Four cards are drawn from a standard 52-card deck without replacement. Find the probability that exactly 3 cards are of the same suit and the remaining card is of a different suit.
What I did:
(4C1)(13C3)(3C1)(13C1) divided by (52C4)
But the final right answer is 0.375.
Can you please tell me how to solve it and why mine is wrong?
Thank you.
Best Answer
There are 4 ways to choose the suit we want three of. There are then 3 ways to choose the remaining suit (we have one of). Then there are ${13 \choose 3}$ ways to choose the three card from the first suit, and $13$ ways to choose the card from the final one.
We divide this by ${52 \choose 4}$ ways to pick four cards, so we have as our probability:
$$\frac{4 \times 3 \times {13 \choose 3} \times 13}{{52 \choose 4}} \approx 0.1648$$