2009 AMC 8 Problem 12

Attempt Problem 12 of the 2009 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 2009 AMC 8 solutions, or check the answer key.

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

12.

The two spinners shown are spun once and each lands on one of the numbered sectors. What is the probability that the sum of the numbers in the two sectors is prime?
12\dfrac{1}{2}
23\dfrac{2}{3}
34\dfrac{3}{4}
79\dfrac{7}{9}
56\dfrac{5}{6}
Answer: D
Concepts:basic probabilityprime
Small hint:
Make a 3×33\times3 table of possible sums.
Big hint:
Only sums of 99 are not prime among the possible outcomes.
Solution:
We can find the sum of the two numbers in every possible outcome.
1+2=31+4=51+6=73+2=53+4=73+6=95+2=75+4=95+6=11. \begin{gather*} 1 + 2 = 3 \\ 1 + 4 = 5 \\ 1 + 6 = 7 \\ 3 + 2 = 5 \\ 3 + 4 = 7 \\ 3 + 6 = 9 \\ 5 + 2 = 7 \\ 5 + 4 = 9 \\ 5 + 6 = 11. \end{gather*}
There are only 22 outcomes where the sum is not prime (the two instances when the sum is 99). Therefore, the probability that the sum is prime is 79.\dfrac{7}{9}.
Thus, D is the correct answer.
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