19218 gas masks of the military were tested, 10322 ere defective. if 2 gas maskare selected, find the probability that they are both defective.

Question

a.assume that the first gas mask is replaced before the next on is selected.

 

b.assume that the first mask is not replaced before he next one is selected.

Answer 1

First, the probability that any one gas mask will be defective is 10322 / 19218.

 

a.  Based on the above fact, the probability that the first gas mask selected will be defective is 10322 / 19218.  If it is put back in the set and another selection is made, the probability that this mask will be defective is again 10322 / 19218.  So the probability that they will BOTH be defective is (10322 / 19218) * (10322 / 19218) = 0.288477.

 

b.  The probability that the first gas mask selected will be defective is 10322 / 19218.  In order for BOTH masks to be defective, we have to assume that the first one was defective.  If it is NOT put back in the set, then there will be 19217 masks left, of which only 10321 will be defective.  So the probability that the second one is defective, assuming the first one was, is 10321 / 19217.  So the probability that they are BOTH defective is (10322 / 19218) * (10321 / 19217) = 0.288464 .