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.
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.
- Find the probability distribution of X.
- 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