Difference between revisions of "2006 Alabama ARML TST Problems/Problem 3"
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===Case 2: Both=== | ===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 | + | 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=== | ===Answer=== | ||
| − | Therefore, <math>S=24+2\cdot | + | Therefore, <math>S=24+2\cdot 24=72</math>. Thus the probability of one spade and one club is <math>\boxed{\dfrac{3}{8}}</math> |
==See also== | ==See also== | ||
{{ARML box|year=2006|state=Alabama|num-b=2|num-a=4}} | {{ARML box|year=2006|state=Alabama|num-b=2|num-a=4}} | ||
Latest revision as of 18:43, 12 April 2012
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 | ||
