Difference between revisions of "1989 AJHSME Problems/Problem 25"
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The first spinner has <math>2</math> odd numbers and <math>2</math> even, so no matter what the second spinner is, there is a <math>1/2</math> chance the first spinner lands on a number with the same parity, so the [[probability]] of an even sum is <math>1/2\rightarrow \boxed{\text{C}}</math>. | The first spinner has <math>2</math> odd numbers and <math>2</math> even, so no matter what the second spinner is, there is a <math>1/2</math> chance the first spinner lands on a number with the same parity, so the [[probability]] of an even sum is <math>1/2\rightarrow \boxed{\text{C}}</math>. | ||
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| + | ==Solution 2 == | ||
| + | |||
| + | For the [[sum]] to be [[even]], the two selected numbers must have the same [[parity]]. | ||
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| + | With this knowledge, we can break down the problem into smaller problems, first, the probability that the first spinner lands on an even number. There is a <math>\frac{1}{2}</math> chance that this happens, and a <math>\frac{1}{3}</math> chance that the second spinner lands on an even number as well, so a <math>\frac{1}{2} \cdot \frac{1}{3} = \frac{1}{6}</math>. If the first spinner lands on an even number, and the second on an odd, there is a <math>\frac{1}{2} \cdot \frac{2}{3} = \frac{2}{6}</math> chance of this happening. Adding these two probabilities together, we get <math>\frac{1}{6} + \frac{2}{6} = \frac{3}{6} = \frac{1}{2}\rightarrow \boxed{\text{C}}</math>. | ||
==See Also== | ==See Also== | ||
Revision as of 13:54, 25 October 2020
Contents
Problem
Every time these two wheels are spun, two numbers are selected by the pointers. What is the probability that the sum of the two selected numbers is even?
![[asy] unitsize(36); draw(circle((-3,0),1)); draw(circle((0,0),1)); draw((0,0)--dir(30)); draw((0,0)--(0,-1)); draw((0,0)--dir(150)); draw((-2.293,.707)--(-3.707,-.707)); draw((-2.293,-.707)--(-3.707,.707)); fill((-2.9,1)--(-2.65,1.25)--(-2.65,1.6)--(-3.35,1.6)--(-3.35,1.25)--(-3.1,1)--cycle,black); fill((.1,1)--(.35,1.25)--(.35,1.6)--(-.35,1.6)--(-.35,1.25)--(-.1,1)--cycle,black); label("$5$",(-3,.2),N); label("$3$",(-3.2,0),W); label("$4$",(-3,-.2),S); label("$8$",(-2.8,0),E); label("$6$",(0,.2),N); label("$9$",(-.2,.1),SW); label("$7$",(.2,.1),SE); [/asy]](http://latex.artofproblemsolving.com/1/7/c/17ccafee32c561156cef7b2bced57d414ca76485.png)
Solution
For the sum to be even, the two selected numbers must have the same parity.
The first spinner has 
odd numbers and even, so no matter what the second spinner is, there is a 
chance the first spinner lands on a number with the same parity, so the probability of an even sum is 
.
Solution 2
For the sum to be even, the two selected numbers must have the same parity.
With this knowledge, we can break down the problem into smaller problems, first, the probability that the first spinner lands on an even number. There is a 
chance that this happens, and a 
chance that the second spinner lands on an even number as well, so a 
. If the first spinner lands on an even number, and the second on an odd, there is a 
chance of this happening. Adding these two probabilities together, we get 
.
See Also
| 1989 AJHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 24 |
Followed by Last Problem | |
| 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 | ||
