# What is the probability of drawing at least one ace on two draws out of a deck of 52 cards, no replacement.?

I posted this once and made a mistake, but people keep answering it over and over with the same answer to the wrong question. So here it is in a new thread.

Again, I suspect the answer is P(A) + P(B) - P(A and B). Which would be 4/52 + 4/51 - 3/51.

Thanks

### 5 Answers

- Clive GLv 510 years agoBest Answer
P(draw ace) = 4/52*48/51 + 48/52*4/51 = 384/2,652

P(draw 2 aces) = 4/52*3/51 = 12/2,652

P(draw 1 or 2 aces) = P(draw ace) + P(draw 2 aces)

= 384/2,652 + 12/2,652

= 396/2,652

= 0.1493

OR

P(draw 0 aces)

= 48/52*47/51

= 2,256/2,652

= 0.8507

P(draw 1 or 2 aces)

= 1 - P(draw 0 aces)

= 1 - 0.8507

= 0.1493

- Mike TLv 710 years ago
The first card drawn has a probability of 4/52 = 1/13, 4 aces out of 52 cards. The probability of not getting an ace is 48/52 = 12/13.

If the first draw is successful, then you can darw any card and be successfully. The probability is 53/53 = 1. Because the two draws are independent, you multiple 1/13 * 1 = 1/13.

If the first draw is unsuccessful, then you need to draw and ace. The probability is 4/51. Again, multuply

12/13 * 4/51 = 48/663

Add the probabilities to get the total probability = 1/13 + 48/663 = 51/663 + 48/663 = 99/663 = 33/221

The other way to think about it is the probability of success is one nminus the probability of failure.

1 - 48/52 * 47/51 = 1 - 12/13 * 47/51 = 1 - 564/663 = 99/663 = 33/221

- MerlynLv 710 years ago
You can use the hypergeometric distribution to find the solution

Let X be the number of aces when drawing two cards from a standard deck of cards. X has the hypergeometric distribution with the following parameters.

K = number of items to be drawn = 2

N = total objects = 52

M = number of objects of a given type = 4

The probability mass function for the hypergeometric distribution is defined as:

P(X = x | N, M, K) = ( M C x ) * ( (N - M) C (K - x) ) / ( N C K )

for x = {0, ..., K}; M - (N - K) ≤ x ≤ K

P(X = 0 | N, M, K) = 0 otherwise

Note that the constraints on x here are very generic and it is possible to have value of K, N and M such that for x in {0, ..., K} P(X = x) = 0.

If you have n objects and chose r of them, the number of combinations is:

n! / ( r! (n-r)! )

this can be written as nCr

the N C K is the total number of possible combinations of K objects drawn from N objects.

the M C x is the number of combinations of getting x objects of the given type

the (N - M) C ( K - x) is the number of combinations of non typed objects to be drawn.

Looking at the PMF you should be able to see that it is the ratio of the number of combination of selecting the X of the items of interest times the number of combinations of choosing K - X items from the remaining items and this is all divided by the total number of combination for choosing K items from N objects.

The expectations of the Hypergeometric distribution is KM / N = 0.1538462

The variance is K (M/N)(1-M/N)(N-K)/(N-1) = 0.1392273

The std dev is the sqrt of the variance: 0.3731317

The Probability Mass Function, PMF,

f(X) = P(X = x) is:

P(X = 0 ) = 0.8506787

P(X = 1 ) = 0.1447964

P(X = 2 ) = 0.004524887

P(X ≥ 1) = P(X = 1) + P(X = 2) = 0.1447964 + 0.004524887 = 0.1493213

P(X ≥ 1) = 1 - P(X = 0) = 1 - 0.8506787 = 0.1493213

- 10 years ago
List all the possibilities and add them togerther

First possibility is:

P(ace and ace):

= P(ace) x P (ace|ace)

=(4/52)(3/51) -----> the first ace is 4/52, then the second time you draw a card, there's only 51 left so its 51 cards and choosing from 3 aces)

=12/2652

Second:

P(ace and not ace)

=P(ace) x P(not ace | ace)

=(4/52)(48/51)--->not ace is 52 - 4 cards you can choose from.

Third:

P(not ace and ace)

=P(not ace) x P(ace|not ace)

=(48/52)(4/51)

Add one, two and three togerther. Try to see the logic behind it. if you can't then probably i did it wrong..

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- sawczakLv 43 years ago
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