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 P(B < Y < R)?

Short answer

a.$\text{Probability that no two dice have same outcome}=\frac{5}{9}=0.55$

role="math" localid="1647620052349" $$\begin{gathered} \text{b.Probability that value of blue die is less than value of yellow die isless} \\ \text{thanvalue of red die}=\frac{5}{54}=0.092 \end{gathered}$$

$$\begin{gathered} \text{c.Probability that value of blue die is less than value of yellow die is less} \\ \text{than value of red die}=\frac{5}{54}=0.092 \end{gathered}$$

Step-by-step solution

Given information

Three 6-sided, a blue, a yellow, a red dices are given.

The value on the blue dice is less than the value on the yellow dice is less than the value on the red dice.

Formula used

$\text{Number of ways of selecting}r\text{distinct objects out of}p\text{distinct objects is}\left(\begin{array}{l}p\\ r\end{array}\right)$

$\text{Number of ways of arranging}r\text{distinct objects is}r\text{!}$

$\text{Probability of an event with discrete outcomes}=\frac{\text{Number of favorable outcomes}}{\text{Number of total outcomes}}$

Calculation (Part a)

Sample space is all the possible combinations of values on three dice $6^3=216$

For no two values to be equal, selecting the value on the first dice in $\frac{6}{1}$ways, then selecting the value for the second dice in $\frac{5}{1}$ways and selecting value for the third dice in $\frac{4}{1}$ways.

Hence, total number of possible outcomes so that no two values of dice are equal $6\times 5\times 4$

Probability that no two dice are equal$=\frac{6\times 5\times 4}{6^3}=\frac{5}{9}=0.55$

Calculation (Part b)

g

Calculation (Part c)

g