We test values for $l$:
- If $l = 1$,$m^2 < 4 \implies m = 1$ ($1$ pair).
- If $l = 2$,$m^2 < 8 \implies m = 1, 2$ ($2$ pairs).
- If $l = 3$,$m^2 < 12 \implies m = 1, 2, 3$ ($3$ pairs).
- If $l = 4$,$m^2 < 16 \implies m = 1, 2, 3$ ($3$ pairs).
- If $l = 5$,$m^2 < 20 \implies m = 1, 2, 3, 4$ ($4$ pairs).
- If $l = 6$,$m^2 < 24 \implies m = 1, 2, 3, 4$ ($4$ pairs).
- If $l = 7$,$m^2 < 28 \implies m = 1, 2, 3, 4, 5$ ($5$ pairs).
Total favorable outcomes = $1 + 2 + 3 + 3 + 4 + 4 + 5 = 22$.
Probability = $\frac{22}{49}$.

(D) The roots of $x^2+x+1=0$ are $\omega$ and $\omega^2$,where $\omega$ is a complex cube root of unity.
We know that $1+\omega+\omega^2=0$.
For $\alpha^a+\alpha^b+\alpha^c=0$,the powers $\alpha^a, \alpha^b, \alpha^c$ must be a permutation of ${1, \omega, \omega^2}$.
This implies that $a, b, c$ must be of the form $3k_1+r_1, 3k_2+r_2, 3k_3+r_3$ where ${r_1, r_2, r_3} = {0, 1, 2}$ modulo $3$.
In a die,the numbers are ${1, 2, 3, 4, 5, 6}$.
Modulo $3$,these are ${1, 2, 0, 1, 2, 0}$.
There are two $1$s,two $2$s,and two $0$s.
To have ${r_1, r_2, r_3} = {0, 1, 2}$,we need to choose one number from each set of residues.
Number of ways to choose $a, b, c$ such that their residues are ${0, 1, 2}$ in any order is $3! \times (2 \times 2 \times 2) = 6 \times 8 = 48$.
Total outcomes $= 6^3 = 216$.
Probability $= \frac{48}{216} = \frac{2}{9}$.

(D) Let $W_A, W_B, W_C$ be the events of drawing a white ball from bags $A, B, C$ respectively,and $R_A, R_B, R_C$ be the events of drawing a red ball from bags $A, B, C$ respectively.
$P(W_A) = \frac{3}{7}, P(R_A) = \frac{4}{7}$
$P(W_B) = \frac{4}{9}, P(R_B) = \frac{5}{9}$
$P(W_C) = \frac{5}{11}, P(R_C) = \frac{6}{11}$
We need one white and two red balls. This can happen in three mutually exclusive cases:
Case $I$: White from $A$,Red from $B$,Red from $C$: $P_1 = \frac{3}{7} \times \frac{5}{9} \times \frac{6}{11} = \frac{90}{693}$
Case $II$: Red from $A$,White from $B$,Red from $C$: $P_2 = \frac{4}{7} \times \frac{4}{9} \times \frac{6}{11} = \frac{96}{693}$
Case $III$: Red from $A$,Red from $B$,White from $C$: $P_3 = \frac{4}{7} \times \frac{5}{9} \times \frac{5}{11} = \frac{100}{693}$
Total probability $= P_1 + P_2 + P_3 = \frac{90+96+100}{693} = \frac{286}{693}$.

(A) Total screws $= 50$.
Defective screws $= 5$.
Non-defective screws $= 45$.
Let $A$ be the event that all $3$ screws drawn are non-defective.
$(a)$ With replacement:
The probability of drawing a non-defective screw in one draw is $P = \frac{45}{50} = \frac{9}{10}$.
Since the drawing is with replacement,the events are independent.
$P(A) = \left(\frac{9}{10}\right) \times \left(\frac{9}{10}\right) \times \left(\frac{9}{10}\right) = \left(\frac{9}{10}\right)^3$.
$(b)$ Without replacement:
The probability is calculated as:
$P(A) = \frac{45}{50} \times \frac{44}{49} \times \frac{43}{48}$.
$P(A) = \frac{9}{10} \times \frac{44}{49} \times \frac{43}{48} = \frac{3}{10} \times \frac{11}{49} \times \frac{43}{4} = \frac{1419}{1960}$.

(B) By the axioms of probability,if $E_1 \subseteq E_2$,then $P(E_1) \leq P(E_2)$.
Statement (iv) states that if $E_1 \subseteq E_2$,then $P(E_2) \leq P(E_1)$,which is incorrect.
Therefore,statement (iv) is not satisfied by $P$.

(A) Let $O$ denote an odd number and $E$ denote an even number. In the set $\{1, 2, \dots, 100\}$,there are $50$ odd and $50$ even numbers. Thus,$P(O) = \frac{50}{100} = \frac{1}{2}$ and $P(E) = \frac{50}{100} = \frac{1}{2}$.
For the sum of three numbers to be odd,we must have an odd number of odd-numbered balls. The possible cases are:
$1$. Three odd balls: $P(O, O, O) = \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{8}$
$2$. One odd and two even balls: The odd ball can be in any of the $3$ positions $(OEE, EOE, EEO)$.
$P(O, E, E) = \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{8}$
$P(E, O, E) = \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{8}$
$P(E, E, O) = \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{8}$
Total probability $= \frac{1}{8} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8} = \frac{4}{8} = \frac{1}{2}$.

(A) Let $R$ be the number of red balls and $B$ be the number of black balls. Initially,$R = 19$ and $B = 19$.
In each step,two balls are removed:
$1$. If two red balls are removed,$R$ decreases by $2$ $(R \to R-2, B \to B)$.
$2$. If two black balls are removed,$B$ decreases by $2$ $(R \to R, B \to B-2)$.
$3$. If one red and one black ball are removed,the black ball is discarded and the red ball is returned $(R \to R, B \to B-1)$.
Notice that in all cases,the number of black balls $B$ decreases by either $0$ or $2$ or $1$. Specifically,the parity of the number of black balls changes if we remove one black ball,but the process continues until no more pairs can be formed.
Crucially,the number of red balls $R$ only changes parity if we remove two red balls. Since we start with $19$ red balls (odd) and $19$ black balls (odd),and the process must terminate,we observe that the number of black balls will eventually reach $0$. Since we cannot remove a red ball without another red ball,and we only discard red balls in pairs,the number of red balls will remain odd. Thus,the process must terminate with $1$ red ball remaining. Therefore,the probability is $1$.

