# In a best of five series (first team to win 3 games wins the series), the probablity of Team A defeating Team B in any particular game is 70%. What is the probablity that all 5 games will be needed...

In a best of five series (first team to win 3 games wins the series), the probablity of Team A defeating Team B in any

particular game is 70%. What is the probablity that all 5 games will be needed to be played in order for a winner to be determined.

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### 1 Answer

In a best of five series, the first team to win 3 games wins the series. The probability of Team A defeating Team B in any

particular game is 70%.

The probability that all 5 games will be needed to be played in order for a winner to emerge has to be determined.

This is the case if the outcome of the first 4 games is that 2 games are won by team A and the other two are won by team B. The probability that team A wins any game is 0.7 and the probability that it loses any game is 1 - 0.7 = 0.3.

5 games have to be played if team A wins the sets of games (1, 2), (1, 3), (1, 4), (2, 3), (2, 4) or (3, 4) and loses the other two. The probability of this happening is 0.7*0.7*0.3*0.3*6 = 0.2646. This is the required probability.

**The probability that all 5 games have to be played before a winner emerges is 26.46%**