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

Revision as of 09:01, 14 June 2024

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 1

Case 1: There are two heads and 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 and 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 $P(2)$ 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

Video Solution

https://www.youtube.com/watch?v=4vLTPszBLeg ~David

2002 AMC 8 (Problems • Answer Key • Resources)
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.

@@ Line 18: / Line 18: @@
 We can easily calculate <math>P(2)</math> because there are <math>\binom{4}{2} = 6</math> ways to get <math>2</math> heads and <math>2</math> tails, and there are <math>2^4=16</math> total ways to flip these coins, giving <math>P(2)=\frac{6}{16}=\frac{3}{8}</math>, and plugging this in gives us <math>\boxed{\text{(E)}\ \frac{11}{16}}</math>.
 ~chrisdiamond10
-==Solution 3==
-You can use complimentary counting to solve this problem. Because we are trying to figure out the probability that Harold gets at least as many heads as tails, when using complimentary counting, we want to find the probability that there are MORE tails than heads. That are the cases when there are <math>3</math> tails and <math>1</math> head, and all tails. For the first case (<math>3</math> tails and <math>1</math> head), there are <math>4</math> cases that it works.
--<math>HTTT</math>
--<math>THTT</math>
--<math>TTHT</math>
--<math>TTTH</math>
-For the second case (all tails), there is <math>1</math> case that works. Which is, of course <math>TTTT</math>.
-So in total, there is <math>4+1=5</math> cases that there are more tails then heads out of <math>2^4=16</math> possibilities. Since we are doing complimentary counting, we need to subtract <math>5/16</math> from one. Hence the answer is <math>\boxed{\text{(E)}\ \frac{11}{16}}</math>.
 ==Video Solution==

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