Difference between revisions of "1989 AJHSME Problems/Problem 25"

Revision as of 20:44, 24 June 2021

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?

$\text{(A)}\ \frac{1}{6} \qquad \text{(B)}\ \frac{3}{7} \qquad \text{(C)}\ \frac{1}{2} \qquad \text{(D)}\ \frac{2}{3} \qquad \text{(E)}\ \frac{5}{7}$

$[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]$

Solution

For the sum to be even, the two selected numbers must have the same parity.

The first spinner has $2$ odd numbers and $2$ even, so no matter what the second spinner is, there is a $1/2$ chance the first spinner lands on a number with the same parity, so the probability of an even sum is $1/2\rightarrow \boxed{\text{C}}$ .

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 $\frac{1}{2}$ chance that this happens, and a $\frac{1}{3}$ chance that the second spinner lands on an even number as well, so a $\frac{1}{2} \cdot \frac{1}{3} = \frac{1}{6}$ . If the first spinner lands on an odd number, and the second on an odd, there is a $\frac{1}{2} \cdot \frac{2}{3} = \frac{2}{6}$ chance of this happening. Adding these two probabilities together, we get $\frac{1}{6} + \frac{2}{6} = \frac{3}{6} = \frac{1}{2}\rightarrow \boxed{\text{C}}$ .

~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 $\frac{1}{2}$ , and the right has probability of $\frac{1}{3}$ for an even number, of $\frac{1}{6}$ .

Case 2: The left spinner is even. That has $\frac{1}{2}$ , and the right has probability of $\frac{2}{3}$ for an odd number, of $\frac{1}{3}$ .

The sum is $\frac{1}{6}+\frac{1}{3} = \frac{1}{2} \implies \boxed{\textbf{(C)}}$

@@ Line 31: / Line 31: @@
 ~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==

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

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