(A) Each of the $12$ balls can be placed in any of the $3$ boxes in $3$ ways. Therefore,the total number of ways to distribute $12$ balls into $3$ boxes is $3^{12}$.
To find the number of favorable outcomes where the first box contains exactly $3$ balls:
$1$. Choose $3$ balls out of $12$ to be placed in the first box,which can be done in ${}^{12}C_3$ ways.
$2$. The remaining $9$ balls must be placed in the other $2$ boxes. Each of these $9$ balls has $2$ choices,so there are $2^9$ ways to distribute them.
Thus,the number of favorable ways is ${}^{12}C_3 \times 2^9$.
The required probability is $\frac{\text{Favorable outcomes}}{\text{Total outcomes}} = \frac{{}^{12}C_3 \times 2^9}{3^{12}}$.

(B) The word $REGULATIONS$ contains $11$ distinct letters.
Since the relative positions of the letters $G, U, L, A, T, I, O, N, S$ remain fixed,we only need to consider the positions of $R$ and $E$ among the $11$ available slots.
The total number of ways to place $R$ and $E$ in $11$ slots is $^{11}P_2 = 11 \times 10 = 110$.
We want exactly $4$ letters between $R$ and $E$. If $R$ is at position $i$ and $E$ is at position $j$,then $|i - j| = 5$.
The possible pairs $(i, j)$ are $(1, 6), (2, 7), (3, 8), (4, 9), (5, 10), (6, 11)$.
Since $R$ and $E$ can be interchanged,we have $6 \times 2 = 12$ favorable outcomes.
The probability is $\frac{12}{110} = \frac{6}{55}$.

(C) When a die is rolled three times,the total number of outcomes is $6 \times 6 \times 6 = 216$.
We are looking for the sum $S$ such that $S$ is a prime number of the form $4n+1$.
The possible sums range from $3$ to $18$.
The prime numbers in this range are $3, 5, 7, 11, 13, 17$.
Among these,the primes of the form $4n+1$ are $5, 13, 17$.
Number of ways to get sum $5$: $(1,1,3)$ in $3$ permutations,$(1,2,2)$ in $3$ permutations. Total = $6$ ways.
Number of ways to get sum $13$: $(1,6,6)$ in $3$ permutations,$(2,5,6)$ in $6$ permutations,$(3,4,6)$ in $6$ permutations,$(3,5,5)$ in $3$ permutations,$(4,4,5)$ in $3$ permutations. Total = $21$ ways.
Number of ways to get sum $17$: $(5,6,6)$ in $3$ permutations. Total = $3$ ways.
Total favorable outcomes = $6 + 21 + 3 = 30$.
Required probability = $\frac{30}{216} = \frac{5}{36}$.

(B) The total number of ways to distribute $5$ different books among $4$ students is $4^5 = 1024$.
To ensure each student gets at least one book,we must distribute the books such that one student gets $2$ books and the other three students get $1$ book each.
The number of ways to choose which student gets $2$ books is $\binom{4}{1} = 4$.
The number of ways to choose $2$ books out of $5$ is $\binom{5}{2} = 10$.
The number of ways to distribute the remaining $3$ books to the remaining $3$ students is $3! = 6$.
Total favorable ways = $4 \times 10 \times 6 = 240$.
The required probability is $\frac{240}{1024} = \frac{15}{64}$.
Thus,option $B$ is correct.

(B) Let $A$ be the event that the sum is $7$ and $B$ be the event that the sum is $11$.
The total number of outcomes when two dice are thrown is $36$.
The number of outcomes for sum $7$ is $n(A) = \{(1,6), (6,1), (2,5), (5,2), (3,4), (4,3)\} = 6$.
The number of outcomes for sum $11$ is $n(B) = \{(5,6), (6,5)\} = 2$.
The probabilities are $P(A) = \frac{6}{36} = \frac{1}{6}$ and $P(B) = \frac{2}{36} = \frac{1}{18}$.
We are interested in the event that $A$ occurs before $B$. This can happen in the first trial,or in the third trial (if the first two trials result in neither $A$ nor $B$),and so on.
Let $p = P(A) = \frac{1}{6}$ and $q = P(B) = \frac{1}{18}$. The probability of neither $A$ nor $B$ occurring is $r = 1 - (p + q) = 1 - (\frac{1}{6} + \frac{1}{18}) = 1 - \frac{4}{18} = 1 - \frac{2}{9} = \frac{7}{9}$.
The probability that $A$ occurs before $B$ is given by the sum of the infinite geometric series:
$P = p + rp + r^2p + \dots = \frac{p}{1-r} = \frac{1/6}{1 - 7/9} = \frac{1/6}{2/9} = \frac{1}{6} \times \frac{9}{2} = \frac{3}{4}$.

(A) The total number of outcomes when three dice are thrown is $6^3 = 216$.
Let $S$ be the sum of the numbers on the three dice. The possible values for $S$ range from $3$ to $18$.
Event $A$ is the sum $S > 14$,i.e.,$S \in \{15, 16, 17, 18\}$.
The number of ways to get these sums are:
$S=15: (6,6,3) \times 3, (6,5,4) \times 6, (5,5,5) \times 1 = 10$ ways.
$S=16: (6,6,4) \times 3, (6,5,5) \times 3 = 6$ ways.
$S=17: (6,6,5) \times 3 = 3$ ways.
$S=18: (6,6,6) \times 1 = 1$ way.
Total outcomes for $A = 10 + 6 + 3 + 1 = 20$.
Event $B$ is the sum $S$ being a multiple of $3$,i.e.,$S \in \{3, 6, 9, 12, 15, 18\}$.
Outcomes for $S=15$ are $10$ and for $S=18$ are $1$.
Thus,$A \cap B$ contains outcomes where $S=15$ or $S=18$,so $n(A \cap B) = 10 + 1 = 11$.
$P(A \cap \overline{B}) = P(A) - P(A \cap B) = \frac{20}{216} - \frac{11}{216} = \frac{9}{216}$.
Now,$P(\overline{A} \cap B) = P(B) - P(A \cap B)$.
Number of outcomes for $B$:
$S=3: 1$,$S=6: 10$,$S=9: 25$,$S=12: 25$,$S=15: 10$,$S=18: 1$.
Total $n(B) = 1 + 10 + 25 + 25 + 10 + 1 = 72$.
$P(\overline{A} \cap B) = \frac{72}{216} - \frac{11}{216} = \frac{61}{216}$.
Finally,$P(A \cap \overline{B}) + P(\overline{A} \cap B) = \frac{9}{216} + \frac{61}{216} = \frac{70}{216} = \frac{35}{108}$.

(C) Let $P(A) = x$, $P(B) = y$, $P(C) = z$, $P(A \cap B) = p$, $P(B \cap C) = q$, $P(C \cap A) = r$, and $P(A \cap B \cap C) = k = \frac{1}{16}$.
Given $P(\text{exactly one of } A \text{ or } B) = P(A) + P(B) - 2P(A \cap B) = x + y - 2p = \frac{1}{4}$.
Similarly, $y + z - 2q = \frac{1}{4}$ and $z + x - 2r = \frac{1}{4}$.
Adding these three equations: $2(x + y + z) - 2(p + q + r) = \frac{3}{4} \implies x + y + z - (p + q + r) = \frac{3}{8}$.
The probability that at least one event occurs is $P(A \cup B \cup C) = (x + y + z) - (p + q + r) + k$.
Substituting the values: $P(A \cup B \cup C) = \frac{3}{8} + \frac{1}{16} = \frac{6+1}{16} = \frac{7}{16}$.

