Question

Identify incorrect option –

• A. $P\left(\frac{Y}{X_5}\right)=\frac{4}{9}$
• B. $P\left(\frac{Z}{X_4}\right)=\frac{13}{81}$
• C. $P\left(X_{2k-1}\right)=P\left(X_{2k}\right) \forall k \in\{1,2,3\}$
• D. $P\left(\frac{Z}{X_1}\right)=\frac{64}{729}$

Answer 1

Answer: (D)

$P\left(\frac{Z}{X_1}\right)=\frac{64}{729}$

Answer 2

The match ends when one player is ahead by 2 points or when 6 sets are played. A wins a set with probability $p = 2/3$ and B wins with probability $q = 1/3$. Let $S_n$ be the score difference after $n$ sets (A's points - B's points). The match ends if $|S_n| \ge 2$ or $n=6$.

Let's trace the possible sequences of set wins (A or B) and the state of the match. $S_0 = 0$. Set 1: A wins (A, $S_1=1$, prob $p$) or B wins (B, $S_1=-1$, prob $q$). Match continues. $P(X_1)=1, P(X_2)=1$. Set 2: AA: $S_2=2$. A wins. Prob $p^2$. Match ends. AB: $S_2=0$. Prob $pq$. Match continues. BA: $S_2=0$. Prob $qp$. Match continues. BB: $S_2=-2$. B wins. Prob $q^2$. Match ends. Match ends in 2 sets with probability $p^2+q^2 = (2/3)^2+(1/3)^2 = 4/9+1/9=5/9$. Match continues after 2 sets if $S_2=0$. Prob $2pq = 2(2/3)(1/3) = 4/9$. $P(X_3) = P(\text{match continues after 2 sets}) = 2pq = 4/9$. $P(X_4) = P(\text{match continues after 2 sets}) = 2pq = 4/9$. (Match cannot end in 3 sets as $S_3$ would be $\pm 1$ if $S_2=0$).

Set 3 (if $S_2=0$): ABA: $S_3=1$. Prob $pqp=p^2q$. Continues. ABB: $S_3=-1$. Prob $pqq=pq^2$. Continues. BAA: $S_3=1$. Prob $qpp=qp^2$. Continues. BAB: $S_3=-1$. Prob $qpq=q^2p$. Continues. $P(S_3=1 | S_2=0) = \frac{p^2q+qp^2}{2pq} = \frac{2p^2q}{2pq} = p$. $P(S_3=-1 | S_2=0) = \frac{pq^2+q^2p}{2pq} = \frac{2pq^2}{2pq} = q$.

Set 4 (if $S_3=\pm 1$ and $S_2=0$): If $S_3=1$ (score 2-1 to A): A wins (score 3-1, $S_4=2$, A wins, ends, prob $p$) or B wins (score 2-2, $S_4=0$, continues, prob $q$). If $S_3=-1$ (score 1-2 to B): A wins (score 2-2, $S_4=0$, continues, prob $p$) or B wins (score 1-3, $S_4=-2$, B wins, ends, prob $q$). Match ends in 4 sets if $S_4=\pm 2$. $P(\text{A wins in 4 sets}) = P(S_2=0) \times P(S_3=1 | S_2=0) \times P(\text{A wins set 4} | S_3=1) = 2pq \times p \times p = 2p^3q$. $P(\text{B wins in 4 sets}) = P(S_2=0) \times P(S_3=-1 | S_2=0) \times P(\text{B wins set 4} | S_3=-1) = 2pq \times q \times q = 2pq^3$. Match continues after 4 sets if $S_4=0$. $P(S_4=0) = P(S_2=0) \times [P(S_3=1|S_2=0)P(\text{B wins set 4}|S_3=1) + P(S_3=-1|S_2=0)P(\text{A wins set 4}|S_3=-1)] = 2pq \times [p \times q + q \times p] = 2pq(2pq) = 4p^2q^2$. $P(X_5) = P(S_4=0) = 4p^2q^2 = 4(2/3)^2(1/3)^2 = 4(4/9)(1/9) = 16/81$. $P(X_6) = P(S_4=0) = 16/81$. (Match cannot end in 5 sets as $S_5$ would be $\pm 1$ if $S_4=0$).

Let's check the third option: $P(X_{2k-1})=P(X_{2k}) \forall k \in\{1,2,3\}$. $k=1$: $P(X_1)=P(X_2)$. $1=1$. Correct. $k=2$: $P(X_3)=P(X_4)$. $4/9=4/9$. Correct. $k=3$: $P(X_5)=P(X_6)$. $16/81=16/81$. Correct. Option 3 is correct.

