Problem:
Forty slips are placed into a hat, each bearing a number , or , with each number entered on four slips. Four slips are drawn from the hat at random and without replacement. Let be the probability that all four slips bear the same number. Let be the probability that two of the slips bear a number and the other two bear a number . What is the value of
Answer Choices:
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Solution:
The total number of ways that the numbers can be chosen is . Exactly of these possibilities result in the four slips having the same number.
Now we need to determine the number of ways that two slips can have a number and the other two slips have a number , with . There are ways to choose the distinct numbers and . For each value of there are to choose the two slips with and for each value of there are to choose the two slips with . Hence the number of ways that two slips have some number and the other two slips have some distinct number is
So the probabilities and are and , respectively, which implies that
Answer: .
The problems on this page are the property of the MAA's American Mathematics Competitions