1991 AJHSME Problems/Problem 22: Difference between revisions
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{{AJHSME box|year=1991|num-b=21|num-a=23}} | {{AJHSME box|year=1991|num-b=21|num-a=23}} | ||
[[Category:Introductory Probability Problems]] | [[Category:Introductory Probability Problems]] | ||
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Latest revision as of 23:08, 4 July 2013
Problem
Each spinner is divided into 
equal parts. The results obtained from spinning the two spinners are multiplied. What is the probability that this product is an even number?
![[asy] draw(circle((0,0),2)); draw(circle((5,0),2)); draw((0,0)--(sqrt(3),1)); draw((0,0)--(-sqrt(3),1)); draw((0,0)--(0,-2)); draw((5,0)--(5+sqrt(3),1)); draw((5,0)--(5-sqrt(3),1)); draw((5,0)--(5,-2)); fill((0,5/3)--(2/3,7/3)--(1/3,7/3)--(1/3,3)--(-1/3,3)--(-1/3,7/3)--(-2/3,7/3)--cycle,black); fill((5,5/3)--(17/3,7/3)--(16/3,7/3)--(16/3,3)--(14/3,3)--(14/3,7/3)--(13/3,7/3)--cycle,black); label("$1$",(0,1/2),N); label("$2$",(sqrt(3)/4,-1/4),ESE); label("$3$",(-sqrt(3)/4,-1/4),WSW); label("$4$",(5,1/2),N); label("$5$",(5+sqrt(3)/4,-1/4),ESE); label("$6$",(5-sqrt(3)/4,-1/4),WSW); [/asy]](http://latex.artofproblemsolving.com/8/b/f/8bf8c01f9a0a54b51e38ee9525e14b08f60035d1.png)
Solution
Instead of computing this probability directly, we can find the probability that the product is odd, and subtract that from 
.
The product of two integers is odd if and only if each of the two integers is odd. The probability the first spinner yields an odd number is 
and the probability the second spinner yields an odd number is 
, so the probability both yield an odd number is 
.
The desired probability is thus 
.
See Also
| 1991 AJHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 21 |
Followed by Problem 23 | |
| 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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