All of the answers above are incorrect. To satisfy the condition of A winning the game and winning the 7th match, A must win at four matches and the 7th match must be won. What this means is, in the first six matches, A must win 3 games or more. This can be expressed in the following probability: P(A wins the game) = P(A wins at least three of the six matches AND A wins the 7th match) = P(A wins at least three matches) * P(A wins the 7th match). This structure follows P(A n B) = P(A) * P(B) where the matches are independent. P(A wins at least 3 of the six matches) = P(A wins exactly 3 of the first six matches) + P(A wins exactly 4 of the first six matches) + P(A wins exactly 5 of the first six matches) + P(A wins exactly 6 of the first six matches) This probability above needs to be multiplied by P(A wins the 7th match) = p Hint: Binomial Probability
P({A wins last match} AND {A wins 4 matches out of 7}) = // definition of conditional prob P({A wins last match} | {A wins 4 matches out of 7}) * P({A wins 4 matches out of 7}) The first element of this product is easy to compute: P({A wins last match} | {A wins 4 matches out of 7}) = 4/7 The second element of this product can be computed using a Binomial(prob=p, n=7) distribution, with k=4. P({A wins 4 matches out of 7}) = binom(7, 4) * p^4 * (1-p)^3 So overall, P({A wins last match} AND {A wins 4 matches out of 7}) = 4/7 * binom(7, 4) * p^4 * (1-p)^3 = binom(6, 3) * p^4 * (1-p)^3 That is, the final answer is: binom(6, 3) * p^4 * (1-p)^3