(C) Let $n-1$ be the number of unfair coins (heads on both sides) and $2n - (n-1) = n+1$ be the number of fair coins.
Total coins = $2n$.
The probability of picking an unfair coin is $\frac{n-1}{2n}$ and a fair coin is $\frac{n+1}{2n}$.
The probability of getting a head is given by:
$P(H) = P(H|\text{unfair})P(\text{unfair}) + P(H|\text{fair})P(\text{fair})$
$\frac{41}{56} = (1) \times \frac{n-1}{2n} + (\frac{1}{2}) \times \frac{n+1}{2n}$
$\frac{41}{56} = \frac{2(n-1) + (n+1)}{4n}$
$\frac{41}{56} = \frac{3n-1}{4n}$
$41 \times 4n = 56 \times (3n-1)$
$164n = 168n - 56$
$4n = 56 \Rightarrow n = 14$.
The number of unfair coins is $n-1 = 14-1 = 13$.

(D) Total number of balls $= 2 + 3 + 5 = 10$.
Let $R_3$ be the event that the third ball drawn is red.
By the symmetry of drawing balls without replacement,the probability that the $k$-th ball drawn is of a certain color is equal to the initial proportion of that color in the bag.
Specifically,for any position $k$ (where $1 \le k \le 10$),the probability that the $k$-th ball is red is given by $P(R_k) = \frac{\text{Number of red balls}}{\text{Total number of balls}}$.
Thus,$P(R_3) = \frac{5}{10} = \frac{1}{2}$.
Alternatively,using the law of total probability:
$P(R_3) = P(R_3|R_1 R_2)P(R_1 R_2) + P(R_3|R_1 R_2^c)P(R_1 R_2^c) + P(R_3|R_1^c R_2)P(R_1^c R_2) + P(R_3|R_1^c R_2^c)P(R_1^c R_2^c) = \frac{1}{2}$.

(C) Let $E_A$ be the event of getting a sum of $6$ and $E_B$ be the event of getting a sum of $7$.
The probability of getting a sum of $6$ is $P(E_A) = \frac{5}{36}$.
The probability of not getting a sum of $6$ is $P(E_A^c) = 1 - \frac{5}{36} = \frac{31}{36}$.
The probability of getting a sum of $7$ is $P(E_B) = \frac{6}{36} = \frac{1}{6}$.
The probability of not getting a sum of $7$ is $P(E_B^c) = 1 - \frac{1}{6} = \frac{5}{6}$.
$A$ wins if he gets $6$ on his $1^{st}$ turn,or if $A$ fails,$B$ fails,and then $A$ gets $6$ on his $2^{nd}$ turn,and so on.
$P(A \text{ wins}) = P(E_A) + P(E_A^c)P(E_B^c)P(E_A) + P(E_A^c)P(E_B^c)P(E_A^c)P(E_B^c)P(E_A) + \dots$
This is an infinite geometric series with first term $a = \frac{5}{36}$ and common ratio $r = P(E_A^c)P(E_B^c) = \frac{31}{36} \times \frac{5}{6} = \frac{155}{216}$.
$P(A \text{ wins}) = \frac{a}{1-r} = \frac{\frac{5}{36}}{1 - \frac{155}{216}} = \frac{\frac{5}{36}}{\frac{61}{216}} = \frac{5}{36} \times \frac{216}{61} = \frac{30}{61}$.

(B) Given that,$P(A) = 0.6$ and $P(B) = 0.8$.
Therefore,$P(A') = 0.4$ and $P(B') = 0.2$.
$A$ wins if $A$ hits on the $1^{st}$ shot,or $A$ misses,$B$ misses,and $A$ hits on the $3^{rd}$ shot,or $A$ misses,$B$ misses,$A$ misses,$B$ misses,and $A$ hits on the $5^{th}$ shot,and so on.
Probability that $A$ wins $= P(A) + P(A')P(B')P(A) + P(A')P(B')P(A')P(B')P(A) + \dots$
$= 0.6 + (0.4)(0.2)(0.6) + (0.4)^2(0.2)^2(0.6) + \dots$
This is an infinite geometric series with first term $a = 0.6$ and common ratio $r = (0.4)(0.2) = 0.08$.
The sum of an infinite geometric series is $S = \frac{a}{1-r}$.
$P(A \text{ wins}) = \frac{0.6}{1 - 0.08} = \frac{0.6}{0.92} = \frac{60}{92} = \frac{15}{23}$.

(C) Let $A$ be the event that aeroplane-$I$ hits the target and $B$ be the event that aeroplane-$II$ hits the target.
Given: $P(A) = 0.3$ and $P(B) = 0.2$.
The probability that aeroplane-$I$ misses is $P(A') = 1 - P(A) = 1 - 0.3 = 0.7$.
The probability that aeroplane-$II$ misses is $P(B') = 1 - P(B) = 1 - 0.2 = 0.8$.
The second plane bombs only if the first misses. The target is hit by the $2^{nd}$ plane if:
$1.$ Plane-$I$ misses $AND$ Plane-$II$ hits: $P(A')P(B) = 0.7 \times 0.2 = 0.14$.
$2.$ Plane-$I$ misses,Plane-$II$ misses,Plane-$I$ misses,Plane-$II$ hits: $P(A')P(B')P(A')P(B) = 0.7 \times 0.8 \times 0.7 \times 0.2 = (0.56) \times 0.14$.
$3.$ This continues in an infinite geometric progression.
The total probability $P(X)$ that the target is hit by the $2^{nd}$ plane is:
$P(X) = 0.14 + 0.14(0.56) + 0.14(0.56)^2 + \dots$
This is an infinite $GP$ with first term $a = 0.14$ and common ratio $r = 0.56$.
Sum $= \frac{a}{1-r} = \frac{0.14}{1-0.56} = \frac{0.14}{0.44} = \frac{14}{44} = \frac{7}{22} \approx 0.31818...$
Note: Based on the provided options,the closest value is $0.32$.

(C) Given,$P(A) = 0.7$.
Probability that exactly one of them is selected is $P(A \cap \overline{B}) + P(\overline{A} \cap B) = 0.6$.
Since $A$ and $B$ are independent events,$P(A \cap B) = P(A) \cdot P(B)$.
The probability that exactly one is selected is given by $P(A) + P(B) - 2P(A \cap B) = 0.6$.
Substituting the values: $0.7 + P(B) - 2(0.7)P(B) = 0.6$.
$0.7 + P(B) - 1.4P(B) = 0.6$.
$-0.4P(B) = 0.6 - 0.7$.
$-0.4P(B) = -0.1$.
$P(B) = \frac{0.1}{0.4} = 0.25$.
Thus,the probability that $B$ is selected is $0.25$.

