Question

Best of three In a best out of three series played between teams A and B, the team that gets two wins first wins the entire series. Assuming that team A has a 70% chance of winning a game and that outcome of games are indepen- dent events, the probability that the series ends after two games is 58%
a. Find the probability distribution of X number of games played in the series.
b. Find the expected number of games to be played in the series.
c. Verify that the probability that the series ends after two games is 58% by writing out the sample space of all possible sequences of wins and losses for team A in a best of three series. (A tree diagram may help.) Find the probability for each sequence and then add up those for which the series ends after two games. (Use what you learned in Section 5.2 about probabilities for intersections of independent events.)

Answer 1

Answer:

a) P (x = 2 games )  =  0.49 + 0.09 = 0.58

    P ( x = 3 games ) = 0.063 + 0.147 + 0.147 + 0.063 = 0.42

b) = 2.42 ≈ 2 games

c) P (x = 2 games )  =  0.49 + 0.09 = 0.58

Step-by-step explanation:

Team A chance of winning a game in the series. P( team A ) = 70% = 0.7

P ( team B ) = 0.3

probability of series ending after two games = 58% = 0.58

A) Determine the probability distribution of X number of games played in the series

First we have to consider the possible combinations that will decide the series and they are

( A,A ) , ( B,B) , ( A,B,B) , ( A,B,A ) , ( B,A,A ), ( B,A,B)  = 6 Combinations

( A,A ) = 0.7 * 0.7 = 0.49

( B,B ) = 0.3 * 0.3 = 0.09

( A,B,B ) = 0.7 * 0.3 *0.3 = 0.063

( A,B,A ) = 0.147

( B,A,A ) = 0.147

( B,A,B ) = 0.063

The distribution of the games in the series can be either game or three games before the end of the series

P (x = 2 games )  =  0.49 + 0.09 = 0.58

P ( x = 3 games ) = 0.063 + 0.147 + 0.147 + 0.063 = 0.42

B) the expected number of games to be played

∑ x(Px) = ( 2 * 0.58 ) + 3 ( 0.42 ) = 2.42 ≈ 2 games

C)  Verify that the probability that series ends after two games = 58%

sample space of all possible sequences of wins and losses

( A,A ) , ( B,B) , ( A,B,B) , ( A,B,A ) , ( B,A,A ), ( B,A,B)  = 6 Combinations

( A,A ) = 0.7 * 0.7 = 0.49

( B,B ) = 0.3 * 0.3 = 0.09

( A,B,B ) = 0.7 * 0.3 *0.3 = 0.063

( A,B,A ) = 0.147

( B,A,A ) = 0.147

( B,A,B ) = 0.063

hence :

P (x = 2 games )  =  0.49 + 0.09 = 0.58