Problem 2 Let X be the number of the cars being repaired at

Problem 2 Let X be the number of the cars being repaired at a repair shop. We have the following information: . At any time, there are at most 3 cars being repaired. . The probability of having 2 cars at the shop is the same as the probability of having one car. . The probability of having no car at the shop is the same as the probability of having 3 cars. . The probability of having 1 or 2 cars is half of the probability of having 0 or 3 cars. Find the PMF of X.

Solution

P (X= 0,1,2,3) are mutually exclusive, independent and exhaustive set of events.

So,

P(X=0) + P(X=1) + P(X=2) + P(X=3) = 1

WE know that,

P(X=0) = P(X=3) and

P(X=1) = P(X=2)

So,

P(X=0) + P(X=1) + P(X=1) + P(X=0) = 1

P(X=0) + P(X=1) = 1 / 2

Now,

P(X=0) = 2 P(X=1)

Thus,

2P(X=1) + P(X=1) = 1/2

P(X=1) = 1/6 = P(X=2)

and P(X=0) = 1/3 = P(X=3)

PMF :

P(X=0) = 1/3

P(X=1) = 1/6

P(X=2) = 1/6

P(X=3) = 1/3

Hope this helps. Ask if you have any doubts.