Question

A gambler faces three different slot machines, A, B and C, from which he randomly picks one and plays the machine only once. The probability he will win if he plays machine A is 0.32. The probability he will win if he plays machine B is 0.05. The probability he will win if he plays machine C is 0.04. Use four decimals in your answers. (a) What is the probability of this gambler winning? (b) If he does win, what is the probability he played machine C? (c) If he did not win, what is the probability he did play machine A?

Answer 1

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a. P(gambler wins) = P(selects A)*P(wins against A)+P(selects B)*P(wins against B)+P(selects C)*P(wins against C)

= (1/3)*(.32) + (1/3)*(.05)+ (1/3)*(.04) = 0.1367

( since, randomly selecting any of the above slot machines is 1/3 each)

b. P( given he wins, that he played C)

= prob he wins through C / prob he wins

= (1/3)*(.04)/ (1/3)*(.32) + (1/3)*(.05)+ (1/3)*(.04) = (1/3)*(.04)/.1367 = 0.0975

c. P( if he doesn't win, he played A) =

= ((1/3)*(1-.32)) / ((1/3)*(1-.32) + (1/3)*(1-.05)+ (1/3)*(1-.04))

= 0.2625

( please note that we have used complementry method to solve this problem, i.e. P(A' ) = 1- P(A))