# 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.

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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.**