2008 AMC 10B Problem 16

Below is the professionally curated solution for Problem 16 of the 2008 AMC 10B, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2008 AMC 10B solutions, or check the answer key.

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Concepts:conditional probability dice (probability)parity

Difficulty rating: 1490

16.

Two fair coins are to be tossed once. For each head that results, one fair die is to be rolled. What is the probability that the sum of the die rolls is odd? (Note that if no die is rolled, the sum is $0.$ )

$\dfrac{3}{8}$

$\dfrac{1}{2}$

$\dfrac{43}{72}$

$\dfrac{5}{8}$

$\dfrac{2}{3}$

Solution:

Whenever at least one die is rolled, by symmetry the sum is odd with probability $\tfrac12.$

No die is rolled only when both coins are tails, with probability $\tfrac14;$ that sum $0$ is even. So the answer is $\left(1-\tfrac14\right)\cdot\tfrac12=\tfrac38.$

Thus, the correct answer is A.

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