Now let's evaluate the conditional probabilities. $P(Y|X_5) = P(\text{A wins} | \text{at least 5 sets played})$. At least 5 sets are played if $S_4=0$. Given $S_4=0$, the match continues to set 5. If $S_4=0$, score is 2-2. Set 5: A wins (score 3-2, $S_5=1$, prob $p$) or B wins (score 2-3, $S_5=-1$, prob $q$). Match continues. Set 6: If $S_5=1$ (score 3-2): A wins (score 4-2, $S_6=2$, A wins, ends, prob $p$) or B wins (score 3-3, $S_6=0$, ends, no winner, prob $q$). If $S_5=-1$ (score 2-3): A wins (score 3-3, $S_6=0$, ends, no winner, prob $p$) or B wins (score 2-4, $S_6=-2$, B wins, ends, prob $q$). Given $X_5$ (i.e., $S_4=0$), A wins the match if A wins set 5 AND A wins set 6 (sequence A, A from $S_4=0$), or if A wins in earlier sets. But the conditioning is on $X_5$, meaning the match reached set 5. So we only consider outcomes from $S_4=0$. Given $S_4=0$, A wins if the sequence of results for sets 5 and 6 is AA. $P(\text{A wins} | S_4=0) = P(\text{A wins set 5}) P(\text{A wins set 6} | S_5=1) = p \times p = p^2 = (2/3)^2 = 4/9$. This is $P(Y|X_5)$. Option 1: $P(Y|X_5) = 4/9$. Correct.

$P(Z|X_4) = P(\text{B wins} | \text{at least 4 sets played})$. At least 4 sets are played if $S_2=0$. Given $S_2=0$, the match continues to set 3. If $S_2=0$, score is 1-1. Set 3: A wins ($S_3=1$, prob $p$) or B wins ($S_3=-1$, prob $q$). Set 4: If $S_3=1$: A wins ($S_4=2$, A wins, ends, prob $p$) or B wins ($S_4=0$, continues, prob $q$). If $S_3=-1$: A wins ($S_4=0$, continues, prob $p$) or B wins ($S_4=-2$, B wins, ends, prob $q$). Set 5 (if $S_4=0$): A wins ($S_5=1$, prob $p$) or B wins ($S_5=-1$, prob $q$). Continues. Set 6 (if $S_5=\pm 1$ and $S_4=0$): If $S_5=1$: A wins ($S_6=2$, A wins, ends, prob $p$) or B wins ($S_6=0$, ends, no winner, prob $q$). If $S_5=-1$: A wins ($S_6=0$, ends, no winner, prob $p$) or B wins ($S_6=-2$, B wins, ends, prob $q$).

Given $X_4$ (i.e., $S_2=0$), B wins the match if:

1. B wins in 4 sets: B wins set 3 AND B wins set 4 (sequence BB from $S_2=0$). Prob $q \times q = q^2$.
2. B wins in 6 sets: Match reaches set 5 ($S_4=0$, prob $2pq$) AND B wins set 5 (prob $q$) AND B wins set 6 (prob $q$). $P(S_4=0 | S_2=0) = P(S_3=1, \text{B wins set 4}) + P(S_3=-1, \text{A wins set 4}) = p q + q p = 2pq$. $P(\text{B wins in 6 sets} | S_2=0) = P(S_4=0 | S_2=0) \times P(S_5=-1 | S_4=0) \times P(S_6=-2 | S_5=-1) = 2pq \times q \times q = 2pq^3$. $P(Z|X_4) = P(\text{B wins in 4 sets} | S_2=0) + P(\text{B wins in 6 sets} | S_2=0) = q^2 + 2pq^3$. $q=1/3, p=2/3$. $P(Z|X_4) = (1/3)^2 + 2(2/3)(1/3)^3 = 1/9 + 4/81 = 9/81 + 4/81 = 13/81$. Option 2: $P(Z|X_4) = 13/81$. Correct.

$P(Z|X_1) = P(\text{B wins} | \text{at least 1 set played})$. $X_1$ is always true, so $P(Z|X_1) = P(Z)$. B can win in 2, 4, or 6 sets. B wins in 2 sets (BB): $q^2 = (1/3)^2 = 1/9$. B wins in 4 sets (from $S_2=0$, then BB): $P(S_2=0) \times q^2 = 2pq \times q^2 = 2pq^3 = 2(2/3)(1/3)^3 = 4/81$. B wins in 6 sets (from $S_4=0$, then BB): $P(S_4=0) \times q^2 = 4p^2q^2 \times q^2 = 4p^2q^4 = 4(2/3)^2(1/3)^4 = 4(4/9)(1/81) = 16/729$. $P(Z) = P(\text{B wins in 2 sets}) + P(\text{B wins in 4 sets}) + P(\text{B wins in 6 sets})$. $P(Z) = q^2 + 2pq^3 + 4p^2q^4 = (1/3)^2 + 2(2/3)(1/3)^3 + 4(2/3)^2(1/3)^4 = 1/9 + 4/81 + 16/729 = 81/729 + 36/729 + 16/729 = 133/729$. Option 4: $P(Z|X_1) = 64/729$. Incorrect.

The incorrect option is $P(Z|X_1)=\frac{64}{729}$.

Final check of the options: Option 1: $P(Y|X_5) = 4/9$. Correct. Option 2: $P(Z|X_4) = 13/81$. Correct. Option 3: $P(X_{2k-1})=P(X_{2k}) \forall k \in\{1,2,3\}$. Correct. Option 4: $P(Z|X_1) = 133/729$. The given value is $64/729$. Incorrect.