2008 AMC 10B Problems/Problem 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.)

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


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

Case 1, 0 heads: The probability of this occurring 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 occurring is $\frac{1} {2} \cdot \frac {1} {2} \cdot 2 = \frac {1} {2}.$ The probability that 1 die results in an odd number is $\frac{1} {2}$.

Case 3, 2 heads: The probability of this occurring 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 $\frac{1} {4} \cdot 0 + \frac{1} {2} \cdot \frac{1} {2} + \frac{1} {4} \cdot \frac{1} {2}=\frac{3} {8}\Rightarrow \boxed{A}$

Solution 2 (possibly slightly faster)

We use complementary counting or subtracting $P(\text{Even})$ from $1$. We use casework now.

Case $1$: $2$ Tails. $2$ tails occur with probability $\frac{1}{4}$, but we will always get an even number, so the overall probability to get an even sum is $\frac{1}{4}$.

Case $2$: $1$ Tail: This event occurs with probability $\frac{1}{2}$ and the probability we get an even is $\frac{1}{2}$, so the overall probability to get an even, in this case, is also $\frac{1}{4}$.

We know $P\text{(Even)}$ is greater than $\frac{1}{2}$, so $P\text{(Odd)}$ is less than $\frac{1}{2}$.

Only $\boxed{\text{(A)}\frac{3}{8}}$ is less than $\frac{1}{2}$.

See also

2008 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 15
Followed by
Problem 17
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png