2006 Alabama ARML TST Problems/Problem 3

Problem

River draws four cards from a standard 52 card deck of playing cards. Exactly 3 of them are 2’s. Find the probability River drew exactly one spade and one club from the deck.

Solution

The probability is equal two the number of successful outcomes( $S$ ) divided by the number of outcomes( $N$ ). $N=4\cdot 48$ , from the 4 ways to choose the 2's and the 48 ways to choose the other card. Now we find $S$ . From the three 2's, there must be at least one spade or club.

Case 1: One but not the other

Whether it's a spade or a club in the 2's, the probability is the same, so we must multiply by two. Now the number of ways to choose a spade but not a club is 12, since after we choose the 3 2's, we must choose a club that is not a 2. $12\cdot 2=24$ .

Case 2: Both

There are two ways to choose the third 2, and then we must choose a heart or a diamond, which there are 25 of.

Answer

Therefore, $S=24+2\cdot 25=74$ . Thus the probability of one spade and one club is $\boxed{\dfrac{37}{96}}$

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