1994 AMC 8 Problem 22

Attempt Problem 22 of the 1994 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 1994 AMC 8 solutions, or check the answer key.

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22.

The two wheels shown at the right are spun and the two resulting numbers are added. The probability that the sum of the two numbers is even is
16\dfrac{1}{6}
14\dfrac{1}{4}
13\dfrac{1}{3}
512\dfrac{5}{12}
49\dfrac{4}{9}
Answer: D
Concepts:basic probabilityparity
Small hint:
On the first wheel 33 covers half and 11 and 22 each cover a quarter; the second wheel is divided into equal thirds
Big hint:
The sum is even when both numbers are even or both are odd
Solution:
On the first wheel, P(1)=14,P(1) = \dfrac14, P(2)=14,P(2) = \dfrac14, and P(3)=12.P(3) = \dfrac12. On the second wheel, each of 4,4, 5,5, 66 has probability 13.\dfrac13.
The sum is even when both numbers are odd or both are even. Both odd: the first is 11 or 33 (34)\left(\dfrac34\right) and the second is 55 (13),\left(\dfrac13\right), giving 34⋅13=14.\dfrac34 \cdot \dfrac13 = \dfrac14. Both even: the first is 22 (14)\left(\dfrac14\right) and the second is 44 or 66 (23),\left(\dfrac23\right), giving 14⋅23=16.\dfrac14 \cdot \dfrac23 = \dfrac16.
The total probability is 14+16=512.\dfrac14 + \dfrac16 = \dfrac{5}{12}.
Thus, the correct answer is D .
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