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CUET · MATHS · PYQ PAPER 2025

Two cards are drawn successively with replacement from a well-shuffled deck of 52 cards.
The probability distribution of number of aces is given by :

  1. A
    \(x\)\(0\)12
    \(P(x)\)\(\frac{24}{169}\)\(\frac{1}{169}\)\(\frac{144}{169}\)
  2. B
    \(x\)\(0\)12
    \(P(x)\)\(\frac{24}{169}\)\(\frac{144}{169}\)\(\frac{1}{169}\)
  3. C
    \(x\)\(0\)12
    \(P(x)\)\(\frac{144}{169}\)\(\frac{24}{169}\)\(\frac{1}{169}\)
  4. D
    \(x\)\(0\)12
    \(P(x)\)\(\frac{1}{169}\)\(\frac{144}{169}\)\(\frac{24}{169}\)
Verified Solution

Answer & Solution

Correct Answer

(C)

\(x\)\(0\)12
\(P(x)\)\(\frac{144}{169}\)\(\frac{24}{169}\)\(\frac{1}{169}\)

Step-by-step Solution

Detailed explanation

\(P(\text{ace}) = \frac{4}{52} = \frac{1}{13}\) \(P(\text{non-ace}) = \frac{48}{52} = \frac{12}{13}\) \(P(x=0) = P(\text{non-ace})^2 = (\frac{12}{13})^2 = \frac{144}{169}\)…
From CUET
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