# Answer to Question #371 in Statistics and Probability for mary

Question #371

The deck of cards (36 pieces) is randomly divided into 2 equal packs. What is the probability that:

a) there will be two aces in each pack;

b) there will be all four aces in one pack;

c) there will be equal amount of red cards in each pack?

a) there will be two aces in each pack;

b) there will be all four aces in one pack;

c) there will be equal amount of red cards in each pack?

Expert's answer

The number of ways to select 18 cards (half of the deck) out of 36 is: N = (С36)^18 = 36!/(18!*18!).

a) Total amount of aces in the deck is 4, the amount of "not aces" is 32.

The "favorable" case is when one half of the deck (18 cards) has got two aces and 16 "not aces".

The number of ways that you can choose 2 out of 4 aces is (C4)^2 = 4!/(2!*2!).

The number of ways that you can select the 16 "not aces" out of 32 cards is (C32)^16 = 32!/(16!*16!).

The number of ways you can achieve "favorable" cards is: m = (C4)^2*(C32)^16 = 4!*32!/(2!*2!*16!*16!).

The probability of “favorable” cards is: P = m/N = ((C4)^2*(C32)^16)/(С36)^18 = 4!*32!*18!*18!/(2!*2!*16!*16!*36!) ≈ 0.795.

b) The "favorable" case is when one half of the deck (18 cards) has got 4 aces and 14 "not aces".

The number of ways that you can choose all 4 aces is (C4)^4=1.

The number of ways that you can select the 14 "not aces" out of 32 cards is (C32)^14=32!/(14!*18!).

The number of ways you can achieve "favorable" cards is: m = (C4)^4*(C32)^14 = 1*32!/(14!*18!).

The probability of “favorable” cards is: P = m/N = ((C4)^4*(C32)^14)/(С36)^18 = 1*32!*18!*18!/(14!*18!*36!) ≈ 0.052.

c) Total amount of red cards in the deck is 18, the amount of black cards is also 18.

The "favorable" case is when one half of the deck (18 cards) has got 9 red and 9 black cards.

The number of ways that you can choose 9 out of 18 red cards is (C18)^9=18!/(9!*9!).

The number of ways that you can choose 9 out of 18 black cards is (C18)^9=18!/(9!*9!).

The number of ways you can achieve "favorable" cards is: m = (C18)^9*(C18)^9 = 18!*18!/(9!*9!*9!*9!).

The probability of “favorable” cards is: P = m/N = ((C18)^9*(C18)^9)/(С36)^18 = 18!*18!* 18!*18!/ (9!*9!*9!*9!*36!) ≈ 0.26.

a) Total amount of aces in the deck is 4, the amount of "not aces" is 32.

The "favorable" case is when one half of the deck (18 cards) has got two aces and 16 "not aces".

The number of ways that you can choose 2 out of 4 aces is (C4)^2 = 4!/(2!*2!).

The number of ways that you can select the 16 "not aces" out of 32 cards is (C32)^16 = 32!/(16!*16!).

The number of ways you can achieve "favorable" cards is: m = (C4)^2*(C32)^16 = 4!*32!/(2!*2!*16!*16!).

The probability of “favorable” cards is: P = m/N = ((C4)^2*(C32)^16)/(С36)^18 = 4!*32!*18!*18!/(2!*2!*16!*16!*36!) ≈ 0.795.

b) The "favorable" case is when one half of the deck (18 cards) has got 4 aces and 14 "not aces".

The number of ways that you can choose all 4 aces is (C4)^4=1.

The number of ways that you can select the 14 "not aces" out of 32 cards is (C32)^14=32!/(14!*18!).

The number of ways you can achieve "favorable" cards is: m = (C4)^4*(C32)^14 = 1*32!/(14!*18!).

The probability of “favorable” cards is: P = m/N = ((C4)^4*(C32)^14)/(С36)^18 = 1*32!*18!*18!/(14!*18!*36!) ≈ 0.052.

c) Total amount of red cards in the deck is 18, the amount of black cards is also 18.

The "favorable" case is when one half of the deck (18 cards) has got 9 red and 9 black cards.

The number of ways that you can choose 9 out of 18 red cards is (C18)^9=18!/(9!*9!).

The number of ways that you can choose 9 out of 18 black cards is (C18)^9=18!/(9!*9!).

The number of ways you can achieve "favorable" cards is: m = (C18)^9*(C18)^9 = 18!*18!/(9!*9!*9!*9!).

The probability of “favorable” cards is: P = m/N = ((C18)^9*(C18)^9)/(С36)^18 = 18!*18!* 18!*18!/ (9!*9!*9!*9!*36!) ≈ 0.26.

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