Question

In a purchase of 100 Bulbs, 4 are defective. Two bulbs are randomly selected for inspection. What is the probability that both bulbs are defective if the first bulb is not replaced after inspecting?

• A. $\text{P(E)}=\frac{4}{825}$
• B. $\text{P(E)}=\frac{2}{285}$
• C. $\text{P(E)}=\frac{1}{825}$
• D. $\text{P(E)}=\frac{3}{582}$

Answer 1

The correct option is C $\text{P(E)}=\frac{1}{825}$

Number of bulbs purchased $=100$
Number of defective bulbs $=4$

Probability of choosing first defective bulb is $\text{P(A)}=\frac{4}{100}=\frac{1}{25}$

Probability of choosing second defective bulb is $\text{P(B)}=\frac{3}{99}=\frac{1}{33}$

So, probablity of choosing 2 defective bulbs at random without replacing is $\text{P(A and B)}=\text{P(E)}=\text{P(A)}\times \text{P(B)}$

$\Rightarrow \text{P(E)}=\frac{1}{25}\times \frac{1}{33}=\frac{1}{825}.$