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:
- A
\(x_i \) \(0\) 1 2 \(p_i\) \(\frac{144}{169}\) \(\frac{24}{169}\) \(\frac{1}{169}\) - B
\(x_i \) \(0\) 1 2 \(p_i\) \(\frac{188}{221}\) \(\frac{32}{221}\) \(\frac{1}{221}\) - C
\(x_i \) \(0\) 1 2 \(p_i\) \(\frac{124}{169}\) \(\frac{20}{169}\) \(\frac{1}{169}\) - D
\(x_i \) \(0\) 1 2 \(p_i\) \(\frac{144}{169}\) \(\frac{20}{169}\) \(\frac{1}{169}\)
Answer & Solution
Correct Answer
(A) \(x_i \) \(0\) 1 2 \(p_i\) \(\frac{144}{169}\) \(\frac{24}{169}\) \(\frac{1}{169}\)
Step-by-step Solution
Detailed explanation
\( P(\text{Ace}) = \frac{4}{52} = \frac{1}{13} \) \( P(\text{Not Ace}) = \frac{48}{52} = \frac{12}{13} \) \( P(X=0) = (\frac{12}{13})^2 = \frac{144}{169} \) \( P(X=1) = 2 \times \frac{1}{13} \times \frac{12}{13} = \frac{24}{169} \) \( P(X=2) = (\frac{1}{13})^2 = \frac{1}{169} \)…
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