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 Mean of X.

Answer 1

We have a lot of $6$ items, 2 are defective $\to 6-2=4$ are non-defective.

P (drawing a defective item) = $\frac{2}{6}=\frac{1}{3}$

P (drawing a non-defective item) = 1 - P (drawing a defective item) $=1-\frac{1}{3}=\frac{2}{3}$

Note$:4$ items can be drawn from a lot of 6 items in ${}^6{C}_4$ 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) = $\frac{{}^4{C}_4}{{}^6{C}_4}=\frac{1}{15}$

$P(X=1)=$ P (one defective item from the lot) = P(one defective, 3 non-defective items) $=(\frac{{}^2{C}_1{}^4{C}_3}{{}^6{C}_4})=\frac{8}{15}$

$P(X=2)=$ P (two defective items from the lot) = P(2 defective, 2 non-defective bulbs) =$\frac{{}^2{C}_2{}^4{C}_2}{{}^6{C}_4}=\frac{6}{15}$

Therefore the required probability distribution is as follows:

$X:012$

$P\left(X\right):\frac{1}{15}\frac{8}{15}\frac{6}{15}$

So Mean of $X$ will be$:{\sum }_{i=0}^2{x}_{i}P(X=x_i)$

$\to 0\times \frac{1}{15}+1\times \frac{8}{15}+2\times \frac{6}{15}=\frac{4}{3}$