A red die, a blue die, and a yellow die (all 6 sided) arerolled. We are interest

Question

A red die, a blue die, and a yellow die (all 6 sided) arerolled. We are interested in the probability that the numberappearing on the blue die is less than that appearing on the yellowdie, which is less than that appearing on the red die. Thatis, with B, Y and R denoting, respectively, the number appearing onthe blue, yellow, and red die, we are interested in P(B < Y <R).

(a) What is the probability that no two of the dice land on thesame number?
(b) Given that no two of the dice land on the same number, what isthe conditional probability that B < Y < R?
(c) What is P(B < Y < R)?

Explanation / Answer

a)    No two dice land on same number. Thefirst die can land on any number P =(6/6), the second die can landon any number except the 1st P = (5/6), and the last die can landon any number except previous two P = (4/6). So P(no two dice are the same) = (6/6)*(5/6)*(4/6)=5/9. b)    P(B<Y<R | no two die are thesame) = P(B<Y<R and No two are same) / P(no two are thesame). From part a, P(no two are the same) = 5/9. P(B<Y<R) isthe same as P(B<Y<R and No two are same) because in order forB<Y<R there can't be two that are the same. So P(B<Y<R)= 5/54. I did this by listing all ways B<Y<R. For B=1, wehave {(1,2,3) (1,2,4) (1,2,5) (1,2,6) (1,3,4) (1,3,5) (1,3,6)(1,4,5) (1,4,6) (1,5,6)}. So for B = 1 there are 10 ways, and forrolling three dice, the sample space is 216(6^3) so P(B<Y<R |B= 1) = 10/216. This logic is used for B=2 (6/216), B= 3 (3/216), and B = 4(1/216). Obviously B can't = 5,6 because we can't have (5,6,7) fora six-sided die. Anyway, add up all these probabilities (10/216) +(6/216) + (3/216) + (1/216) and we get 5/54 (this isanswer to part c). To finish part b we use [5/54]/[5/9] =1/6 Hope that helps! a)    No two dice land on same number. Thefirst die can land on any number P =(6/6), the second die can landon any number except the 1st P = (5/6), and the last die can landon any number except previous two P = (4/6). So P(no two dice are the same) = (6/6)*(5/6)*(4/6)=5/9. b)    P(B<Y<R | no two die are thesame) = P(B<Y<R and No two are same) / P(no two are thesame). From part a, P(no two are the same) = 5/9. P(B<Y<R) isthe same as P(B<Y<R and No two are same) because in order forB<Y<R there can't be two that are the same. So P(B<Y<R)= 5/54. I did this by listing all ways B<Y<R. For B=1, wehave {(1,2,3) (1,2,4) (1,2,5) (1,2,6) (1,3,4) (1,3,5) (1,3,6)(1,4,5) (1,4,6) (1,5,6)}. So for B = 1 there are 10 ways, and forrolling three dice, the sample space is 216(6^3) so P(B<Y<R |B= 1) = 10/216. This logic is used for B=2 (6/216), B= 3 (3/216), and B = 4(1/216). Obviously B can't = 5,6 because we can't have (5,6,7) fora six-sided die. Anyway, add up all these probabilities (10/216) +(6/216) + (3/216) + (1/216) and we get 5/54 (this isanswer to part c). To finish part b we use [5/54]/[5/9] =1/6 Hope that helps!

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