A box contains 10 bulbs, of which 3 are defective. Two bulbs are drawn at random, one after the other, without replacement. If it is known that at least one of the two bulbs is defective, then the probability that both bulbs are defective is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\displaystyle\frac{1}{8}\)
Let D be the event that both bulbs are defective and L the event that at least one bulb is defective. By the multiplication theorem:
\( \displaystyle P(D) = \frac{3}{10}\times\frac{2}{9} = \frac{1}{15} \)
L fails only when both bulbs are good:
\( \displaystyle P(\text{both good}) = \frac{7}{10}\times\frac{6}{9} = \frac{7}{15} \)
\( \displaystyle P(L) = 1-\frac{7}{15} = \frac{8}{15} \)
If both are defective, then at least one is defective, so \(D\cap L = D\). By conditional probability:
\( \displaystyle P(D|L) = \frac{P(D)}{P(L)} = \frac{1/15}{8/15} = \frac{1}{8} \)
Why the other options are incorrect
Option (1) \(\displaystyle\frac{1}{15}\) — Incorrect. This is P(D), the probability of both defective without the given condition.
Option (2) \(\displaystyle\frac{8}{15}\) — Incorrect. This is P(L), the probability of the condition itself.
Option (4) \(\displaystyle\frac{2}{9}\) — Incorrect. This is the probability that the second bulb is defective given that the first one is; the condition here is only that at least one bulb is defective.