We need to find What is the probability of rolling three fair dice and having at least one of the three dice show an even number?

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

Let's solve the problem using two methods:

Method 1:

Now there are 8 outcomes possible
(Odd, Odd, Odd), (Odd, Odd, Even), (Odd, Even Odd), (Odd, Even, Even), (Even, Odd, Odd), (Even, Odd, Even), (Even, Even Odd), (Even, Even, Even) and there is an equal chance of each of them happening

=> P(At least one number is Even) = \(\frac{7}{8}\) (all cases except Odd, Odd, Odd)

Method 2:

Out of the 216 comes lets eliminate all options in which all three outcomes are odd
All three can be odd in 3*3*3 (=27 ways), as in each roll we can get any number out of 1, 3, and 5 => 3 choices in each roll

=> P(At least one number is Even) = \(\frac{216 - 27}{216}\) = \(\frac{189}{216}\) = \(\frac{7}{8}\)

So, Answer will be B
Hope it helps!

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