2005 AMC 10B Problems/Problem 9

Problem

One fair die has faces $1$, $1$, $2$, $2$, $3$, $3$ and another has faces $4$, $4$, $5$, $5$, $6$, $6$. The dice are rolled and the numbers on the top faces are added. What is the probability that the sum will be odd?

$\mathrm{(A)} \frac{1}{3} \qquad \mathrm{(B)} \frac{4}{9} \qquad \mathrm{(C)} \frac{1}{2} \qquad \mathrm{(D)} \frac{5}{9} \qquad \mathrm{(E)} \frac{2}{3}$

Solution

In order to obtain an odd sum, exactly one out of the two dice must have an odd number. We can easily find the total probability using casework.

Case 1: The first die is odd and the second die is even.

The probability of this happening is $\dfrac{4}{6}\times\dfrac{4}{6}=\dfrac{16}{36}=\dfrac{4}{9}$

Case 2: The first die is even and the second die is odd.

The probability of this happening is $\dfrac{2}{6}\times\dfrac{2}{6}=\dfrac{4}{36}=\dfrac{1}{9}$

Adding these two probabilities will give us our final answer. $\dfrac{4}{9}+\dfrac{1}{9}=\boxed{\mathrm{(D)}\ \dfrac{5}{9}}$

If you run out of time, you can see it is obviously greater than 1 so it narrows guesses.

Solution 2 (Complementary Counting)

We see can see that there are $2$ ways to get an even number, and other ways get us an odd. Therefore, if we subtract the $P(\text{even})$, we will get $P(\text{odd})$. We can get an even either by getting $2$ evens or $2$ odds. Both cases give $\frac{2}{9}$, so $P(\text{odd})$ is $1-2 \cdot \frac{2}{9}=\boxed{\text{(D)} \frac{5}{9}}$

See Also

2005 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 8
Followed by
Problem 10
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

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