1991 AMC 8 Problem 22

Attempt Problem 22 of the 1991 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 1991 AMC 8 solutions, or check the answer key.

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

22.

Each spinner is divided into 33 equal parts. The results obtained from spinning the two spinners are multiplied. What is the probability that this product is an even number?
13\dfrac13
12\dfrac12
23\dfrac23
79\dfrac79
11
Answer: D
Concepts:complementary probabilityparity
Small hint:
A product is even unless both spun numbers are odd
Big hint:
Find the probability of an odd product (odd times odd), then subtract from 11
Solution:
The product is odd only when both numbers are odd. The first spinner is odd (11 or 33) with probability 23,\dfrac23, and the second is odd (55) with probability 13.\dfrac13.
So the product is odd with probability 23×13=29,\dfrac23 \times \dfrac13 = \dfrac29, and even with probability 1−29=79.1 - \dfrac29 = \dfrac79.
Thus, the correct answer is D .
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