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Daniel plays two games in the casino. The first game he believes he has 60% chance to win. If he wins the first game, he will win the second game with 35% chance, if he loses the first game, he will win the second game with 65% chance. The winning prices for the first and second game are $1 and $2 respectively. a. If he wins the second game, what is the probability that he lost the first game

Respuesta :

ogorwyne

Answer:

0.5532

Step-by-step explanation:

We have these events

WF = Wins the first game

LF = lose the first game

WS = Wins the second game

LS = lose the second game

Then

Prob(WF) = 0.6

Prob(LF) = 0.4

Prob(WS|WF) = 0.35

Prob(WS|LF) = 0.65

1. If he wins the second game, the probability that he lost the first =

P(Lf|WS)

= 0.4x0.65/0.4x0.65+0.6x0.35

= 0.21/0.47

= 0.5532

0.5532 is the probability that he lost the first game given that he wins the second

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