Question

In a group of five items, two are defective. Find the dis need to find the first defective item, the mean and the variance ofN if, a) draws are with replacement, b) draws are without replacement. #7. tribution of N, the number of draws we

Answer 1

a)

If balls are drawn with replacement, then probability of selecting a defective item is 2/5 on each draw.

Then N ~ Geometric(p = 2/5)

and E(N) = 1/p = 5/2 = 2.5

Var(N) = (1 - p) / p² = (1 - 2/5) / (2/5)² = 15/4 = 3.75

b)

If balls are drawn without replacement, then the possible values of N are 1, 2, 3, 4. (After 3 draws, if all selected items are non-defective then it is obvious that defective item will be selected in 4th draw)

P(N = 1) = 2/5 (First item selected is defective)

P(N = 2) = (3/5) * (2/4) = 3/10 (First item selected is non-defective and second item selected is defective)

P(N = 3) = (3/5) * (2/4) * (2/3) = 1/5 (First two items selected are non-defective and third item selected is defective)

P(N = 4) = (3/5) * (2/4) * (1/3) * (2/2) = 1/10 (First 3 items selected are non-defective and 4th item selected is defective)

E[N] = 1 * (2/5) + 2 * (3/10) + 3 * (1/5) + 4 * (1/10) = 2

E[N²] = 1² * (2/5) + 2² * (3/10) + 3² * (1/5) + 4² * (1/10) = 5

Var[N] = E[N²] - E[N]²  = 5 - 2² = 1