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If 3 cards are drawn from a pack of 52 cards at random, then the probability of getting 2 cards from one suit and one card from another suit is

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$\frac{^{4} P_{2} \times^{13} C_{2} \times^{13} C_{1}}{^{52} C_{3}}$

$\frac{^{13} C_{2} \times^{13} C_{1}}{^{52} C_{3}}$

$\frac{^{4} C_{2} \times^{13} C_{2} \times^{13} C_{1}}{^{52} C_{3}}$

answer is B.

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Detailed Solution

$n (S) = 52 C_{3}$

$E =$ 2 cards are one suit and

1 card is another suit

$n (E) = 4 P_{2} \times 13 C_{2} \times 13 C_{1} P (E) = 4 P_{2} \times 13 C_{2} \times 13 C_{1} 52 C_{3}$

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