2006 Alabama ARML TST Problems/Problem 3
Contents
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 to the number of successful outcomes() divided by the number of outcomes(). , from the 4 ways to choose the 2's and the 48 ways to choose the other card. Now we find . 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. .
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).
Answer
Therefore, . Thus the probability of one spade and one club is
See also
2006 Alabama ARML TST (Problems) | ||
Preceded by: Problem 2 |
Followed by: Problem 4 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 |