CUET Preparation Today
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: |
$\frac{1}{2}$ $\frac{1}{3}$ $\frac{1}{6}$ $\frac{1}{4}$ |
$\frac{1}{6}$ |
The correct answer is Option (3) → $\frac{1}{6}$ |
