Difference between revisions of "1989 AJHSME Problems/Problem 25"
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==Solution== | ==Solution== | ||
| − | For the [[sum]] to be [[even]], the two selected numbers must have the same [[parity]]. | + | For the [[sum]] to be [[even]], the two selected numbers must have the same [[parity]] (must be either odd and odd or even and even). |
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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For the [[sum]] to be [[even]], the two selected numbers must have the same [[parity]]. | 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 <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 | + | 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 odd 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>. |
| + | ~Shadow-18 | ||
| + | |||
| + | ==Solution 3 == | ||
| + | |||
| + | We can complement count this, so we are looking for odd results. This happens when they have different [[parity]]. We can find the probability of an odd sum by using casework. | ||
| + | |||
| + | Case 1: The left spinner is odd. That has <math>\frac{1}{2}</math>, and the right has probability of <math>\frac{1}{3}</math> for an even number, of <math>\frac{1}{6}</math>. | ||
| + | |||
| + | Case 2: The left spinner is even. That has <math>\frac{1}{2}</math>, and the right has probability of <math>\frac{2}{3}</math> for an odd number, of <math>\frac{1}{3}</math>. | ||
| + | |||
| + | The sum is <math>\frac{1}{6}+\frac{1}{3} = \frac{1}{2} \implies \boxed{\textbf{(C)}}</math> | ||
==See Also== | ==See Also== | ||
Latest revision as of 12:57, 16 October 2023
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 (must be either odd and odd or even and even).
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 odd number, and the second on an odd, there is a 
chance of this happening. Adding these two probabilities together, we get 
.
~Shadow-18
Solution 3
We can complement count this, so we are looking for odd results. This happens when they have different parity. We can find the probability of an odd sum by using casework.
Case 1: The left spinner is odd. That has , and the right has probability of for an even number, of 
.
Case 2: The left spinner is even. That has , and the right has probability of 
for an odd number, of .
The sum is 
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 | ||
