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A deck of cards includes 40 different cards. There are 8 cards in each of 5 suits. The cards are shuffled and a player receives 3 (different) cards.

The probability that exactly 2 of these cards have the same suit is in

$(A) (0.46, 0.48].\\ (B) (0.44, 0.46].\\(C) (0.48, 0.50].\\(D) (A) (C) false.$

So, my approach is:

$P(\text{ 2 out of 3 are same suit })=3 P(\text{ picking card from any suit })P(\text{ picking card from the suit of the first card })P(\text{ pciking card of a different suit than the first 2 })=3(1)\left(\frac{7}{39}\right)\left(\frac{32}{38}\right)=3(0.151)=0.453$

So $B$ is correct?

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2 Answers 2

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Yes your answer is correct.

Here is another way to look at it. There are $5$ suits and there are exactly $2$ cards of the same suit. So we first choose the suit that the player gets two cards of. That is $5 \choose 1$. Now we choose $2$ cards from $8$ cards of that suit and the third card from remaining $32$ cards of other suits.

So the probability is,

$\displaystyle P = {5 \choose 1} {8 \choose 2} {32 \choose 1} \Big / {40 \choose 3} = \frac{7\cdot3\cdot32}{39\cdot38} = \frac{112}{247}$

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I got B, with P = 112/247, by going through the scenario one card at a time. The suit of the first card is irrelevant. The second card has a chance of p = 7/39 of matching the first card's suit. From there, we want to subtract the chance of the third card also having that suit: (7/39) - (7/39)(6/38) = 112/741 If instead, the second card has a different suit, with p = 32/39, the third card has two equal probabilities for matching the first card or the second card: (32/39)(7/38) + (32/39)(7/38) = 224/741 Add that to the previous probability to get 112/247.

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