Respuesta :
Answer:
(a) 1.6197
(b) 1.3803
Step-by-step explanation:
(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]
