2002 AMC 8 Problems/Problem 21

Revision as of 11:39, 28 December 2020 by Chrisdiamond10 (talk | contribs)

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

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 $P(2)$ similarly as above solution, so plugging this in gives us $\boxed{\text{(E)}\ \frac{11}{16}}$. ~chrisdiamond10

See Also

2002 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 20
Followed by
Problem 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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png