Question

4. Three fair dice have different colors: red, blue, and yellow. These three dice are rolled and the face value of each is recorded as R, B, Y, respectively. (a) Compute the probability that B< Y < R, given all the numbers are different; (b) Compute the probability that B< Y< R.

Answer 1

a) Given that all numbers are different. Let these 3 numbers be: K, L and M. Now there are 3! = 6 permutations for the 3 numbers to be allotted to 3 dice colours. Out of these 6 cases, only 1 will be such that the number on blue dice < number on yellow dice < number on red dice. Therefore 1/6 is the required probability here.

b) Without any prior condition, Total cases = 6*6*6 = 216 for numbers on the 3 dice.

Now, number of cases such that: B < Y < R are computed as: 6c3 = 20 ways as if we select 3 distinct numbers from 6 numbers ( we can arrange them in 1 way to get the desired result )

Therefore the probability here is computed as:

$= \frac{20}{216} = \frac{5}{54}$

Therefore 5/54 is the required probability here.