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The following cards are split into three piles at random, so that every pile contains the same number of cards.  What is the probability that every pile contains an Ace?

 

Ace of spades

Ace of hearts

Ace of diamonds

Ace of clubs

Two of spades

Two of hearts

 Jun 29, 2024
 #1
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Let's consider the card labeled as follows:

Aces: \(A_1,A_2,A_3,A_4\)

Non-Aces: \(B_1,B_2 \)

The total number of ways to choose 2 cards out of 6 for the first pile is \(\binom{6}{2} \). After choosing the first pile, we choose 2 cards out of the remaining 4 for the second pile, which is \(\binom{4}{2} \). The remaining 2 cards automatically form the third pile, and this can be done in \(\binom{2}{2} \) ways.

Total ways is: \(\binom{6}{2}\times\binom{4}{2}\times\binom{2}{2}\)

But we have to divide 3!, since the order of the piles doesn't matter.

So we get \(\frac{\binom{6}{2}\times\binom{4}{2}\times\binom{2}{2}}{6}\) = 15

Next, we count the number of favorable outcomes where each pile contains exactly one Ace. Each pile must have one Ace and one non-Ace. There are 4 Aces (\(A_1,A_2,A_3,A_4\)) and 2 Non-Aces (\(B_1,B_2 \)).

The assignment of Aces to each pile can be done in 4 ways for first pile 3 ways for second 2 for final simplifying: \(4\times3\times 2\) = 24 ways

Every non-ace gets included ensuring simplified piles count as above:

\(\boxed{\frac{4}{15}}\)

smileycoolsmiley

 Jun 29, 2024

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