Question

A carton contains $20$ bulbs, $5$ of which are defective. The probability that, if a sample of $3$ bulbs is chosen at random (with replacement) from the carton, exactly $2$ will be defective is

• A. $\frac{3}{64}$
• B. $\frac{1}{16}$
• C. $\frac{9}{64}$
• D. $\frac{2}{3}$

Answer 1

Answer: (C)

$\frac{9}{64}$

Let $D$ be the event in which defective bulbs are chosen and $N$ be the event in which non-defective bulbs are chosen.
Total number of bulbs $=20$
$n\left(D\right)=5,n\left(N\right)=15$
Now, required probability $=P(D,D,N)+\left)P\right(D,N,D)+P(N,D,D)$
$=\frac{5}{20}\times \frac{5}{20}\times \frac{15}{20}+\frac{5}{20}\times \frac{15}{20}\times \frac{5}{20}+\frac{15}{20}\times \frac{5}{20}\times \frac{5}{20}$
$=3(\frac{5}{20}\times \frac{5}{20}\times \frac{15}{20})$
$=\frac{9}{64}$