2005 AMC 10B Problems/Problem 21
Contents
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 ?
Video Solution
https://youtu.be/wopflrvUN2c?t=252
~ pi_is_3.14
Solution 1 (where the order of drawing slips matters)
There are ways to determine which number to pick. There are ways to then draw those four slips with that number, and total ways to draw four slips. Thus .
There are ways to determine which two numbers to pick for the second probability. There are ways to arrange the order which we draw the non-equal slips, and in each order there are ways to pick the slips, so .
Hence, the answer is .
Solution 2 (order does not matter)
For probability , there are ways to choose the card you want to show up times.
Hence, the probability is .
For probability , there are ways to choose the numbers you want to show up twice. There are to pick which cards you want out of the of each.
Hence, the probability is
Hence, .
See Also
2005 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 20 |
Followed by Problem 22 | |
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All AMC 10 Problems and Solutions |
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