(B) Let $A$ and $B$ select numbers $x$ and $y$ respectively from the set $\{1, 2, 3, \ldots, n\}$. The total number of possible outcomes is $n \times n = n^2$.
The number of ways such that $x < y$ is given by the number of ways to choose $2$ distinct numbers from $n$,which is $\binom{n}{2} = \frac{n(n-1)}{2}$.
Given the probability $P(x < y) = \frac{n(n-1)}{2n^2} = \frac{n-1}{2n} = \frac{1009}{2019}$.
Solving for $n$: $2019(n-1) = 2018n \Rightarrow 2019n - 2019 = 2018n \Rightarrow n = 2019$.
Now,we need the probability that $y = x + 1$. The possible pairs $(x, y)$ such that $y = x + 1$ are $(1, 2), (2, 3), \ldots, (n-1, n)$. There are $n-1$ such pairs.
Thus,the required probability is $\frac{n-1}{n^2} = \frac{2019-1}{(2019)^2} = \frac{2018}{(2019)^2}$.

(D) Total number of balls = $7 + 6 + 8 = 21$.
Number of white balls = $7$.
Number of non-white balls = $6 + 8 = 14$.
We need to find the probability that at least one ball is white.
It is easier to calculate the probability of the complement event: $P(\text{at least one white}) = 1 - P(\text{no white balls})$.
If no white balls are drawn,both balls must be non-white.
Probability of drawing the first non-white ball = $\frac{14}{21}$.
After drawing one non-white ball,$13$ non-white balls remain out of $20$ total balls.
Probability of drawing the second non-white ball = $\frac{13}{20}$.
$P(\text{no white}) = \frac{14}{21} \times \frac{13}{20} = \frac{2}{3} \times \frac{13}{20} = \frac{26}{60} = \frac{13}{30}$.
Therefore,$P(\text{at least one white}) = 1 - \frac{13}{30} = \frac{17}{30}$.

(B) Let the four machines be $M_1, M_2, F_1, F_2$,where $F_1, F_2$ are faulty and $M_1, M_2$ are non-faulty.
We need to identify both faulty machines. The total number of ways to pick two machines out of four in a specific order is $4 \times 3 = 12$.
For only two tests to be needed,the first two machines tested must be the two faulty ones ($F_1, F_2$ or $F_2, F_1$).
The probability of picking a faulty machine in the first test is $\frac{2}{4}$.
Given the first was faulty,the probability of picking the second faulty machine in the second test is $\frac{1}{3}$.
Therefore,the probability that both faulty machines are identified in exactly two tests is $\frac{2}{4} \times \frac{1}{3} = \frac{2}{12} = \frac{1}{6}$.

(B) Let $P(A) = x$ and $P(B) = y$. Since $A$ and $B$ are independent,$P(A \cap B) = P(A)P(B) = xy = \frac{1}{6}$.
Also,$P(A^c \cap B^c) = P(A^c)P(B^c) = (1-x)(1-y) = \frac{1}{3}$.
Expanding the second equation: $1 - x - y + xy = \frac{1}{3}$.
Substituting $xy = \frac{1}{6}$: $1 - (x+y) + \frac{1}{6} = \frac{1}{3}$.
$x+y = 1 + \frac{1}{6} - \frac{1}{3} = 1 + \frac{1-2}{6} = 1 - \frac{1}{6} = \frac{5}{6}$.
We have $x+y = \frac{5}{6}$ and $xy = \frac{1}{6}$.
These are the roots of the quadratic equation $t^2 - (x+y)t + xy = 0$,which is $t^2 - \frac{5}{6}t + \frac{1}{6} = 0$.
Multiplying by $6$: $6t^2 - 5t + 1 = 0$.
$(2t-1)(3t-1) = 0$.
So,$t = \frac{1}{2}$ or $t = \frac{1}{3}$.
Since $P(A) > P(B)$,we have $P(A) = \frac{1}{2}$ and $P(B) = \frac{1}{3}$.
Thus,the probability of the occurrence of $B$ is $\frac{1}{3}$.

(A) $I$. $A, B$ are mutually exclusive events $\Rightarrow P(A \cap B) = 0$. Therefore,$P(A \cup B) = P(A) + P(B)$. This matches $(iv)$.
$II$. $A, B$ are independent events $\Rightarrow P(A \cap B) = P(A) \cdot P(B)$. Then $P(\frac{\bar{A}}{B}) = \frac{P(\bar{A} \cap B)}{P(B)} = \frac{P(\bar{A}) \cdot P(B)}{P(B)} = P(\bar{A}) = 1 - P(A)$. This matches $(iii)$.
$III$. $A \cap B = A \Rightarrow A \subset B \Rightarrow P(A) \leq P(B)$. This matches $(ii)$.
$IV$. $A \cup B = S \Rightarrow P(A \cup B) = 1$. Since $P(A \cup B) = P(A) + P(B) - P(A \cap B) = 1$,we have $P(A \cap B) = P(A) + P(B) - 1$. Since $P(A) = 1 - P(\bar{A})$,we get $P(A \cap B) = 1 - P(\bar{A}) + P(B) - 1 = P(B) - P(\bar{A})$. This matches $(i)$.
Thus,the correct matching is $(I)$-$(iv)$,$(II)$-$(iii)$,$(III)$-$(ii)$,$(IV)$-$(i)$.

(C) $P(A) = \frac{75}{100} = \frac{3}{4}$,so $P(\bar{A}) = 1 - \frac{3}{4} = \frac{1}{4}$.
$P(B) = \frac{80}{100} = \frac{4}{5}$,so $P(\bar{B}) = 1 - \frac{4}{5} = \frac{1}{5}$.
They contradict each other if one speaks the truth and the other lies.
This can happen in two ways: ($A$ speaks truth and $B$ lies) $OR$ ($A$ lies and $B$ speaks truth).
$P(\text{contradict}) = P(A \cap \bar{B}) + P(\bar{A} \cap B)$
$P(\text{contradict}) = (P(A) \times P(\bar{B})) + (P(\bar{A}) \times P(B))$
$P(\text{contradict}) = (\frac{3}{4} \times \frac{1}{5}) + (\frac{1}{4} \times \frac{4}{5})$
$P(\text{contradict}) = \frac{3}{20} + \frac{4}{20} = \frac{7}{20}$.

