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Difference between revisions of "1995 AJHSME Problems/Problem 20"

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Note that the probability of Diana rolling a number larger than Apollo's is the same as the probability of Apollo's being more than Diana's. If we denote this common probability <math>D</math>, then <math>2D+P(</math>Apollo=Diana<math>)=1</math>. Now all we need to do is find <math>P(</math>Apollo=Diana<math>)</math>. There are <math>6(6)=36</math> possibilities total, and 6 of those have Apollo=Diana, so <math>P(</math>Apollo=Diana<math>)=\frac{6}{36}=\frac{1}{6}</math>. Going back to our first equation and solving for D, we get <cmath>2D+\frac{1}{6}=1</cmath> <cmath>2D=\frac{5}{6}</cmath> <cmath>D=\frac{5}{12} \Rightarrow \mathrm{(B)}</cmath>
 
Note that the probability of Diana rolling a number larger than Apollo's is the same as the probability of Apollo's being more than Diana's. If we denote this common probability <math>D</math>, then <math>2D+P(</math>Apollo=Diana<math>)=1</math>. Now all we need to do is find <math>P(</math>Apollo=Diana<math>)</math>. There are <math>6(6)=36</math> possibilities total, and 6 of those have Apollo=Diana, so <math>P(</math>Apollo=Diana<math>)=\frac{6}{36}=\frac{1}{6}</math>. Going back to our first equation and solving for D, we get <cmath>2D+\frac{1}{6}=1</cmath> <cmath>2D=\frac{5}{6}</cmath> <cmath>D=\frac{5}{12} \Rightarrow \mathrm{(B)}</cmath>
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==See Also==
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{{AJHSME box|year=1995|num-b=19|num-a=21}}

Revision as of 02:20, 23 December 2012

Problem

Diana and Apollo each roll a standard die obtaining a number at random from $1$ to $6$. What is the probability that Diana's number is larger than Apollo's number?

$\text{(A)}\ \dfrac{1}{3} \qquad \text{(B)}\ \dfrac{5}{12} \qquad \text{(C)}\ \dfrac{4}{9} \qquad \text{(D)}\ \dfrac{17}{36} \qquad \text{(E)}\ \dfrac{1}{2}$

Solution

Note that the probability of Diana rolling a number larger than Apollo's is the same as the probability of Apollo's being more than Diana's. If we denote this common probability $D$, then $2D+P($Apollo=Diana$)=1$. Now all we need to do is find $P($Apollo=Diana$)$. There are $6(6)=36$ possibilities total, and 6 of those have Apollo=Diana, so $P($Apollo=Diana$)=\frac{6}{36}=\frac{1}{6}$. Going back to our first equation and solving for D, we get \[2D+\frac{1}{6}=1\] \[2D=\frac{5}{6}\] \[D=\frac{5}{12} \Rightarrow \mathrm{(B)}\]

See Also

1995 AJHSME (Problems • Answer Key • Resources)
Preceded by
Problem 19 		Followed by
Problem 21
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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