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CUET · MATHS · PYQ PAPER 2023

Two cards are drawn with replacement from a pack of 52 cards.
The probability distribution of number of aces is:

  1. A
    \(x_i \)\(0\)12
    \(p_i\)\(\frac{144}{169}\)\(\frac{24}{169}\)\(\frac{1}{169}\)
  2. B
    \(x_i \)\(0\)12
    \(p_i\)\(\frac{188}{221}\)\(\frac{32}{221}\)\(\frac{1}{221}\)
  3. C
    \(x_i \)\(0\)12
    \(p_i\)\(\frac{124}{169}\)\(\frac{20}{169}\)\(\frac{1}{169}\)
  4. D
    \(x_i \)\(0\)12
    \(p_i\)\(\frac{144}{169}\)\(\frac{20}{169}\)\(\frac{1}{169}\)
Verified Solution

Answer & Solution

Correct Answer

(A)

\(x_i \)\(0\)12
\(p_i\)\(\frac{144}{169}\)\(\frac{24}{169}\)\(\frac{1}{169}\)

Step-by-step Solution

Detailed explanation

\( P(\text{Ace}) = \frac{4}{52} = \frac{1}{13} \) \( P(\text{Not Ace}) = \frac{48}{52} = \frac{12}{13} \) \( P(X=0) = (\frac{12}{13})^2 = \frac{144}{169} \) \( P(X=1) = 2 \times \frac{1}{13} \times \frac{12}{13} = \frac{24}{169} \) \( P(X=2) = (\frac{1}{13})^2 = \frac{1}{169} \)…
From CUET
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