Difference between revisions of "1995 AJHSME Problems/Problem 20"
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<math>\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}</math> | <math>\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}</math> | ||
− | ==Solution== | + | ==Solution 1== |
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> | ||
+ | |||
+ | ==Solution 2== | ||
+ | |||
+ | We can use simple casework to solve this problem too. There are six cases based on Apollo's Roll. | ||
+ | Apollo Rolls a 1: Diana could roll a <math>2</math>, <math>3</math>, <math>4</math>, <math>5</math>, or <math>6</math>. | ||
+ | Apollo Rolls a 2: Diana could roll a <math>3</math>, <math>4</math>, <math>5</math>, or <math>6</math>. | ||
+ | Apollo Rolls a 3: Diana could roll a <math>4</math>, <math>5</math>, or <math>6</math>. | ||
+ | Apollo Rolls a 4: Diana could roll a <math>5</math> or <math>6</math>. | ||
+ | Apollo Rolls a 5: Diana could roll a <math>6</math>. | ||
+ | Apollo Rolls a 6: There are no successful outcomes. | ||
+ | The total amount of successful cases is <math>5+4+3+2+1 = 15</math>. The total amount of possible cases is <math>6(6) = 36</math>. Therefore, the probability of Diana rolling a bigger number is <math>\frac{15}{36} = \frac{5}{12} \Rightarrow \mathrm{(B)}</math> | ||
==See Also== | ==See Also== | ||
{{AJHSME box|year=1995|num-b=19|num-a=21}} | {{AJHSME box|year=1995|num-b=19|num-a=21}} |
Latest revision as of 23:28, 11 February 2021
Contents
Problem
Diana and Apollo each roll a standard die obtaining a number at random from to . What is the probability that Diana's number is larger than Apollo's number?
Solution 1
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 , then Apollo=Diana. Now all we need to do is find Apollo=Diana. There are possibilities total, and 6 of those have Apollo=Diana, so Apollo=Diana. Going back to our first equation and solving for D, we get
Solution 2
We can use simple casework to solve this problem too. There are six cases based on Apollo's Roll. Apollo Rolls a 1: Diana could roll a , , , , or . Apollo Rolls a 2: Diana could roll a , , , or . Apollo Rolls a 3: Diana could roll a , , or . Apollo Rolls a 4: Diana could roll a or . Apollo Rolls a 5: Diana could roll a . Apollo Rolls a 6: There are no successful outcomes. The total amount of successful cases is . The total amount of possible cases is . Therefore, the probability of Diana rolling a bigger number is
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 |