Ben rolls 5 fair 12-sided dice.  The 12 faces of each die are numbered from 1 to 12. What is the probability that exactly two of the dice show an even number?
There is a $\frac{1}{2}$ probability that a 12-sided die will show an even number and a $\frac{1}{2}$ probability that it will show an odd number. We can choose which dice will show the even numbers in $\binom{5}{2}=10$ ways. For each way, there is a $\left( \frac{1}{2} \right) ^5=\frac{1}{32}$ probability that the chosen dice actually roll even numbers and the other dice roll odd numbers. Therefore, the probability that exactly two of the dice show an even number is $10\cdot \frac{1}{32}=\boxed{\frac{5}{16}}$.
\frac{5}{16}
Level 3
Counting & Probability
Problem: Ben rolls 5 fair 12-sided dice.  The 12 faces of each die are numbered from 1 to 12. What is the probability that exactly two of the dice show an even number?
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find the probability that exactly two of the five dice show an even number, we first determine the total number of outcomes when rolling 5 dice. Each die has 12 faces, so the total number of outcomes is 12^5.

Next, we identify the even numbers on a 12-sided die. The even numbers are 2, 4, 6, 8, 10, and 12, totaling 6 even numbers. Therefore, the probability of rolling an even number on one die is 6/12, which simplifies to 1/2. Conversely, the probability of rolling an odd number is also 1/2.

We need to calculate the probability of getting exactly 2 even numbers and 3 odd numbers. This scenario can be modeled using the binomial probability formula, which is given by:

P(X = k) = C(n, k) * p^k * (1-p)^(n-k)

where:
- n is the total number of trials (dice rolled), which is 5.
- k is the number of successful trials (even numbers), which is 2.
- p is the probability of success on each trial (rolling an even number), which is 1/2.
- C(n, k) is the binomial coefficient, representing the number of ways to choose k successes from n trials.

Calculating the binomial coefficient C(5, 2):

C(5, 2) = 5! / (2!(5-2)!) = 5! / (2!3!) = (5 * 4) / (2 * 1) = 10.

Now we substitute into the binomial probability formula:

P(X = 2) = C(5, 2) * (1/2)^2 * (1/2)^(5-2)
= 10 * (1/2)^2 * (1/2)^3
= 10 * (1/2)^5
= 10 * 1/32
= 10/32
= 5/16.

Thus, the final answer is 5/16.<|END_OF_LATENT|> The final answer is $\frac{5}{16}$.