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
<u>A) Determine the probability distribution of X number of games played in the series</u>
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
<u>B) the expected number of games to be played </u>
∑ x(Px) = ( 2 * 0.58 ) + 3 ( 0.42 ) = 2.42 ≈ 2 games
<u>C) Verify that the probability that series ends after two games = 58%</u>
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
