Of five items, two are defective. Find the distribution of N, the number of draws to find the first defective item. What is the mean and variance.

Given is the probability of dective item = 2/5 = 0.4 = p
say

So the probability of not getting the deffective item
in a draw = 1-2/5 = 0.6 = q = (1-p) say .  Also 0 < p , q <
1

The probability of getting the 1st defective in the 1st
draw = 0.4.

Probability drawing the 1st defective in the
2nd draw = P(not getting defective  in the 1st draw)*P(getting a defective in the 2nd
draw) = qp.

Probability of getting the dective in the 3rd
draw = P(failure of getting a defective  in the 1st 2 draws)*P(sucess of getting a
defective  in the 3rd  draw.) = q^2*p

Let  x  denote the
number of draws to get the first defetive item.Then the probability of getting a
defetive in  the nth  draw = P(not getting a defective inall x-1 draw)*P(getting the 1st
defective in the nth draw) = q^(x-1)*p.

Thus the
distribution of  getting the 1st defective item in the xth draw =
q^(x-1)*p.

Mean of the distribution m = E(x) =  {sum
(x*q^x*p) for x = 1,2,3,.... } =
1*p+2*qp+3*q^2p+4*q^3*p+......

E(x) = p(1
+2q+3q^2+4q^3+.....)

E(x) =p(1-q)^(-2) = p/p^2 = 1/p
.

Therefore mean m = E(x) =
1/p = 1/(2/3) = 3/2.

E(x^2) =
1^2p+2^2*(qp)+3^2*(q^2p)+4^2(q^3p)+...

E(x^2) = p {
1+4*q+9q^2+16q^3+..}

E(x^2) = p {
(2-1)+(2*3-2)q+(3*4-2)q^2+(4*5-4)q^4+...}

E(x^2) = 2p
{1+2*3q+3*4q^2+4*5q^3+...}- p{1+2q+3q^2+4q^3+...}

E(x^2) =
2p(1-q)^(-3) - p(1-q)^2 = 2/p^2 -1/p

Therefore variance =
E(x^2) - (E(x))^2 = (2/p^2-1/p)-(1/p)^2

Variance =
2/p^2-1/p -1/p^2 = 1/p^2-1/p =  (1-p)/p^2.

But p =
2/3.

Therefore variance =
(1-2/3)/(2/3)^2 =0.75.