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Question

From a lot of 6 items containing 2 defective items, a sample of 4 items are drawn at random. Let the random variable X denote the number of defective items in the sample. If the sample is drawn without replacement, find Variance of X.

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Solution

We have a lot of 6 items, 2 are defective 62=4 are non-defective.
P (drawing a defective item) = 26=13
P (drawing a non-defective item) = 1 - P (drawing a defective item) =113=23
Note:4 items can be drawn from a lot of 6 items in 6C4 ways.
Let X be the random variable of drawing 2 defective items from the lot.
P(X=0)= P (no defective item in the sample) = 4C46C4=115
P(X=1)= P (one defective item from the lot) = P(one defective, 3 non-defective items) =(2C14C36C4)=815
P(X=2)= P (two defective items from the lot) = P(2 defective, 2 non-defective bulbs) =2C24C26C4=615
Therefore the required probability distribution is as follows:
X: 0 1 2
P(X):115 815 615

So Variance of X will be :E(X2)(E(X))2
E(X)=2i=0xiP(X=xi)
0×115+1×815+2×615=2015
E(X2)=2i=0x2iP(X=xi)
02×115+12×815+22×615=3215
Variance of X=3215(2015)2=1645

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