Question

$A$ and $B$ are playing a badminton match with the agreement that winner of each set will get 1 point and the loser 0 point. The match ends as soon as one of them is ahead by 2 points or number of sets reaches six. It is supposed that the probabilities of $A$ and $B$ winning a set are $\frac{2}{3}$ and $\frac{1}{3}$ respectively and each set is independent.

Let $X_i$ denotes the event that atleast $i$ sets are played and $Y$ and $Z$ denotes the event that match has won by $A$ and $B$ respectively.

Question:

Identify incorrect option -

Answer 1

D

Answer 2

The problem describes a badminton match where the winner of each set gets 1 point. The match ends when one player is ahead by 2 points or when 6 sets are completed. Let $p = P(\text{A wins a set}) = \frac{2}{3}$ and $q = P(\text{B wins a set}) = \frac{1}{3}$. Let $d_n$ be the score difference (A's score - B's score) after $n$ sets. Initially $d_0 = 0$. The match ends if $|d_n| = 2$ or $n=6$.

Let's analyze the possible number of sets played.

• 2 sets: Match ends if $|d_2|=2$. This happens with scores 2-0 (AA) or 0-2 (BB).
  $P(\text{2 sets}) = P(AA) + P(BB) = p^2 + q^2 = (\frac{2}{3})^2 + (\frac{1}{3})^2 = \frac{4}{9} + \frac{1}{9} = \frac{5}{9}$.
  A wins in 2 sets (AA): $p^2 = \frac{4}{9}$.
  B wins in 2 sets (BB): $q^2 = \frac{1}{9}$.
• 3 sets: Match ends if $|d_3|=2$ and $|d_k|<2$ for $k<3$. This is not possible. After 2 sets, the score difference can be 0 (AB or BA, prob $2pq$) or $\pm 2$. If the match continues after 2 sets, $d_2=0$. Then $d_3$ can be $1$ (A wins 3rd) or $-1$ (B wins 3rd). In either case $|d_3|=1$, so the match does not end in 3 sets.
• 4 sets: Match ends if $|d_4|=2$ and $|d_k|<2$ for $k<4$. Match continues after 3 sets if $d_2=0$ and $d_3=\pm 1$.
  $P(d_2=0) = 2pq = 2(\frac{2}{3})(\frac{1}{3}) = \frac{4}{9}$.
  If $d_2=0$, $d_3=1$ with prob $p$ or $d_3=-1$ with prob $q$.
  Match ends in 4 sets if $d_3=1$ and A wins 4th set ($d_4=2$) or $d_3=-1$ and B wins 4th set ($d_4=-2$).
  $P(\text{ends in 4 sets}) = P(d_3=1)P(A) + P(d_3=-1)P(B) = (2pq \cdot p)p + (2pq \cdot q)q = 2p^3q + 2pq^3 = 2pq(p^2+q^2) = \frac{4}{9} \cdot \frac{5}{9} = \frac{20}{81}$.
  A wins in 4 sets ($d_3=1$ and A wins 4th): $2p^2q \cdot p = 2p^3q = 2(\frac{2}{3})^3(\frac{1}{3}) = 2 \cdot \frac{8}{27} \cdot \frac{1}{3} = \frac{16}{81}$.
  B wins in 4 sets ($d_3=-1$ and B wins 4th): $2pq^2 \cdot q = 2pq^3 = 2(\frac{2}{3})(\frac{1}{3})^3 = 2 \cdot \frac{2}{3} \cdot \frac{1}{27} = \frac{4}{81}$.
• 5 sets: Match ends if $|d_5|=2$ and $|d_k|<2$ for $k<5$. Match continues after 4 sets if $d_4=0$. This happens if $d_3=1$ and B wins 4th, or $d_3=-1$ and A wins 4th.
  $P(d_4=0) = P(d_3=1)q + P(d_3=-1)p = (2p^2q)q + (2pq^2)p = 2p^2q^2 + 2p^2q^2 = 4p^2q^2 = 4(\frac{2}{3})^2(\frac{1}{3})^2 = 4 \cdot \frac{4}{9} \cdot \frac{1}{9} = \frac{16}{81}$.
  If $d_4=0$, $d_5=1$ (A wins 5th) or $d_5=-1$ (B wins 5th). Match does not end in 5 sets by the 2-point rule.
• 6 sets: Match ends if $|d_6|=2$ or $n=6$. If the match has not ended before 6 sets, it must end at 6 sets. The match reaches 6 sets if $|d_k|<2$ for $k=2,4$ and $|d_k|=1$ for $k=3,5$. This means $d_2=0, d_3=\pm 1, d_4=0, d_5=\pm 1, d_6$ is some value. The match ends at set 6 regardless of the score difference.
  The match reaches 6 sets if it hasn't ended in 2 or 4 sets.
