In a manufacturing unit inspection from a lot of 20 baskets which include 6 defectives, a sample of 2 baskets is drawn at random without replacement. Prepare the probability distribution of the number of defective baskets. Also, calculate E(X) for the random variable X.

0.6

The correct answer is Option (1) → 0.6

Let the random variable X be defined as the number of defective baskets in a sample of two baskets, then X can take values 0, 1, 2.

Total number of baskets = 20.

Two baskets can be drawn simultaneously in ${^{20}C}_2$ ways

$P(X = 0)$ = P(drawing no defective basket)

$=\frac{{^{14}C}_2}{{^{20}C}_2}=\frac{14×13}{20×19}=\frac{91}{190}$

$P(X = 1)$ = P(drawing one defective basket)

$\frac{{^{14}C}_1 × {^6C}_1}{{^{20}C}_2}=\frac{2 × 14 × 6}{20 × 19}=\frac{42}{95}$

$P(X = 2)$ = P(drawing two defective baskets)

$=\frac{{^6C}_1}{{^{20}C}_2}=\frac{6×5}{20×19}=\frac{3}{38}$

Now, $E(X)=∑p_ix_i$

$=0×\frac{91}{190}+1×\frac{42}{95}+2×\frac{3}{38}$

$=\frac{84+30}{190}=\frac{114}{190}=0.6$