# Difference between revisions of "2002 AMC 8 Problems/Problem 21"

## Problem

Harold tosses me 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 $_4 C _2 = 6$, and the other two must be tails.

Case 2: There are three heads, one tail. There are $_4 C _1 = 4$ 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 $\frac{6+4+1}{16} = \boxed{\text{(E)}\ \frac{11}{16}}$.