2004 AMC 8 Problem 21

Attempt Problem 21 of the 2004 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 2004 AMC 8 solutions, or check the answer key.

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

21.

Spinners AA and BB are spun. On each spinner, the arrow is equally likely to land on each number. What is the probability that the product of the two spinners’ numbers is even?
14\dfrac{1}{4}
13\dfrac{1}{3}
12\dfrac{1}{2}
23\dfrac{2}{3}
34\dfrac{3}{4}
Answer: D
Concepts:basic probabilitycomplementary probabilityparity
Small hint:
It is easier to count the complement: an odd product.
Big hint:
An odd product requires both spinner results to be odd.
Solution:
For the product to be even, at least one of the spinners must land on an even number.
We can use complementary counting and calculate the probability of both spinners landing on odds.
This happens with a probability of 12⋅23=13. \dfrac{1}{2} \cdot \dfrac{2}{3} = \dfrac{1}{3}.
Then the probability of landing on at least one even is 1−13=23.1 - \dfrac{1}{3} = \dfrac{2}{3}.
Thus, D is the correct answer.
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