A vending machine has four buttons A B C and D which you use

A vending machine has four buttons A, B, C and D which you use to select different kinds of candy. Unfortunately, the buttons do not work well. In particular, the probability that you will actually get candy if you press a specific button is: Probability of candy from pressing A = 1 Probability of candy from pressing B = 3/4 Probability of candy from pressing C = 2/3 Probability of candy from pressing D = 1/2 What is the probability that you choose any button at random and do not receive candy? Suppose you chose either button B or C at random. What is the probability that you do not receive candy? If you chose a button randomly and did receive candy, what is the probability that you pressed button A?

Solution

Let

A, B, C, D = the choice of button

R =event that you received a candy

Thus,

a)

P(R\') = P(A) P(R\'|A) + P(B) P(R\'|B) + P(C) P(R\'|C) + P(D) P(R\'|D) =

As all buttons have a probability of 1/4 each,

P(R\') = (1/4)(1) + (1/4)(3/4) + (1/4)(2/3) + (1/4)(1/2)

P(R\') = 35/48 = 0.729166667   [answer]

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b)

P(R\'|B or C) = P(R\' and (B or C)) / P(B or C)

As

P(R\' and (B or C)) = (1/4)(1-3/4) + (1/4)(1-2/3) = 7/48

and

P(B or C) = (1/4) + (1/4) = 1/2

Then

P(R\'|B or C) = P(R\' and (B or C)) / P(B or C) = (7/48)/(1/2) = 7/24 = 0.291666667 [answer]

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c)

P(A|R\') = P(A and R\') / P(R\')

As

P(A and R\') = P(A) P(R\'|A) = (1/4)(1) = 1/4

Then

P(A|R\') = P(A and R\') / P(R\') = (1/4)/(35/48) = 12/35 =0.342857143 [answer]