A vending machine has four buttons A, B, C and I) which you use to select differ
ID: 3202500 • Letter: A
Question
A vending machine has four buttons A, B, C and I) 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?Explanation / Answer
Let A, B, C, D be the choice of buttons.
Probability to choose each one button is same that is
P(A) = P(B) = P(C) = P(D) = 1/4
R : event that you received a candy
R' : Event that you do not received a candy.
We have given P(R'|A) = 1
P(R'|B) = 3/4
P(R'|C) = 2/3
P(R'|D) = 1/2
a) In this part we have to find probability that you choose any button at random and do not receive candy.
P(R') = P(A) P(R'|A) + P(B) P(R'|B) + P(C) P(R'|C) + P(D) P(R'|D)
P(R') = (1/4)(1) + (1/4)(3/4) + (1/4)(2/3) + (1/4)(1/2)
P(R') = 35/48 = 0.7292
b) We have choose either button B or C at random, and we have to find the probability that you do not receive candy.
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 = 0.29167
c) If you choose a button randomly and did receive candy, and we have to find probability that you pressed button.
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 =0.3429