CUET · MATHS · PYQ PAPER 2023

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 bulbs is:

A \(\frac{1}{2}\)
B \(\frac{1}{3}\)
C \(\frac{1}{6}\)
D \(\frac{1}{4}\)

Verified Solution

Answer & Solution

Correct Answer

(C) \(\frac{1}{6}\)

Step-by-step Solution

Detailed explanation

\(P(\text{first is D}) = \frac{2}{4}\) \(P(\text{second is D | first is D}) = \frac{1}{3}\) \(P(\text{both are D}) = \frac{2}{4} \times \frac{1}{3} = \frac{1}{6}\)

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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X	0	1	2
\(P ( X )\)	\(1 / 4\)	\(1 / 2\)	\(1 / 4\)