probability

mathe25

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Oct 13, 2011
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im not very good at english but i need help.

take 4 cards from a deck of 52
a) without replacement
b) with replacement

i have to find the probability that those 4 cards are 2 ace and 2 king.
 
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im not very good at english but i need help.

By (package) bundles with 32 letters(cards) are issued 4 letters(cards) of
a) no return
b) with return
i have to find the probability that those 4 cards are 2 ace and 2 king.
I think you mean that you take 4 cards from a deck of 32
a) without replacement
b) with replacement

A standard pinochle has 48 cards so I am going to assume that is what you mean, of which 8 are aces and 8 are kings.

The easier problem is (b) since at each draw we have 8 aces and 8 kings out of 48 cards. That means that the probablity of an ace on any one draw is 8/48= 1/6 and the probability of a king on any one draw is 1/6. The probability of "two kings and two aces" in any one order is \(\displaystyle (1/6)(1/6)(1/6)(1/6)= 1/6^4= 1/1296\). But there are \(\displaystyle \frac{4!}{2!2!}= 6\) different orders. The probability of "two kings and two aces" in any order is \(\displaystyle 6/6^4= 1/6^3= 1/216.

"Without replacement" is harder because the situation changes on every draw. For example, there are initially 8 ace or 8 kings in 48 cards. The probability that the first card is a king is 8/48. If the first card drawn is a king, there are 7 kings left out of 47 so the probability the second card drawn is a king is 7/47. There are now 46 cards left with 8 aces. The probability the third card drawn is an ace is 8/46. Then there are 45 cards left, 7 of which are aces. The probablility the fourth card drawn is an ace is 7/45. So the probability of "two kings, then two aces" in that order is (8/48)(7/47)(8/46)(7/45). Now start again. The probability the first card drawn is an ace is 8/48, if the first card drawn is an ace, there are 8 kings in the remaining 47 cards- the probability the second card drawn is a king is 8/47, the probability the third card drawn is an ace is 7/46, and the probability the fourth card is an king is 7/45 so the probability of "ace, king, ace, king" is (8/48)(8/47)(7/46)(7/45). Those are not the same fractions as before but the numerators are all the same and the denominators are all the same so the product is the same. We just need to multiply (8/48)(7/47)(8/46)(7/45) by the number of orders which is again \(\displaystyle \frac{4!}{2!2!}= 6\). The probability is 6(8/48)(7/47)(8/46)(7/45).

(The original post has been edited to change the "48 card deck" to "52 cards".)\)
 
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Hello, mathe25!

Take 4 cards from a deck of 52
. . a) without replacement
. . b) with replacement

Find the probability that those 4 cards are 2 Aces and 2 Kings.

(a) Without replacement

There are \(\displaystyle {52\choose4} \:=\:\frac{52!}{4!\,48!} \:=\:270,\!725 \) possible outcomes.

. . There are \(\displaystyle {4\choose2}{4\choose2} \:=\:36\) ways to get 2 Aces and 2 Kings.

Therefore: .\(\displaystyle P(\text{2 A's, 2 K's}) \:=\:\dfrac{36}{270,\!725}\)


(b) With replacement

There are \(\displaystyle {4\choose2}\,=\,6 \) orders for 2 Aces and 2 Kings.

Then: .\(\displaystyle P(\text{Ace}) \,=\, \frac{4}{52} \,=\,\frac{1}{13};\;\;P(\text{King}) \,=\,\frac{1}{13}\)

Therefore: .\(\displaystyle P(\text{2 A's, 2 K's}) \:=\:6\left(\frac{1}{13}\right)^2\left(\frac{1}{13}\right)^2 \:=\: \dfrac{6}{28,\!561} \)
 
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