Chapter 12.8, Problem 44PPE

a.

To determine

To find $P(\text{red1 and yellow1)}$

Answer to Problem 44PPE

  $P(\text{red1 and yellow1)}=\frac{1}{36}$

Explanation of Solution

Given:

One red cube and one yellow cube.

Calculation:

Suppose roll a red number cube and yellow number cube.

Objective is to find the probability of $P(\text{red1 and yellow1)}$

The rolling of two number cubes is an independent event, because the outcome of first event does not affect the outcomes of second event.

Use the formula for the probability of independent events:

  $\begin{array}{l}P(\text{red1 and yellow1)}=P(\text{red1)}\times \text{P(yellow1)}\\ \text{=}\frac{1}{6}\times \frac{1}{6}\\ =\frac{1}{36}\end{array}$

b.

To determine

To find $P(\text{red2 and yellow2)}$

Answer to Problem 44PPE

  $P(\text{red2 and yellow2)}=\frac{1}{36}$

Explanation of Solution

Given:

One red cube and one yellow cube.

Calculation:

Suppose roll a red number cube and yellow number cube.

Objective is to find the probability of $P(\text{red2 and yellow2)}$

The rolling of two number cubes is an independent event, because the outcome of first event does not affect the outcomes of second event.

Use the formula for the probability of independent events:

  $\begin{array}{l}P(\text{red2 and yellow2)}=P(\text{red2)}\times \text{P(yellow2)}\\ \text{=}\frac{1}{6}\times \frac{1}{6}\\ =\frac{1}{36}\end{array}$

c.

To determine

To find the probability of any matching pair.

Answer to Problem 44PPE

  $\frac{1}{6}$

Explanation of Solution

Given:

One red cube and one yellow cube.

Calculation:

To find the probability of any matching pair, first find the probability of all the matching pair, that is the probability of $P(\text{red3 and yellow3)}$ , $P(\text{red4 and yellow4)}$ , $P(\text{red5 and yellow5)}$ and $P(\text{red6 and yellow6)}$ and then add them.

The probability of $P(\text{red3 and yellow3)}$ is $\frac{1}{36}$

The probability of $P(\text{red4 and yellow4)}$ is $\frac{1}{36}$

The probability of $P(\text{red5 and yellow5)}$ is $\frac{1}{36}$

The probability of $P(\text{red6 and yellow6)}$ is $\frac{1}{36}$

Add them,

  $\frac{1}{36}+\frac{1}{36}+\frac{1}{36}+\frac{1}{36}+\frac{1}{36}+\frac{1}{36}=\frac{1}{6}$