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
Case 1: There are two heads, two tails. The number of ways to choose which two tosses are heads is , 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, no tails. This can only happen way.
There are a total of possible configurations, giving a probability of .
Solution 2 (fastest)
We want the probability of at least two heads out of . We can do this a faster way by noticing that the probabilities are symmetric around two heads. Define as the probability of getting heads on rolls. Now our desired probability is . We can easily calculate because there are ways to get heads and tails, and there are total ways to flip these coins, giving , and plugging this in gives us . ~chrisdiamond10
See Also
2002 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 20 |
Followed by Problem 22 | |
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