2006 Alabama ARML TST Problems/Problem 3


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.


The probability is equal to 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 24 of (We can't choose another 2 of Hearts or Diamonds because that would give us four 2's instead of three).


Therefore, $S=24+2\cdot 24=72$. Thus the probability of one spade and one club is $\boxed{\dfrac{3}{8}}$

See also

2006 Alabama ARML TST (Problems)
Preceded by:
Problem 2
Followed by:
Problem 4
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