  $P(\text{at least 3 sets}) = 1 - P(\text{ends in 2 sets}) = 1 - \frac{5}{9} = \frac{4}{9}$.
  $P(\text{at least 5 sets}) = P(\text{match continues after 4 sets}) = P(d_4=0) = \frac{16}{81}$.
  If the match reaches 6 sets (prob 16/81), the winner is determined by the scores after 6 sets. Let $w_A$ and $w_B$ be the number of sets won by A and B respectively. $w_A + w_B = 6$. $d_6 = w_A - w_B$.
  The match reaches 6 sets if the score difference is 0 after 2 sets (prob $2pq$), then $\pm 1$ after 3 sets (always happens if $d_2=0$), then 0 after 4 sets (prob $2pq$), then $\pm 1$ after 5 sets (always happens if $d_4=0$). So the sequence of score differences for the match to reach 6 sets is $0 \to \pm 1 \to 0 \to \pm 1$.
  $P(\text{reaches 4 sets}) = P(d_2=0) = 2pq = 4/9$.
  $P(\text{reaches 6 sets}) = P(d_4=0 \text{ given } d_2=0) P(d_2=0) = 4p^2q^2 \cdot 2pq = 8p^3q^3$. This is incorrect.
  $P(\text{reaches 6 sets}) = P(d_2=0 \text{ and } d_4=0) = P(d_2=0) P(d_4=0 | d_2=0)$.
  $P(d_4=0 | d_2=0)$: From $d_2=0$, the sequence of differences is $0 \to \pm 1 \to 0$. This happens with prob $p q + q p = 2pq$. So $P(d_4=0 | d_2=0) = 2pq$.
  $P(\text{reaches 4 sets}) = P(d_2=0) = 2pq = 4/9$.
  $P(\text{reaches 6 sets}) = P(d_4=0) = 4p^2q^2 = 16/81$.
  In the case of reaching 6 sets, the score difference is 0 after 4 sets. Let $w_A', w_B'$ be the sets won by A and B in the last 2 sets (sets 5 and 6). The score difference after 6 sets is $d_6 = d_4 + w_A' - w_B' = 0 + w_A' - w_B'$. The possible outcomes for sets 5 and 6 are AA, AB, BA, BB with probabilities $p^2, pq, qp, q^2$.
  If AA: $d_6 = 2$. A wins.
  If AB: $d_6 = 0$. Match ends, winner determined by score.
  If BA: $d_6 = 0$. Match ends, winner determined by score.
  If BB: $d_6 = -2$. B wins.
  If the score is tied after 6 sets (3-3), the match is a draw? The problem says "the match ends as soon as... or number of sets reaches six". This implies the match ends, and there is a winner or loser based on the scores. If the score is 3-3, the difference is 0, so neither is ahead by 2. In this case, the match is likely drawn, or the rules imply the player with more sets wins, but the 2-point rule overrides this unless 6 sets are reached. If the score is tied at 3-3 after 6 sets, neither A nor B wins the match. Let's assume the match ends with a draw, or the winner is determined by some other rule not specified, or if tied, there is no winner. Let's assume if the score is 3-3 after 6 sets, neither A nor B wins the match.
  If the score after 4 sets is 2-2 (prob $4p^2q^2 = 16/81$), then sets 5 and 6 are played.
  Score after 6 sets can be: 4-2 (AA), 3-3 (AB, BA), 2-4 (BB).
  A wins in 6 sets: score 4-2 (AA in sets 5,6). Prob $4p^2q^2 \cdot p^2 = 4p^4q^2 = 4(\frac{2}{3})^4(\frac{1}{3})^2 = 4 \cdot \frac{16}{81} \cdot \frac{1}{9} = \frac{64}{729}$.
  B wins in 6 sets: score 2-4 (BB in sets 5,6). Prob $4p^2q^2 \cdot q^2 = 4p^2q^4 = 4(\frac{2}{3})^2(\frac{1}{3})^4 = 4 \cdot \frac{4}{9} \cdot \frac{1}{81} = \frac{16}{729}$.
  Score 3-3 after 6 sets (AB or BA in sets 5,6). Prob $4p^2q^2 \cdot 2pq = 8p^3q^3 = 8(\frac{2}{3})^3(\frac{1}{3})^3 = 8 \cdot \frac{8}{27} \cdot \frac{1}{27} = \frac{64}{729}$. If this is a draw, then $P(Y \text{ in 6 sets}) = \frac{64}{729}$ and $P(Z \text{ in 6 sets}) = \frac{16}{729}$.