(C) Let $P(A) = x, P(B) = y, P(C) = z$. Since $A, B, C$ are independent,$A^c, B^c, C^c$ are also independent.
Given $P(A \cap B^{c} \cap C^{c}) = P(A)P(B^c)P(C^c) = x(1-y)(1-z) = \frac{1}{4} \dots (i)$
Given $P(A^{c} \cap B \cap C^{c}) = P(A^c)P(B)P(C^c) = (1-x)y(1-z) = \frac{1}{8} \dots (ii)$
Given $P(A^{c} \cap B^{c} \cap C^{c}) = P(A^c)P(B^c)P(C^c) = (1-x)(1-y)(1-z) = \frac{1}{4} \dots (iii)$
Dividing $(i)$ by $(iii)$: $\frac{x(1-y)(1-z)}{(1-x)(1-y)(1-z)} = \frac{1/4}{1/4} \Rightarrow \frac{x}{1-x} = 1 \Rightarrow x = 1-x \Rightarrow 2x = 1 \Rightarrow x = \frac{1}{2}$.
Dividing $(ii)$ by $(iii)$: $\frac{(1-x)y(1-z)}{(1-x)(1-y)(1-z)} = \frac{1/8}{1/4} \Rightarrow \frac{y}{1-y} = \frac{1}{2} \Rightarrow 2y = 1-y \Rightarrow 3y = 1 \Rightarrow y = \frac{1}{3}$.
Substituting $x = \frac{1}{2}$ and $y = \frac{1}{3}$ into $(i)$: $\frac{1}{2} \times (1 - \frac{1}{3}) \times (1-z) = \frac{1}{4} \Rightarrow \frac{1}{2} \times \frac{2}{3} \times (1-z) = \frac{1}{4} \Rightarrow \frac{1}{3}(1-z) = \frac{1}{4} \Rightarrow 1-z = \frac{3}{4} \Rightarrow z = 1 - \frac{3}{4} = \frac{1}{4}$.
Thus,$P(A) = \frac{1}{2}, P(B) = \frac{1}{3}, P(C) = \frac{1}{4}$.

(D) Let $A$ and $B$ denote the events that $P$ speaks the truth and $Q$ speaks the truth,respectively.
Given $P(A) = \frac{70}{100} = 0.7$ and $P(B) = \frac{80}{100} = 0.8$.
Hence,the probabilities of them speaking falsely are $P(\bar{A}) = 1 - 0.7 = 0.3$ and $P(\bar{B}) = 1 - 0.8 = 0.2$.
They agree in stating the same fact if both speak the truth or both speak falsely.
The required probability is $P(A \cap B) + P(\bar{A} \cap \bar{B}) = P(A)P(B) + P(\bar{A})P(\bar{B})$.
$= (0.7 \times 0.8) + (0.3 \times 0.2)$
$= 0.56 + 0.06 = 0.62$.
Converting to percentage,$0.62 \times 100 = 62\%$.
Therefore,they are likely to agree in $62\%$ of the cases.

(B) Let the two events be $A$ and $B$. Given,$P(A) = \frac{2}{5}$ and $P(B') = \frac{3}{10}$.
Since $A$ and $B$ are independent events,their complements $A'$ and $B$ are also independent.
We find the probability of occurrence of $B$ as $P(B) = 1 - P(B') = 1 - \frac{3}{10} = \frac{7}{10}$.
We find the probability of non-occurrence of $A$ as $P(A') = 1 - P(A) = 1 - \frac{2}{5} = \frac{3}{5}$.
The probability that only one of the two events occurs is given by $P(A \cap B') + P(A' \cap B)$.
Since the events are independent,this is $P(A)P(B') + P(A')P(B)$.
Substituting the values: $\frac{2}{5} \times \frac{3}{10} + \frac{3}{5} \times \frac{7}{10} = \frac{6}{50} + \frac{21}{50} = \frac{27}{50}$.

(B) Given that $A$ and $B$ are independent events,we have $P(A \cap B) = P(A)P(B) = \frac{1}{12}$.
Also,the probability that neither $A$ nor $B$ happens is $P(A^c \cap B^c) = P(A^c)P(B^c) = (1 - P(A))(1 - P(B)) = \frac{1}{2}$.
Expanding the second equation: $1 - P(A) - P(B) + P(A)P(B) = \frac{1}{2}$.
Substituting $P(A)P(B) = \frac{1}{12}$: $1 - (P(A) + P(B)) + \frac{1}{12} = \frac{1}{2}$.
$P(A) + P(B) = 1 + \frac{1}{12} - \frac{1}{2} = \frac{12 + 1 - 6}{12} = \frac{7}{12}$.
Let $x = P(A)$ and $y = P(B)$. We have $x + y = \frac{7}{12}$ and $xy = \frac{1}{12}$.
The quadratic equation with roots $x$ and $y$ is $t^2 - (x+y)t + xy = 0$,which is $t^2 - \frac{7}{12}t + \frac{1}{12} = 0$.
Multiplying by $12$: $12t^2 - 7t + 1 = 0$.
$12t^2 - 4t - 3t + 1 = 0 \implies 4t(3t - 1) - 1(3t - 1) = 0$.
$(4t - 1)(3t - 1) = 0$,so $t = \frac{1}{4}$ or $t = \frac{1}{3}$.
Since $P(B) > P(A)$,we have $P(A) = \frac{1}{4}$ and $P(B) = \frac{1}{3}$.

(D) Given that $A$ and $B$ are independent events,$P(A \cap B) = P(A)P(B) = \frac{1}{6}$.
Also,$P(\bar{A} \cap \bar{B}) = P(\bar{A})P(\bar{B}) = (1 - P(A))(1 - P(B)) = \frac{1}{3}$.
Expanding the second equation: $1 - P(A) - P(B) + P(A)P(B) = \frac{1}{3}$.
Substituting $P(A)P(B) = \frac{1}{6}$: $1 - (P(A) + P(B)) + \frac{1}{6} = \frac{1}{3}$.
$P(A) + P(B) = 1 + \frac{1}{6} - \frac{1}{3} = \frac{6 + 1 - 2}{6} = \frac{5}{6}$.
Let $x = P(A)$ and $y = P(B)$. We have $x + y = \frac{5}{6}$ and $xy = \frac{1}{6}$.
These are roots of the quadratic equation $t^2 - (x+y)t + xy = 0$,which is $t^2 - \frac{5}{6}t + \frac{1}{6} = 0$.
Multiplying by $6$: $6t^2 - 5t + 1 = 0$.
$(2t - 1)(3t - 1) = 0$.
So,$t = \frac{1}{2}$ or $t = \frac{1}{3}$.
Thus,$P(A)$ can be $\frac{1}{2}$ or $\frac{1}{3}$.
Since neither $\frac{1}{2}$ nor $\frac{1}{3}$ are uniquely listed as the only correct option,the answer is $D$.

