Question

There is a box containing $30$ bulbs of which $5$ are defective. If two bulbs are chosen at random from the box in succession without replacing the first, what is the probability that both the bulbs chosen are defective?

Answer 1

Given Number of defective bulbs is $5$

Total Number of bulbs is $30$

Let $A$ be the event of picking defective bulb in the first trial

Probability of event $A$ to happen $P\left(A\right)=\frac{5}{30}=\frac{1}{6}$

After picking one defective bulb in the first trial, as the bulb is not replaced

Total Number of bulbs is $29$

Number of defective bulbs is $4$

Let $B$ be the event of picking defective bulb in the second trial

Probability of event $B$ to happen $P\left(B\right)=\frac{4}{29}$

As the two trials does not depend on each other in our case

$P(A\cap B)=P\left(A\right)\times P\left(B\right)=\frac{1}{6}\times \frac{4}{29}=\frac{2}{87}$

Probability that the both bulbs chosen are defective is $\frac{2}{87}$