Ten percent of the tools produced in a certain manufacturing process turn out to be defective. Find the probability that in a sample of 10 tools chosen at random, exactly 2 will be defective, by using the binomial distribution.

0.1937

The correct answer is Option (3) → 0.1937

Here $n = 10, p =\frac{10}{100}=0.1$, so $λ = np = (10) (0.1) = 1$.

$P(X = 2) = {^nC}_2p^2q^{n-2} = {^{10}C}_2p^2q^8=\frac{10.9}{1.2}(0.1)^2 (0.9)^8 = 0.1937$