# What is the probability that you will NOT draw a queen from a deck of 52 cards, if you are to draw 4 cards?

if 4 cards are chosen randomly froma deck of 52 cards, there are 270,725 difference combinations. what do you think the probability is that out of these sets of 4 cards, you will NOT get a queen

Update:

the chances of getting 4 aces, are...

1 x 1 x 1 x 1 / 270,725.

there are 4 queens in the deck of cards, is it logical to say that..... if the chances of getting 4 queens in one draw are 1 / 270,725 that the chances of NOT drawing this combination would be 270,724. But that wouldn't make sense because you can get King, King, King, Queen. 3 Kings, 1 queen. tough question..

Update 2:

the correct answer has been established.

(48/52)*(47/51)*(46/50)*(45/49) = 38916/54145

or a 71.87 percent chance.

Relevance
• Merlyn
Lv 7

Let X be the number of queens drawn. X has the hypergeometric distribution with the following parameters.

K = number of items to be drawn = 4

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}

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 Probability Mass Function, PDF,

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

P(X = 0 ) = 0.7187367 ← no queens ← ANSWER

P(X = 1 ) = 0.2555508 ← one queen

P(X = 2 ) = 0.02499954 ←two queens

P(X = 3 ) = 0.0007092068 ← three queens

P(X = 4 ) = 3.693785e-06 ← four queens

There are two ways to solve this:

a. There is a 48/52 probability that the first card is not a Queen. If the first card is not a Queen, then there is a 47/51 probability that the second card is not also not a Queen. If the first and second cards are both not Queens, then there is a 46/50 probability that the third card is not also not a Queen. If the first, second, and third cards are all not Queens, then there is a 45/49 probability that the 4th card is not a Queen. Therefore, the probability that none of the first four cards chosen at random is a Queen is (48/52)*(47/51)*(46/50)*(45/49).

b. Compute how possible combinations of four cards can be drawn from a deck of 48 different cards. This is the number of possible combinations of cards that include no Queens. Divide this number by the total number of possible combinations of cards (which you computed to be 270,725).

• Pearl
Lv 4
4 years ago

Since there are 4 queens and 52 cards, the chances of drawing a queen every time of your 4 draws would be P = (1/52)(1/51)(1/50)(1/49).

1 in 14

1 in 4 chance

1out of 13 cards drawn should draw a queen or any other card for that matter.So it is 7.69% to draw one and 92.31% that you don't

sorry . wcowell got it right also.Also there are 52 cards and 4 into 52 is 13 and not 14 as Mary Jo, responded.

Source(s): Majored in math.

Approx 92%