Difference between revisions of "2006 Alabama ARML TST Problems/Problem 3"
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| − | The probability is equal | + | The probability is equal to the number of successful outcomes(<math>S</math>) divided by the number of outcomes(<math>N</math>). <math>N=4\cdot 48</math>, from the 4 ways to choose the 2's and the 48 ways to choose the other card. Now we find <math>S</math>. From the three 2's, there must be at least one spade or club. |
===Case 1: One but not the other=== | ===Case 1: One but not the other=== | ||
Revision as of 00:14, 12 December 2011
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 25 of.
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 | ||
