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Difference between revisions of "1997 AJHSME Problems/Problem 20"

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Thus, the answer is <math>\frac{10}{64} = \frac{5}{32}</math>, and the answer is <math>\boxed{A}</math>
 
Thus, the answer is <math>\frac{10}{64} = \frac{5}{32}</math>, and the answer is <math>\boxed{A}</math>
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But  this  may  be  slightly  complicated,  so  you  can  also  do  solution  2 (scroll to bottom).
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== Solution 2 ==
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Make a chart with all products. 10 work out of 64. SImplify for 5/32 or <math> \boxed{A} </math>.
  
 
== See also ==
 
== See also ==
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* [[Mathematics competition resources]]
 
* [[Mathematics competition resources]]
  
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{{MAA Notice}}
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== Solution 2 ==
 
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Make a chart with all products. 10 work out of 64. SImplify for 5/32 or <math> \boxed{A} </math>.
 

Latest revision as of 19:15, 23 October 2015

Problem

A pair of 8-sided dice have sides numbered 1 through 8. Each side has the same probability (chance) of landing face up. The probability that the product of the two numbers that land face-up exceeds 36 is

$\text{(A)}\ \dfrac{5}{32} \qquad \text{(B)}\ \dfrac{11}{64} \qquad \text{(C)}\ \dfrac{3}{16} \qquad \text{(D)}\ \dfrac{1}{4} \qquad \text{(E)}\ \dfrac{1}{2}$

Solution

There are $8 \times 8 = 64$ combinations to examine.

If one die is $1$, then even with an $8$ on the other die, no combinations will work.

If one die is $2$, then even with an $8$ on the other die, no combinations will work.

If one die is $3$, then even with an $8$ on the other die, no combinations will work.

If one die is $4$, then even with an $8$ on the other die, no combinations will work.

If one die is $5$, then the other die must be an $8$ to have a product over $36$. Thus, $(5,8)$ works.

If one die is $6$, then the other die must be either $7$ or $8$ to have a product over $36$. Thus, $(6, 7)$ and $(6, 8)$ both work.

If one die is $7$, then the other die can be $6, 7,$ or $8$ to have a product over $36$. Thus, $(7, 6)$, $(7, 7)$, and $(7, 8)$ all work.

If one die is $8$, then the other die can be $5, 6, 7,$ or $8$ to have a product over $36$. Thus, $(8, 5), (8,6), (8,7),$ and $(8,8)$ work.

There are a total of $1 + 2 + 3 + 4 = 10$ combinations that work out of a total of $64$ possibilities.

Thus, the answer is $\frac{10}{64} = \frac{5}{32}$, and the answer is $\boxed{A}$

But this may be slightly complicated, so you can also do solution 2 (scroll to bottom).

Solution 2

Make a chart with all products. 10 work out of 64. SImplify for 5/32 or $\boxed{A}$.

See also

1997 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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