MHT CET · Maths · Probability

Two cards are drawn successively with replacement from a well- shuffled pack of 52. cards. Let X denote the random variable of number of kings obtained in the two drawn cards. Then \(\mathrm{P}(x=1)+\mathrm{P}(x=2)\) equals

A \(\frac{49}{169}\)
B \(\frac{24}{169}\)
C \(\frac{52}{169}\)
D \(\frac{25}{169}\)

Verified Solution

Answer & Solution

Correct Answer

(D) \(\frac{25}{169}\)

Step-by-step Solution

Detailed explanation

Probability of getting an king card is \(\frac{4}{52}\)
For \(\mathrm{X}=1\), the outcome of king can be either in first draw of the second draw.
\(\begin{aligned}
& \therefore \quad \mathrm{P}(\mathrm{X}=1)=\frac{4}{52} \times \frac{48}{52}+\frac{48}{52} \times \frac{4}{52} \\
& =2 \times \frac{4}{52} \times \frac{48}{52}=\frac{24}{169} \\
& \therefore \quad \mathrm{P}(\mathrm{X}=2)=\frac{4}{52} \times \frac{4}{52}=\frac{1}{169} \\
& \therefore \quad \mathrm{P}(\mathrm{X}=1)+\mathrm{P}(\mathrm{X}=2)=\frac{24}{109}+\frac{1}{169}=\frac{25}{169}
\end{aligned}\)

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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