A tennis match is played between two players, $A$ and $B$. The match consists of at most three sets. Each set is won by either $A$ or $B$, and the match is won by the first player to win two sets. The probability that $A$ wins the first set is $p$. For each set after the first, the conditional probability that $A$ wins the set, given that $A$ won the preceding set, is 0.7. For each set after the first, the conditional probability that $B$ wins the set, given that $B$ won the preceding set, is 0.6. Construct a probability tree showing this information.

1. Calculate the probability that player A wins the match in exactly three sets.

A match won by player $A$ in exactly 3 sets must follow one of two paths of set winners: $A-B-A$ or $B-A-A$.
Using the conditional probabilities given in the tree structure:
$\text{P}(A \text{ wins in 3}) = \text{P}(A_1)\text{P}(B_2 \mid A_1)\text{P}(A_3 \mid B_2) + \text{P}(B_1)\text{P}(A_2 \mid B_1)\text{P}(A_3 \mid A_2)$
$= p(0.3)(0.4) + (1 - p)(0.4)(0.7)$
$= 0.12p + 0.28(1 - p)$
$= 0.28 - 0.16p$

2. Calculate the probability that player B wins the match in exactly three sets.

A match won by player $B$ in exactly 3 sets must follow the paths $A-B-B$ or $B-A-B$.
$\text{P}(B \text{ wins in 3}) = \text{P}(A_1)\text{P}(B_2 \mid A_1)\text{P}(B_3 \mid B_2) + \text{P}(B_1)\text{P}(A_2 \mid B_1)\text{P}(B_3 \mid A_2)$
$= p(0.3)(0.6) + (1 - p)(0.4)(0.3)$
$= 0.18p + 0.12(1 - p)$
$= 0.12 + 0.06p$

3. Equate the probabilities and solve for $p$.

Since the probabilities of winning in 3 sets are equal for both players:
$\text{P}(A \text{ wins in 3}) = \text{P}(B \text{ wins in 3})$
$0.28 - 0.16p = 0.12 + 0.06p$
$0.28 - 0.12 = 0.06p + 0.16p$
$0.16 = 0.22p$
$p = \frac{0.16}{0.22} = \frac{16}{22} = \frac{8}{11} \text{ (shown)}$

1. Calculate the probability of the sequence where $B$ wins the first set and $A$ wins the match.

For $A$ to win the match after $B$ wins the first set, the sequence of wins must be $B$ then $A$ then $A$.
$P(B, A, A) = P(B_1) \times P(A_2 \mid B_1) \times P(A_3 \mid A_2)$
$P(B, A, A) = \left(1 - \frac{8}{11}\right) \times (1 - 0.6) \times 0.7$
$P(B, A, A) = \frac{3}{11} \times 0.4 \times 0.7 = \frac{0.84}{11} = \frac{21}{275}$

2. Calculate the overall probability that $A$ wins the match.

$A$ can win the match through three mutually exclusive sequences of set wins: $A-A$, $A-B-A$, or $B-A-A$.
$P(A, A) = \frac{8}{11} \times 0.7 = \frac{5.6}{11} = \frac{140}{275}$
$P(A, B, A) = \frac{8}{11} \times (1 - 0.7) \times (1 - 0.6) = \frac{8}{11} \times 0.3 \times 0.4 = \frac{0.96}{11} = \frac{24}{275}$
Total probability $A$ wins the match is the sum of these outcomes:
$P(\text{A wins}) = P(A, A) + P(A, B, A) + P(B, A, A)$
$P(\text{A wins}) = \frac{140}{275} + \frac{24}{275} + \frac{21}{275} = \frac{185}{275}$

3. Use the conditional probability formula to find the required probability.

We want to find $P(B \text{ wins 1st} \mid A \text{ wins match})$.
$P(B \text{ wins 1st} \mid A \text{ wins match}) = \frac{P(B \text{ wins 1st} \cap A \text{ wins match})}{P(A \text{ wins match})}$
$P(B \text{ wins 1st} \mid A \text{ wins match}) = \frac{21/275}{185/275} = \frac{21}{185}$