A red die, a blue die, and a yellow die (all six-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 that 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 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 $P(B<Y<R)$?

Given the value that,

$P(B<Y<R)$

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

The outcome space of equally likely events is :

$S=\left\{\right(b,y,r);b,y,r\text{is the number on the blue, yellow and red die, respectively}\}$

$\text{The number of elements in}S\text{, namely}\left|S\right|\text{is}{6}^3$

we consider two events:

D - no two dice land on the same number.

$B<Y<R$- the number on the blue ball is less than the number on the yellow die which is less than the number on the red die.

compute:

a) $P\left(D\right)=$?

b) $P(B<Y<R/D)=$?

c) $P(B<Y<R)=$?

For $P\left(D\right)$count the number of ways in which the dice can land on different numbers.

Choose the number on the blue die in $6$ways, the number on the yellow die in $5$ways(not the number on the blue die), and any of the $4$remaining numbers on the red die.

The formula for the probability on the equally likely set of events yields.

$\text{a)}P\left(D\right)=\frac{6\cdot 5\cdot 4}{6^3}=\frac{5}{9}$

The probability that no two of the dice land on the same number is

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

If the dice all landed on different numbers, i.e. D happened, they can be permuted in $3$! different ways on the blue, yellow, and red die. Only one of those $6$puts the minimal number on the blue die, the middle number on the yellow die, and the maximal number on the red die, that is $B<Y<R$

Thus,

$P(B<Y<R\mid D)=\frac{1}{6}$

The conditional probability that $B<Y<R$is $\frac{1}{6}$

What is$P(B<Y<R)$

First note that:

$B<Y<R\subseteq D\Rightarrow (B<Y<R)\cap D=B<Y<R$

From the definition of a conditional probability ,we obtain multiplication rule for two events.

$P(B<Y<R)=P\left(\right(B<Y<R)\cap D)$

$=P(B<Y<R\mid D)P\left(D\right)$

$=\frac{1}{6}\cdot \frac{5}{9}$

$\begin{array}{rl}& =\frac{5}{54}\\ & \approx 0.09259\end{array}$

$P(B<Y<R)=P(B<Y<R\mid D)P\left(D\right)\approx 0.09259$