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

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==Solution==
 
==Solution==
Case 1: There are two heads, two tails. The number of ways to choose which two tosses are tails is <math>_4 C _2 = 6</math>, and the other two must be tails.
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Case 1: There are two heads, two tails. The number of ways to choose which two tosses are heads is <math>_4 C _2 = 6</math>, and the other two must be tails.
  
 
Case 2: There are three heads, one tail. There are <math>_4 C _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>_4 C _1 = 4</math> ways to choose which of the four tosses is a tail.

Revision as of 12:08, 26 December 2013

Problem

Harold tosses a nickel 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 $\frac{6+4+1}{16} = \boxed{\text{(E)}\ \frac{11}{16}}$.

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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