Two sports teams, Team A and Team B, will play a series of games that works as f

Question

Two sports teams, Team A and Team B, will play a series of games that works as follows. 1. There will be at most 5 games. 2. The series will stop as soon as one of the teams has won three games. The team that wins three games wins the series.
Assume that each game played will be independent of the other games.
For each gamed played, the probability Team A will win is .45
For each game played, the probability Team B will win is .55.

A)Calculate the probability that Team A will win the series by winning the first three games. ( A formula with the exact numbers in it is more important than a single number)

B)Calculate the probability Team B will win the series by winning the first three games. ( A formula with the exact numbers in it is more important than a single number)

C)Calculate the probability that the series will be exactly three games long. ( A formula with exact numbers in it is more important than a single number.

D)Calculate the probability that team B will win exactly 2 of the first three games and then win the series by winning the fourth game. ( A formula with exact numbers is more important than a single number)

E)Calculate the probability that the series will be exactly four games long. ( A formula with exact numbers is more important than a single number)

Explanation / Answer

Let us call p1 = Probability that team A wins a game, which is: p1 = 0.45 Let us call p2 = Probability that team B wins a game, which is: p2 = 0.55 (A) Probability that team A wins the first 3 games is p1 and p1 and p1: P(Team A wins in 3) = p1.p1.p1 P(Team A wins in 3) = 0.45 x 0.45 x 0.45 P(Team A wins in 3) = p1.p1.p1 = 0.091125 (approx 9.11 %) (B) Probability that team B wins the first 3 games is p2 and p2 and p2: P(Team B wins in 3) = p2.p2.p2 P(Team B wins in 3) = 0.55 x 0.55 x 0.55 P(Team B wins in 3) = p2.p2.p2 = 0.166375 (approx 16.64 %) (C) Probablity that the series lasts only 3 games is the same as probability A wins in 3 OR (plus) probability B wins in 3: P(Series lasts 3 games only) = P(Team A wins in 3) + P(Team B wins 3) P(Series lasts 3 games only) = p1.p1.p1 + p2.p2.p2 P(Series lasts 3 games only) = 0.0091125 + 0.166375 P(Series lasts 3 games only) = 0.2575 (25.75 %) (D) Probablity that team B wins in 4 by winning games 1, 2 and 4: P(B wins games 1,2 and 4) = P(B wins g1) and P(B wins g2) and P(A wins g3) and P(B wins g4) P(B wins games 1,2 and 4) = p2 times p2 times p1 times p2 P(B wins games 1,2 and 4) = p2.p2.p1.p2 P(B wins games 1,2 and 4) = 0.55 x 0.55 x 0.45 x 0.55 P(B wins games 1,2 and 4) = 0.0748685 (approx 7.487 %) (e) Series last exactly 4 games. This will occur 6 possible ways. Team A can win in 4 games 3 different ways: (AABA, ABAA, an BAAA), each with probability p1.p1.p1.p2. Team B can win in 4 games also 3 different ways: (BABB, BBAB, and ABBB), each with probability p2.p2.p2.p1 So for the series to last exactly 4 games the probability is: P(series end in 4) = 3(p1.p1.p1.p2) + 3(p2.p2.p2.p1) P(series end in 4) = 3(0.05011875) + 3(0.0748685) P(series end in 4) = 0.15035625 + 0.22460625 P(series end in 4) = 0.3749625 You can trust this I verified all cases (including series lasts 5) add up to 1.

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