# 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.

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?

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