(B) Let $E_1, E_2, E_3$ be the events of choosing urns $U_1, U_2, U_3$ respectively. Since the urn is chosen at random,$P(E_1) = P(E_2) = P(E_3) = \frac{1}{3}$.
Let $B$ be the event of drawing a black ball. We want to find the probability of not getting a black ball,which is $P(B^c) = 1 - P(B)$.
The probability of drawing a black ball from each urn is:
$P(B|E_1) = \frac{2}{5+3+2} = \frac{2}{10} = \frac{1}{5}$
$P(B|E_2) = \frac{2}{4+4+2} = \frac{2}{10} = \frac{1}{5}$
$P(B|E_3) = \frac{3}{3+4+3} = \frac{3}{10}$
Using the law of total probability,$P(B) = P(E_1)P(B|E_1) + P(E_2)P(B|E_2) + P(E_3)P(B|E_3)$
$P(B) = \frac{1}{3} \times \frac{1}{5} + \frac{1}{3} \times \frac{1}{5} + \frac{1}{3} \times \frac{3}{10} = \frac{1}{15} + \frac{1}{15} + \frac{1}{10} = \frac{2}{15} + \frac{1}{10} = \frac{4+3}{30} = \frac{7}{30}$.
Therefore,the probability of not getting a black ball is $P(B^c) = 1 - \frac{7}{30} = \frac{23}{30}$.

(C) Let $W_A, B_A$ be the events of drawing a white or black ball from $A$. $P(W_A) = \frac{4}{5}, P(B_A) = \frac{1}{5}$.
After transferring from $A$ to $B$,urn $B$ has $6$ balls.
Case $1$: If $W_A$ is transferred,$B$ has $4$ white,$2$ black. $P(W_{B|W_A}) = \frac{4}{6}, P(B_{B|W_A}) = \frac{2}{6}$.
Case $2$: If $B_A$ is transferred,$B$ has $3$ white,$3$ black. $P(W_{B|B_A}) = \frac{3}{6}, P(B_{B|B_A}) = \frac{3}{6}$.
Urn $C$ initially has $2$ white,$3$ black. After transfer from $B$,it has $6$ balls.
If $W_B$ is transferred,$C$ has $3$ white,$3$ black. $P(B_C|W_B) = \frac{3}{6}$.
If $B_B$ is transferred,$C$ has $2$ white,$4$ black. $P(B_C|B_B) = \frac{4}{6}$.
Total probability $P(B_C) = P(B_C|W_B)P(W_B) + P(B_C|B_B)P(B_B)$.
$P(W_B) = P(W_B|W_A)P(W_A) + P(W_B|B_A)P(B_A) = (\frac{4}{6} \times \frac{4}{5}) + (\frac{3}{6} \times \frac{1}{5}) = \frac{16+3}{30} = \frac{19}{30}$.
$P(B_B) = 1 - \frac{19}{30} = \frac{11}{30}$.
$P(B_C) = (\frac{3}{6} \times \frac{19}{30}) + (\frac{4}{6} \times \frac{11}{30}) = \frac{57 + 44}{180} = \frac{101}{180}$.

(D) Let $R_P$ be the event of drawing a red ball from bag $P$ and $B_P$ be the event of drawing a black ball from bag $P$. Similarly,let $R_Q$ and $B_Q$ be the events for bag $Q$.
Bag $P$ has $4$ red and $5$ black balls (Total $9$).
Bag $Q$ has $3$ red and $6$ black balls (Total $9$).
We draw $1$ ball from $P$ and $2$ balls from $Q$.
The total number of ways to draw $1$ ball from $P$ and $2$ balls from $Q$ is $\binom{9}{1} \times \binom{9}{2} = 9 \times 36 = 324$.
We want $2$ black and $1$ red ball. This can happen in two mutually exclusive cases:
Case $1$: $1$ red from $P$ and $2$ black from $Q$.
Probability $= P(R_P) \times P(2B_Q) = \frac{4}{9} \times \frac{\binom{6}{2}}{\binom{9}{2}} = \frac{4}{9} \times \frac{15}{36} = \frac{4}{9} \times \frac{5}{12} = \frac{20}{108} = \frac{5}{27}$.
Case $2$: $1$ black from $P$ and $1$ red,$1$ black from $Q$.
Probability $= P(B_P) \times P(1R_Q, 1B_Q) = \frac{5}{9} \times \frac{\binom{3}{1} \times \binom{6}{1}}{\binom{9}{2}} = \frac{5}{9} \times \frac{3 \times 6}{36} = \frac{5}{9} \times \frac{18}{36} = \frac{5}{9} \times \frac{1}{2} = \frac{5}{18}$.
Total Probability $= \frac{5}{27} + \frac{5}{18} = \frac{10 + 15}{54} = \frac{25}{54}$.

(C) Total number of ways to choose $2$ numbers from $100$ is $N = {}^{100}C_2 = \frac{100 \times 99}{2} = 4950$.
Event $A$: The product is even. This happens if at least one number is even. The complement is that both numbers are odd. Number of odd numbers is $50$.
$n(A) = {}^{100}C_2 - {}^{50}C_2 = 4950 - 1225 = 3725$.
Event $B$: The product is divisible by $4$. This happens if (both are even and at least one is a multiple of $4$) or (one is odd and one is a multiple of $4$).
Let $E$ be the set of even numbers ${2, 4, \dots, 100}$ ($50$ numbers) and $O$ be the set of odd numbers ${1, 3, \dots, 99}$ ($50$ numbers).
Let $M_4$ be the set of multiples of $4$ ${4, 8, \dots, 100}$ ($25$ numbers).
Let $E_2$ be the set of even numbers not divisible by $4$ ${2, 6, \dots, 98}$ ($25$ numbers).
$n(A \cap B)$ is the number of pairs whose product is divisible by $4$.
Pairs are: (one from $M_4$,any other) or (both from $E_2$).
$n(A \cap B) = {}^{25}C_1 \times {}^{75}C_1 + {}^{25}C_2 = 25 \times 75 + 300 = 1875 + 300 = 2175$.
$n(A \cap \bar{B}) = n(A) - n(A \cap B) = 3725 - 2175 = 1550$.
$P(A \cap \bar{B}) = \frac{1550}{4950} = \frac{155}{495} = \frac{31}{99}$.
Re-evaluating: The product is even but not divisible by $4$ if one number is from $E_2$ and the other is from $O$.
$n(A \cap \bar{B}) = {}^{25}C_1 \times {}^{50}C_1 = 25 \times 50 = 1250$.
$P(A \cap \bar{B}) = \frac{1250}{4950} = \frac{125}{495} = \frac{25}{99}$.

