2002 AMC 8 Problems/Problem 21
Problem
Harold tosses a coin four times. The probability that he gets at least as many heads as tails is
Solution 1
Case 1: There are two heads and two tails. There are ways to choose which two tosses are heads, and the other two must be tails.
Case 2: There are three heads, one tail. There are ways to choose which of the four tosses is a tail.
Case 3: There are four heads and no tails. This can only happen way.
There are a total of possible configurations, giving a probability of .
Video Solution
https://www.youtube.com/watch?v=4vLTPszBLeg ~David
Video Solution by WhyMath
See Also
2002 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 20 |
Followed by Problem 22 | |
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