MHT CET · Maths · Probability
Two cards are drawn simultaneously from a well shuffled pack of 52 cards. If \(X\) is the random variable of getting queens, then the value of \(2 E(X)+3 E\left(X^2\right)\) for the number of queens is
- A \(\frac{132}{221}\)
- B \(\frac{108}{221}\)
- C \(\frac{176}{221}\)
- D \(\frac{68}{221}\)
Answer & Solution
Correct Answer
(C) \(\frac{176}{221}\)
Step-by-step Solution
Detailed explanation
\( P(X=0) = \frac{\binom{4}{0}\binom{48}{2}}{\binom{52}{2}} = \frac{1 \times 1128}{1326} = \frac{1128}{1326} \) \( P(X=1) = \frac{\binom{4}{1}\binom{48}{1}}{\binom{52}{2}} = \frac{4 \times 48}{1326} = \frac{192}{1326} \)
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