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

Revision as of 19:23, 23 December 2012

Harold tosses a nickel four times. The probability that he gets at least as many heads as tails is

$\text{(A)}\ \frac{5}{16}\qquad\text{(B)}\ \frac{3}{8}\qquad\text{(C)}\ \frac{1}{2}\qquad\text{(D)}\ \frac{5}{8}\qquad\text{(E)}\ \frac{11}{16}$

Case 1: There are two heads, two tails. The number of ways to choose which two tosses are tails 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 $\frac{6+4+1}{16} = \boxed{\text{(E)}\ \frac{11}{16}}$ .

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

@@ Line 1: / Line 1: @@
+==Problem==
 Harold tosses a nickel four times. The probability that he gets at least as many heads as tails is
 <math> \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} </math>
+==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.
+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 3: There are four heads, no tails. This can only happen <math>1</math> way.
+There are a total of <math>2^4=16</math> possible configurations, giving a probability of <math>\frac{6+4+1}{16} = \boxed{\text{(E)}\ \frac{11}{16}}</math>.
+==See Also==
+{{AMC8 box|year=2002|num-b=20|num-a=22}}