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2)Three cards are drawn at random from a standard deck without replacement. What is the probability that all three cards are hearts?

3)Three cards are drawn at random from a standard deck with replacement. What is the probability that exactly two of the three cards are red?

4)Three cards are drawn at random from a standard deck without replacement. What is the probability that exactly two of the three cards are red?

5)Three cards are drawn at random from a standard deck with replacement. What is the probability that at least two of the three cards are red?

I am having trouble finding the correct answer when I am having to do the math to figure out the with replacement and without replacement. And I also don't understand the difference in the math I need to do to figure out the exactly two of the three cars and the at least two cards.

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Can you show us what you have written? –  The Substitute Nov 12 '12 at 4:51
    
For number 2 i figured you would do 13/26 x 12/25 x 11/24 –  susie q Nov 12 '12 at 5:01
    
and then for number 3 I thought youd just do 13/26x13/26x13/26 since you are replacing the card each time –  susie q Nov 12 '12 at 5:02
    
where are you getting the 26, 25, 24 for number 2? There are 52 cards in the deck, and there are 13 hearts. Therefore, the probability of drawing a heart on the first draw is $\frac{13}{52}$. –  The Substitute Nov 12 '12 at 21:49

3 Answers 3

The term without replacement means that the deck is different for each draw so the probability of getting a heart on the first draw is $\frac{13}{52}$ as there are 13 hearts in a full deck of 52 cards.

Now given you have already drawn a heart on your first draw the probability of getting a heart on your second draw is $\frac{12}{51}$ as there are now only 12 hearts left in the pack and the pack has 51 cards remaining.

So the probability of hearts on both your first and second draw is $\frac{13}{52} \cdot \frac{12}{51} = \frac{156}{2652} = \frac{1}{17}$

I'm sure you can see from here how you would calculate three hearts.

The term with replacement means that the card is put back and mixed up again after each draw so the probability of drawing a heart on the second draw is $\frac{13}{52}$ because you are still drawing from a full pack of cards.

For two hearts the probability is $\frac{13}{52} \cdot \frac{13}{52} = \frac{169}{2704} = \frac{1}{16}$ and for three hearts ...

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2) $P(\text{3 hearts})=P(\text{heart on first draw})\cdot P(\text{heart on second draw given that we drew a heart on first draw})\cdot P(\text{heart on third draw given that we drew two hearts})=\frac{13}{52}\cdot \frac{12}{51}\cdot \frac{11}{50}=\frac{11}{850}$

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Here's a related example to show how you can approach the replacement question.

  • Two cards are drawn at random from a standard deck without replacement. What is the probability that both are even?

The total number of ways to draw two cards is $C(52, 2)$ (also known as "52 choose 2"). The total number of ways to draw two cards with both even is $C(20, 2)$, since you're drawing from the 20 even cards only. So the probability is $C(20, 2) / C(52, 2) = (20 \times 19) / (52 \times 51)$, which is about $.143

  • Two cards are drawn at random from a standard deck with replacement. What is the probability that both are even?

Each time you draw a card, the probability is $1 / 2$. Each draw is independent. So the probability of doing it twice is $1/2 * 1/2 = 1 / 4$.

See the difference? In the first case you have to think about drawing a pair of cards, but in the second case the two draws are independent. In the first case, notice that if one card is even, that makes it slightly harder to get an even card the second time (since there's one fewer even cards to choose from), so the probability is slightly lower for the first case than for the second.

I hope that helps you apply the idea to red cards or hearts now.

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Thank you! It just seems so much easier when you only have two cards. Hopefully I'll get this figured out –  susie q Nov 12 '12 at 4:59
    
In your with-replacement calculation, the probability at each draw should be $20/52$, not $1/2$, to bring it in line with the without-replacement calculation. That makes the answer for doing it twice $(20/52)^2=25/169\approx0.148$. –  Barry Cipra Mar 4 at 12:45

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