A lot had 10% defective items. Ten items are chosen randomly from this lot. Find the probability that exactly two of the chosen items are defective.

A lot had 10% defective items. Ten items are chosen randomly from this lot. The probability that exactly two of the chosen items are defective is 

(A) 0.0036 

(B) 0.1937 

(C) 0.2234 

(D) 0.3874 

(GATE 2005) 

Answer: (B) 0.1937 

Explanation: 

Let $X$ be a random variable that denotes the number of defective items chosen.  

Here, the number of 'trials', $n = 10$ and 

the probability of 'success', $p = \dfrac{10}{100} = 0.10$ 

Therefore, the required probability is 

$= P(X = 2)$

$ = ^{10}C_2 \times (0.10)^2 \times (1 - 0.10)^{10 - 2}$ 

$= 0.1937$