There are 4 bulbs out of which 2 are defective. Each bulb is tested in random order till both the defective bulbs are identified. The probability that only two tests are required to identify the defective bulb is:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → $\frac{1}{6}$
P (first defective) = $\frac{2}{4}=\frac{1}{2}$
P (second defective | first defective) = $\frac{1}{3}$
P (Both defect in 2 test) = $\frac{1}{2}×\frac{1}{3}=\frac{1}{6}$