You randomly select a card from a 52-card deck. Find the probability of selecting a black seven OR a black three.

Question

matt101 · Answer

There are two black sevens (clubs or spades) and two black threes (clubs or spades). The probability of selecting a specific card from the deck is 1/52. The probability of selecting a second specific card right after it is 1/51 (since you've removed one of the cards. Therefore, the probability of selecting two specific cards in a row is (1/52)(1/51).

The probability of getting two cards in two ways is (1/52)(1/51) + (1/52)(1/51).

anonymous · Answer

I am still a little confused..

matt101 · Answer

What don't you get?

anonymous · Answer

I guess I am not understanding the OR part

matt101 · Answer

Ok so you understand first how I got (1/52)(1/51) for selecting two specific cards in a row?

anonymous · Answer

Yes

matt101 · Answer

So the OR means one situation OR the other. The probability of selecting any two cards in a row is (1/52)(1/51). But your saying its OK for there to be two different pairs selected - either the threes or the sevens. Since you're allowing two combinations, the probability is doubled, i.e. (1/52)(1/51) + (1/52)(1/51) or more simply 2*(1/52)(1/51)

anonymous · Answer

@matt101: I gave you a medal, but  your answer doesn't seem correct to me now.

matt101 · Answer

Haha actually you're right I misread the question. You're not selecting two cards in a row...this is actually much easier than the question I thought it was.

anonymous · Answer

It seems $ \frac 1 {13} $ is the right answer.

matt101 · Answer

It would be 1/13 if you were only selecting one card. However, because you're selecting two, your probability of getting the card you want is greater.

matt101 · Answer

Let me revise my answer...

There are two black sevens and two black threes...four cards possible that you want. There are also four scenarios if you draw twice: you get a card you want each time, you get a card you want the first time but not the second, you get a card you want the second time but not the first, or you don't get a card you want either time.

This means the probability of getting at least one card you want is 1 minus the probability you don't get any of the cards you want either time. Since there are 4/52 cards you do want, there are 48/52 cards you don't want. After you draw one card, the probability of again getting a card you don't want is 47/51.

1 - (48/52)(47/51) is the answer, whatever it works out to. That number is greater than 1/13 too.