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

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==Solution==
 
==Solution==
Case 1: There are two heads, two tails. There are <math>\binom{4}{2} = 6</math> to choose which two tosses are heads, and the other two must be tails.
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Case 1: There are two heads, two tails. There are <math>\binom{4}{2} = 6</math> 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 <math>\binom{4}{1} = 4</math> ways to choose which of the four tosses is a tail.
 
Case 2: There are three heads, one tail. There are <math>\binom{4}{1} = 4</math> ways to choose which of the four tosses is a tail.

Revision as of 12:21, 5 July 2021

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. There are $\binom{4}{2} = 6$ 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 $\binom{4}{1} = 4$ 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 (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

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