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.
Answer & explanation
Correct answer: option 1
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, $E(X)=∑p_ix_i$
$=0×\frac{91}{190}+1×\frac{42}{95}+2×\frac{3}{38}$
$=\frac{84+30}{190}=\frac{114}{190}=0.6$