Supposed three cellphones are tested at random. Let D represent the defective cell phone and N let represent the non-defective cell phone. If we let X be the random variable for the number of defective cell phone, construct the probability distribution of the random variable X.

Given D represents defective cell phone while N represents non-defective cellphone

Then if:

1) All 3 non defective cell phones occur; {NNN}, then, X=0

2) 2 non-defective and 1 defective and cell phones occur; {DNN, NDN, NND}, then, X = 1

3) 1 non-defective and 2 defective occur; {DDN, DND,NDD} , then, X = 2

4) All 3 defective cell phone occur; {DDD} , then, X = 3

Therefore, the only possible values of the random variable X are 0, 1, 2, 3 with the probability distribution will be given by:

X 0 1 2 3

Possible outcomes {NNN} {DNN, NDN, NND} {DDN, DND,NDD} {DDD}

P(X) 1/8 3/8 3/8 1/8