Given that a fair 6-sided die is rolled three times and We need to find What is the probability that the result of exactly one of the rolls will be an even number?

As we are rolling three dice => Number of cases = \(6^3\) = 216

Now we have three places in these 3 tosses _ _ _

First of all let's find the toss in which we will get an even number. This can happen by selecting 1 place out of 3 where even number will come.
=> 3C1 = 3 ways

Now, Probability of getting an even number in any toss = \(\frac{3}{6}\) = \(\frac{1}{2}\) (As 3 out of the 6 possible outcomes are even)
Similarly, P of getting an odd number = \(\frac{1}{2}\)

=> Probability that exactly one of the rolls will be an even number = Place of that even number * Probability of getting an even number * Probability of getting an odd number * Probability of getting an odd number = 3 * \(\frac{1}{2}\) * \(\frac{1}{2}\) * \(\frac{1}{2}\) = \(\frac{3}{8}\) = 0.375

So, Answer will be D
Hope it helps!

Watch the following video to learn How to Solve Dice Rolling Probability Problems

Hello from the GRE Prep Club BumpBot!

Thanks to another GRE Prep Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.