2008 AMC 10B Problems/Problem 16

Problem

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 is no die is rolled, the sum is 0.)

$\mathrm{(A)}\ {{{\frac{3} {8}}}} \qquad \mathrm{(B)}\ {{{\frac{1} {2}}}} \qquad \mathrm{(C)}\ {{{\frac{43} {72}}}} \qquad \mathrm{(D)}\ {{{\frac{5} {8}}}} \qquad \mathrm{(E)}\ {{{\frac{2} {3}}}}$

Solution

We consider 3 cases based on the outcome of the coin:

Case 1, 0 heads: The probability of this occuring on the coin flip is $\frac{1} {4}$ . The probability that 0 rolls of a die will result in an odd sum is $0$ .

Case 2, 1 head: The probability of this case occuring is $\frac{1} {2}$ . The proability that one die results as an odd number is $\frac{1} {2}$ .

Case 3, 2 heads: The probability of this occuring is $\frac{1} {4}$ . The probability that 2 dice result in an odd sum is $\frac{1} {2}$ , because regardless of what we throw on the first die, we have $\frac{1} {2}$ probability that the second die will have the opposite parity.

Thus, the probability of having an odd sum rolled is $\frac{1} {4} \cdot 0 + \frac{1} {2} \cdot \frac{1} {2} + \frac{1} {4} \cdot \frac{1} {2}=\frac{3} {8}\Rightarrow \boxed{A}$

2008 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 15	Followed by Problem 17
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2008 AMC 10B Problems/Problem 16

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