# 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 $\text{(A)}\ \frac{5}{16}\qquad\text{(B)}\ \frac{3}{8}\qquad\text{(C)}\ \frac{1}{2}\qquad\text{(D)}\ \frac{5}{8}\qquad\text{(E)}\ \frac{11}{16}$

## Solution

Case 1: There are two heads, two tails. The number of ways to choose which two tosses are heads is $\binom{4}{2} = 6$, and the other two must be tails.

Case 2: There are three heads, one tail. There are $\binom{4}{1} = 4$ ways to choose which of the four tosses is a tail.

Case 3: There are four heads, no tails. This can only happen $1$ way.

There are a total of $2^4=16$ possible configurations, giving a probability of $\frac{6+4+1}{16} = \boxed{\text{(E)}\ \frac{11}{16}}$.

## Solution 2 (fastest)

We want the probability of at least two heads out of $4$. We can do this a faster way by noticing that the probabilities are symmetric around two heads. Define $P(n)$ as the probability of getting $n$ heads on $4$ rolls. Now our desired probability is $\frac{1-P(2)}{2} +P(2)$. We can easily calculate $P(2)$ because there are $\binom{4}{2} = 6$ ways to get $2$ heads and $2$ tails, and there are $2^4=16$ total ways to flip these coins, giving $P(2)=\frac{6}{16}=\frac{3}{8}$, and plugging this in gives us $\boxed{\text{(E)}\ \frac{11}{16}}$. ~chrisdiamond10

## See Also

 2002 AMC 8 (Problems • Answer Key • Resources) Preceded byProblem 20 Followed byProblem 22 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AJHSME/AMC 8 Problems and Solutions

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