(C) Let $B_1$ be the first bag and $B_2$ be the second bag.
Bag $B_1$ contains $4$ red and $5$ black balls (Total = $9$).
Bag $B_2$ contains $3$ red and $6$ black balls (Total = $9$).
We draw $1$ ball from $B_1$ and $2$ balls from $B_2$. Total balls drawn = $3$.
We need $2$ black and $1$ red ball. There are two cases:
Case $I$: The ball from $B_1$ is red and the two balls from $B_2$ are black.
$P(Case I) = P(R_1) \times P(B_2, B_2) = \frac{4}{9} \times \frac{\binom{6}{2}}{\binom{9}{2}} = \frac{4}{9} \times \frac{15}{36} = \frac{4}{9} \times \frac{5}{12} = \frac{20}{108} = \frac{5}{27}$.
Case $II$: The ball from $B_1$ is black and the two balls from $B_2$ are one red and one black.
$P(Case II) = P(B_1) \times P(R_2, B_2) = \frac{5}{9} \times \frac{\binom{3}{1} \times \binom{6}{1}}{\binom{9}{2}} = \frac{5}{9} \times \frac{3 \times 6}{36} = \frac{5}{9} \times \frac{18}{36} = \frac{5}{9} \times \frac{1}{2} = \frac{5}{18}$.
Total Probability = $P(Case I) + P(Case II) = \frac{5}{27} + \frac{5}{18} = \frac{10 + 15}{54} = \frac{25}{54}$.

(B) Let $O_i$ and $E_i$ denote the events of drawing an odd and an even number from Box-$i$ respectively.
For Box-$I$ $(1, 2, 3)$: $P(O_1) = \frac{2}{3}$,$P(E_1) = \frac{1}{3}$.
For Box-$II$ $(1, 2, 3, 4, 5)$: $P(O_2) = \frac{3}{5}$,$P(E_2) = \frac{2}{5}$.
For Box-$III$ $(1, 2, 3, 4, 5, 6, 7)$: $P(O_3) = \frac{4}{7}$,$P(E_3) = \frac{3}{7}$.
The sum $x_1+x_2+x_3$ is odd if we have either three odd numbers or one odd and two even numbers.
Case $1$: All three are odd: $P(O_1 \cap O_2 \cap O_3) = \frac{2}{3} \times \frac{3}{5} \times \frac{4}{7} = \frac{24}{105}$.
Case $2$: One odd and two even:
- $O_1, E_2, E_3$: $\frac{2}{3} \times \frac{2}{5} \times \frac{3}{7} = \frac{12}{105}$.
- $E_1, O_2, E_3$: $\frac{1}{3} \times \frac{3}{5} \times \frac{3}{7} = \frac{9}{105}$.
- $E_1, E_2, O_3$: $\frac{1}{3} \times \frac{2}{5} \times \frac{4}{7} = \frac{8}{105}$.
Total Probability $= \frac{24+12+9+8}{105} = \frac{53}{105}$.

(A) Let $A$ be the event that $A$ speaks the truth and $B$ be the event that $B$ speaks the truth.
Given probabilities are:
$P(A) = 75 \% = \frac{75}{100} = \frac{3}{4}$
$P(B) = 80 \% = \frac{80}{100} = \frac{4}{5}$
Their statements do not match if one speaks the truth and the other lies.
This can happen in two ways: ($A$ speaks the truth $AND$ $B$ lies) $OR$ ($A$ lies $AND$ $B$ speaks the truth).
Required probability $= P(A) \times P(\bar{B}) + P(\bar{A}) \times P(B)$
$= P(A) \times (1 - P(B)) + (1 - P(A)) \times P(B)$
$= \frac{3}{4} \times (1 - \frac{4}{5}) + (1 - \frac{3}{4}) \times \frac{4}{5}$
$= \frac{3}{4} \times \frac{1}{5} + \frac{1}{4} \times \frac{4}{5}$
$= \frac{3}{20} + \frac{4}{20} = \frac{7}{20}$

(A) The probability of getting a $3$ in a single throw is $p = \frac{1}{6}$.
The probability of not getting a $3$ is $q = 1 - \frac{1}{6} = \frac{5}{6}$.
$A$ starts the game. $A$ wins if $A$ gets a $3$ on the $1^{st}, 3^{rd}, 5^{th}, \dots$ throw.
$P(A) = p + q^2p + q^4p + \dots = \frac{p}{1 - q^2}$.
Substituting the values: $P(A) = \frac{1/6}{1 - (5/6)^2} = \frac{1/6}{1 - 25/36} = \frac{1/6}{11/36} = \frac{6}{11}$.
Since the total probability is $1$,$P(B) = 1 - P(A) = 1 - \frac{6}{11} = \frac{5}{11}$.
Thus,the probabilities are $\frac{6}{11}$ and $\frac{5}{11}$.

(C) The outcome of each coin toss is an independent event.
For a fair coin,the probability of getting a head in any single toss is always $P(H) = \frac{1}{2}$.
Since the tosses are independent,the result of the $1947^{\text{th}}$ toss does not depend on the results of any other tosses.
Therefore,the probability of getting a head on the $1947^{\text{th}}$ toss remains $\frac{1}{2}$.

(C) Let $W$ denote winning (getting a number $> 4$,probability $p = \frac{2}{6} = \frac{1}{3}$) and $L$ denote losing (getting a number $\le 4$,probability $q = \frac{4}{6} = \frac{2}{3}$). The player quits when $W$ occurs or after $3$ throws. The possible outcomes are:
$(i)$ $W$: Win $5$ rupees. Probability $P_1 = \frac{1}{3}$.
(ii) $LW$: Lose $1$ rupee,then win $5$ rupees. Net gain $= 4$ rupees. Probability $P_2 = \frac{2}{3} \times \frac{1}{3} = \frac{2}{9}$.
(iii) $LLW$: Lose $1$ rupee,lose $1$ rupee,then win $5$ rupees. Net gain $= 3$ rupees. Probability $P_3 = \frac{2}{3} \times \frac{2}{3} \times \frac{1}{3} = \frac{4}{27}$.
(iv) $LLL$: Lose $1$ rupee,lose $1$ rupee,lose $1$ rupee. Net gain $= -3$ rupees. Probability $P_4 = \frac{2}{3} \times \frac{2}{3} \times \frac{2}{3} = \frac{8}{27}$.
The expected value $E = \sum x_i P_i = 5(\frac{1}{3}) + 4(\frac{2}{9}) + 3(\frac{4}{27}) + (-3)(\frac{8}{27})$.
$E = \frac{5}{3} + \frac{8}{9} + \frac{12}{27} - \frac{24}{27} = \frac{45 + 24 + 12 - 24}{27} = \frac{57}{27} = \frac{19}{9}$.

(D) The total number of functions from set $A$ to set $B$ is given by $|B|^{|A|}$.
Here,$|A| = 3$ and $|B| = 5$.
Total number of functions = $5^3 = 125$.
$A$ function is one-one if each element in $A$ is mapped to a distinct element in $B$.
The number of one-one functions is given by the number of permutations of $5$ elements taken $3$ at a time,which is $P(5, 3) = 5 \times 4 \times 3 = 60$.
The probability that a randomly selected function is one-one is the ratio of the number of one-one functions to the total number of functions.
Probability = $\frac{60}{125} = \frac{12}{25}$.

