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 DHope it helps!
Playlist on Solved Problems on Probability hereWatch the following video to MASTER Dice Rolling Probability Problems