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A red die, a blue die, and a yellow die (all 6-sided) are rolled. We are interested in the probability that the number appearing on the blue die is less than that appearing on the yellow die, which is less than the number appearing on the red die. That is, with B, Y, and R denoting, respectively, the number appearing on the blue, yellow and red die, we are interest in P(B

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Answer:

The question has some details  missing, here is the other details ; we are interested in P(B<Y<R).

a). What is the probability that no two of the dice land on the same number?

b). Given that no two of the dice land on the same number, what is the conditional probability that B<Y<R?

c). What is P(B<Y<R)?

Step-by-step explanation:

To find the probability that no two of the dice land on the same number?

the die has 6 faces and as such first die will have an outcome of 6, second die an outcome of 5 and third die will have 4 possible outcomes.

  • applying combinatorics ; nCr = n!/r!(n-r)!
  • therefore; the probability that no two of the dice land on the same number?
  • = 6C1 X 5C1 X 4C1 /6C1 X 6C1 X 6C1 = 0.5556

b) To find the conditional probability that B<Y<R ;

= P(B<Y<R | no two dice land on same number )

= P(B<Y<R N no two dice land on same number)/ P(no two dice land on same number)

  • = After the third die with an outcome of 4, the fourth die will have an outcome of 3, fifth die with 2 outcomes and sixth die will 1 outcome.
  • As such, total number of all possible outcomes = 216
  • P(B<Y<R) = 4! or 3! or 2! or 1! /216
  • = 0.09259

hence P(B<Y<R N no two dice land on same number)/ P(no two dice land on same number)

= 0.09259/0.5556

c) P(B<Y<R) = 4! or 3! or 2! or 1! /216

= 0.09259

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