Question

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 B<Y<R?$B<Y<R?$

Short answer

Calculate a) and b) as probabilities on the outcome space of events that are equally likely.

The event$B<Y<R$can occur only if the dice are rolled on different numbers.

a)The probability that no two dice of the dice land on the same number islocalid="1649409200689" $\frac{5}{9}$

b)The conditional probability $B<Y<R$is localid="1649410071370" $\frac{1}{6}$

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

Step-by-step solution

Step1:The outcome space of events that are equally likely to occur (part a)

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

The number of elements in S, namely |S| is$6^3$

We consider two events:

$B<Y<R$The blue ball's value is less than the yellow die's value. which is less than the number of the red die

Compute:

$\text{a)}P\left(D\right)=\text{?}$

localid="1649408084900" $\text{b)}P(B<Y<R\mid D)=\text{?}$

$\text{c)}P(B<Y<R)=\text{?}$

For P(D)

Count the number of possible outcomes for the dice. Choose the blue die number six times, the yellow die number five times (but not the blue die number), and any of the four remaining red die numbers.

The probability formula produces for an equally likely set of events:

localid="1649409546971" $$\begin{gathered} \text{a)}P\left(D\right)=\frac{6\times 5\times 4}{6^3} \\ =\frac{5}{9} \end{gathered}$$

Step2:Conditionally probability(part b)

If the dice all land on the same number, such as D, they can be permuted in a variety of ways on the blue, yellow, and red die. Only one of those six rolls the blue die, the yellow die, and the red die, with the lowest number on the blue die, the middle number on the yellow die, and the highest number on the red die.$B<Y<R.$

$$\begin{gathered} \text{b)}P(B<Y<R\mid D) \\ =\frac{1}{6} \end{gathered}$$

Step3:Find P(B<Y<R)(part c)

First note that:

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

We can derive a multiplication rule for two events from the definition of conditional probability.

localid="1649408390569" $$\begin{gathered} P(B<Y<R) \\ =P\left(\right(B<Y<R)\cap D) \end{gathered}$$
$=P(B<Y<R\mid D)P\left(D\right)$

localid="1649409564502" $=\frac{1}{6}\times \frac{5}{9}$

$=\frac{5}{54}$

localid="1649656032093" $\approx 0.09259$