Total probability of A winning $P(Y) = P(\text{A wins in 2}) + P(\text{A wins in 4}) + P(\text{A wins in 6})$
$P(Y) = p^2 + 2p^3q + 4p^4q^2 = \frac{4}{9} + \frac{16}{81} + \frac{64}{729} = \frac{4 \cdot 81 + 16 \cdot 9 + 64}{729} = \frac{324 + 144 + 64}{729} = \frac{532}{729}$.
Total probability of B winning $P(Z) = P(\text{B wins in 2}) + P(\text{B wins in 4}) + P(\text{B wins in 6})$
$P(Z) = q^2 + 2pq^3 + 4p^2q^4 = \frac{1}{9} + \frac{4}{81} + \frac{16}{729} = \frac{81 + 36 + 16}{729} = \frac{133}{729}$.
Total probability of match ending with a winner = $P(Y) + P(Z) = \frac{532+133}{729} = \frac{665}{729}$.
The remaining probability is for a 3-3 tie after 6 sets: $1 - \frac{665}{729} = \frac{64}{729}$. This confirms the probability of a 3-3 tie.

Let's evaluate the options.
$X_i$: at least $i$ sets are played.
$X_1$: always true, $P(X_1) = 1$.
$X_2$: Match does not end in 1 set (always true). $P(X_2)=1$.
$X_3$: Match does not end in 2 sets. $P(X_3) = 1 - P(\text{ends in 2 sets}) = 1 - \frac{5}{9} = \frac{4}{9}$.
$X_4$: Match does not end in 2 or 3 sets. Since it never ends in 3 sets, $P(X_4) = P(X_3) = \frac{4}{9}$.
$X_5$: Match does not end in 2, 3, or 4 sets. $P(X_5) = P(\text{match continues after 4 sets}) = P(d_4=0) = \frac{16}{81}$.
$X_6$: Match does not end before 6 sets. $P(X_6) = P(X_5) = \frac{16}{81}$.
$X_7$: Match does not end in 6 sets. $P(X_7) = 0$.

Option A: $P(Y | X_5) = \frac{P(Y \cap X_5)}{P(X_5)}$. $Y \cap X_5$ is the event that A wins and at least 5 sets are played. A can win in 2, 4, or 6 sets. If A wins in 2 or 4 sets, the number of sets is less than 5, so $X_5$ does not occur. If A wins in 6 sets, then at least 5 sets are played, so $X_5$ occurs. Thus, $Y \cap X_5 = \{\text{A wins in 6 sets}\}$.
$P(Y \cap X_5) = P(\text{A wins in 6 sets}) = \frac{64}{729}$.
$P(X_5) = \frac{16}{81}$.
$P(Y | X_5) = \frac{64/729}{16/81} = \frac{64}{729} \cdot \frac{81}{16} = \frac{4}{9}$.
Option A is correct.

Option B: $P(Z | X_4) = \frac{P(Z \cap X_4)}{P(X_4)}$. $Z \cap X_4$ is the event that B wins and at least 4 sets are played. B can win in 2, 4, or 6 sets. If B wins in 2 sets, $X_4$ does not occur. If B wins in 4 or 6 sets, $X_4$ occurs. Thus, $Z \cap X_4 = \{\text{B wins in 4 sets or B wins in 6 sets}\}$.
$P(Z \cap X_4) = P(\text{B wins in 4 sets}) + P(\text{B wins in 6 sets}) = \frac{4}{81} + \frac{16}{729} = \frac{36+16}{729} = \frac{52}{729}$.
$P(X_4) = \frac{4}{9}$.
$P(Z | X_4) = \frac{52/729}{4/9} = \frac{52}{729} \cdot \frac{9}{4} = \frac{13}{81}$.
Option B is correct.

Option C: $P(X_{2k-1}) = P(X_{2k}) \forall k \in \{1, 2, 3\}$.
For $k=1$: $P(X_1) = P(X_2)$. We found $P(X_1)=1, P(X_2)=1$. So $1=1$, true.
For $k=2$: $P(X_3) = P(X_4)$. We found $P(X_3)=4/9, P(X_4)=4/9$. So $4/9=4/9$, true.
For $k=3$: $P(X_5) = P(X_6)$. We found $P(X_5)=16/81, P(X_6)=16/81$. So $16/81=16/81$, true.
Option C is correct.

Option D: $P(Z | X_1) = \frac{64}{729}$.
$P(Z | X_1) = \frac{P(Z \cap X_1)}{P(X_1)}$. $X_1$ is the event that at least 1 set is played, which is always true. So $P(X_1)=1$.
$Z \cap X_1 = Z$, the event that B wins the match.
$P(Z | X_1) = P(Z) = \frac{133}{729}$.
The option states $P(Z | X_1) = \frac{64}{729}$.
$\frac{133}{729} \neq \frac{64}{729}$.
Option D is incorrect.

The final answer is $\boxed{D}$.