(B) When three dice are thrown,the total number of outcomes is $6^3 = 216$.
The sum of the numbers on the three dice can range from $3$ to $18$.
The prime numbers in this range are $3, 5, 7, 11, 13, 17$.
We count the number of ways to get each sum:
Sum $= 3$: $(1,1,1) \rightarrow 1$ way.
Sum $= 5$: $(1,1,3) \times 3, (1,2,2) \times 3 \rightarrow 6$ ways.
Sum $= 7$: $(1,1,5) \times 3, (1,2,4) \times 6, (1,3,3) \times 3, (2,2,3) \times 3 \rightarrow 15$ ways.
Sum $= 11$: $(1,4,6) \times 6, (1,5,5) \times 3, (2,3,6) \times 6, (2,4,5) \times 6, (3,3,5) \times 3, (3,4,4) \times 3 \rightarrow 27$ ways.
Sum $= 13$: $(1,6,6) \times 3, (2,5,6) \times 6, (3,4,6) \times 6, (3,5,5) \times 3, (4,4,5) \times 3 \rightarrow 21$ ways.
Sum $= 17$: $(5,6,6) \times 3 \rightarrow 3$ ways.
Total favorable outcomes $= 1 + 6 + 15 + 27 + 21 + 3 = 73$.
The probability is $\frac{73}{216}$.

(C) chessboard has $8 \times 8 = 64$ squares. The total number of ways to choose two squares is $\binom{64}{2} = \frac{64 \times 63}{2} = 2016$.
Two squares have a side in common if they are adjacent horizontally or vertically.
Number of horizontal adjacent pairs: Each row has $7$ pairs,and there are $8$ rows,so $8 \times 7 = 56$.
Number of vertical adjacent pairs: Each column has $7$ pairs,and there are $8$ columns,so $8 \times 7 = 56$.
Total adjacent pairs = $56 + 56 = 112$.
The number of pairs that do not have a side in common is $2016 - 112 = 1904$.
The probability is $\frac{1904}{2016}$.
Dividing both by $112$,we get $\frac{1904 \div 112}{2016 \div 112} = \frac{17}{18}$.

(B) Total number of ways to select three numbers from $50$ is $^{50}C_3 = \frac{50 \times 49 \times 48}{3 \times 2 \times 1} = 19,600$.
For three numbers $a, b, c$ to be in arithmetic progression,$2b = a + c$.
This implies $a + c$ must be even,which occurs if both $a$ and $c$ are odd or both are even.
Number of odd numbers in the set is $25$ and number of even numbers is $25$.
Number of ways to choose two odd numbers is $^{25}C_2 = \frac{25 \times 24}{2} = 300$.
Number of ways to choose two even numbers is $^{25}C_2 = \frac{25 \times 24}{2} = 300$.
Total favorable cases = $300 + 300 = 600$.
Probability = $\frac{600}{19600} = \frac{6}{196} = \frac{3}{98}$.

(B) Total number of ways to draw $2$ cards from $52$ is $C(52, 2) = \frac{52 \times 51}{2} = 1326$.
There are $12$ face cards in a deck ($J, Q, K$ of each suit).
There are $13$ spade cards in total.
Face cards that are spades are $3$ ($J, Q, K$ of spades).
Spade cards that are $NOT$ face cards are $13 - 3 = 10$.
We need to select $1$ face card and $1$ spade card that is not a face card.
Case $1$: Select $1$ face card from the $3$ spade face cards and $1$ spade card from the $10$ non-face spade cards. Number of ways = $C(3, 1) \times C(10, 1) = 3 \times 10 = 30$.
Case $2$: Select $1$ face card from the $9$ non-spade face cards and $1$ spade card from the $10$ non-face spade cards. Number of ways = $C(9, 1) \times C(10, 1) = 9 \times 10 = 90$.
Total favorable outcomes = $30 + 90 = 120$.
Probability = $\frac{120}{1326} = \frac{20}{221}$.

(C) Given that $A, B, C$ are pairwise independent events,we have $P(A \cap B) = P(A)P(B)$,$P(B \cap C) = P(B)P(C)$,and $P(C \cap A) = P(C)P(A)$.
We are given $P(\bar{B} \cup \bar{C}) = \frac{1}{2}$.
By De Morgan's Law,$P(\bar{B} \cup \bar{C}) = 1 - P(B \cap C) = \frac{1}{2}$,which implies $P(B \cap C) = \frac{1}{2}$.
Since $B$ and $C$ are independent,$P(B)P(C) = bc = \frac{1}{2}$.
We need to find $P((\bar{B} \cap \bar{C}) \mid A) = \frac{P((\bar{B} \cap \bar{C}) \cap A)}{P(A)}$.
Using the property of sets,$(\bar{B} \cap \bar{C}) \cap A = A \setminus (A \cap (B \cup C)) = A \setminus ((A \cap B) \cup (A \cap C))$.
Thus,$P((\bar{B} \cap \bar{C}) \cap A) = P(A) - P((A \cap B) \cup (A \cap C)) = P(A) - [P(A \cap B) + P(A \cap C) - P(A \cap B \cap C)]$.
Since the events are pairwise independent,$P(A \cap B \cap C)$ is not necessarily $P(A)P(B)P(C)$ unless they are mutually independent. However,the question implies pairwise independence. Assuming mutual independence for the intersection term:
$P((\bar{B} \cap \bar{C}) \cap A) = P(A) - P(A)P(B) - P(A)P(C) + P(A)P(B)P(C) = P(A)(1 - b - c + bc)$.
Substituting $bc = \frac{1}{2}$,we get $P((\bar{B} \cap \bar{C}) \cap A) = P(A)(1 - b - c + \frac{1}{2}) = P(A)(\frac{3}{2} - b - c)$.
Therefore,$P((\bar{B} \cap \bar{C}) \mid A) = \frac{3}{2} - b - c$.

(B) Let $p$ be the probability of getting a head and a $6$ in a single throw. The probability of getting a head is $\frac{1}{2}$ and the probability of getting a $6$ is $\frac{1}{6}$. Since these are independent events,$p = \frac{1}{2} \times \frac{1}{6} = \frac{1}{12}$. The probability of not winning in a single throw is $q = 1 - p = \frac{11}{12}$.
$A$ starts the game. $B$ wins if $A$ fails,then $B$ succeeds,or $A$ fails,$B$ fails,$A$ fails,$B$ succeeds,and so on.
The probability that $B$ wins is:
$P(B) = qp + q^3p + q^5p + \dots$
This is an infinite geometric series with first term $a = qp = \frac{11}{12} \times \frac{1}{12} = \frac{11}{144}$ and common ratio $r = q^2 = (\frac{11}{12})^2 = \frac{121}{144}$.
$P(B) = \frac{a}{1-r} = \frac{\frac{11}{144}}{1 - \frac{121}{144}} = \frac{\frac{11}{144}}{\frac{23}{144}} = \frac{11}{23}$.