The Reds and the Cubs are playing 3 games. In each game the probability that the Reds win is 0.54. The probability of the Reds winning is not affected by who has won any previous games.

(a) What is the expected value for the number of games the Reds will win? (Give your answer correct to two decimal places.) _

(b) What is the expected value for the number of games the Cubs will win? (Give your answer correct to two decimal places.) _

(a) Let X denote the number of games the Reds win. If Reds win exactly one it means that in the other two games Reds lost i.e Cubs won. Now, the probability of the Reds losing is determined by [tex] 1 -\text{probability of winning} = 1-0.54 = 0.46 [/tex]. So, if X wins exactly [tex]i[/tex] times, then we should multiply the probabilities associated with

[tex]P(X=0) = \binom{3}{0} \times 0.46 \times 0.46 \times 0.46 = 0.0973[/tex]

[tex] P(X=1) = \binom{3}{1} 0.54 \times 0.46 \times 0.46 = 0.3429[/tex]

[tex] P(X=2) = \binom{3}{2} 0.54 \times 0.54 \times 0.46 = 0.4023[/tex]

[tex] P(X=3) = \binom{3}{3} 0.54 \times 0.54 \times 0.54 = 0.1574[/tex]

Now,

[tex] E(X) = \sum_{i=0}^{3} i \times P(X=i) = 1.6197[/tex]

(b) Let Y denote the number of games won by Cubs. Then by the similar logic as above,

[tex] P(Y=3) = \binom{3}{0} 0.46 \times 0.46 \times 0.46 = 0.0973[/tex]

[tex] P(Y=2) = \binom{3}{1} 0.54 \times 0.46 \times 0.46 = 0.3429[/tex]

[tex] P(Y=1) = \binom{3}{2} 0.54 \times 0.54 \times 0.46 = 0.4023[/tex]

[tex] P(Y=0) = \binom{3}{3} 0.54 \times 0.54 \times 0.54 = 0.1574[/tex]

Now

[tex] E(Y) = \sum_{j=0}^{3} j \times P(Y=j) = 1.3803[/tex]