Thread: Probability Distribution

A box of 10 flashbulbs contains 3 defective bulbs. A random sample of 2 is selected and tested. Let X be the random variable associated with the number of defective bulbs in the sample.

1. Find the probability distribution of X.
2. Find the expected number of defective bulbs in a sample.

The distribution I created is:


xi 0 1 2
pi 7/15 7/15 1/15


0 defective is 7/10*6/9 = 7/15



1 defective is 7/10*3/9 + 3/10*7/9 = 7/15





2 defective is 3/10*2/9=7/15

E(x) = 0*7/15 + 1*7/15 + 2*1/15 = 9/15=.6

hypergeometric, where



The mean is the same as a binomial, np where n=2 and p=.3, giving you .6 quickly
NOW the variance if the hyper isn't the same as a binomial but is is asymptotically equivalent.

Terrific, but I have to specifically complete the table and solve E(x). Is my work correct?