1989 AMC 8 Problem 25

Attempt Problem 25 of the 1989 AMC 8 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1989 AMC 8 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

25.

Every time the two wheels shown are spun, two numbers are selected by the pointers. What is the probability that the sum of the two selected numbers is even?
16\dfrac{1}{6}
37\dfrac{3}{7}
12\dfrac{1}{2}
23\dfrac{2}{3}
57\dfrac{5}{7}
Answer: C
Concepts:basic probabilityparity
Small hint:
A sum is even exactly when both numbers are even or both are odd
Big hint:
The first wheel has 22 even and 22 odd numbers; the second has 11 even and 22 odd
Solution:
The sum is even when both numbers are even or both are odd. The first wheel has evens {4,8}\{4, 8\} and odds {3,5},\{3, 5\}, each with probability 24=12.\frac{2}{4} = \frac{1}{2}. The second wheel has even {6}\{6\} with probability 13\frac{1}{3} and odds {7,9}\{7, 9\} with probability 23.\frac{2}{3}.
So the probability of an even sum is 12⋅13+12⋅23=16+26=12.\frac{1}{2} \cdot \frac{1}{3} + \frac{1}{2} \cdot \frac{2}{3} = \frac{1}{6} + \frac{2}{6} = \frac{1}{2}.
Thus, the correct answer is C .
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