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

(Solution 2)
(Solution 2)
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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 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>.
+
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
 
~Shadow-18

Revision as of 01:54, 30 November 2020

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

See Also

1989 AJHSME (ProblemsAnswer KeyResources)
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