Chapter 12, Problem 47CR

To determine

Find probability $P\left(oddtheneven\right)$ .

Answer to Problem 47CR

  $P\left(oddtheneven\right)=\frac{3}{11}$

Explanation of Solution

Given information:

You roll a number cube twice.

Calculation:

Consider the given data,

Roll a number cube twice.

It is formulate as,

  $P\left(AthenB\right)=P\left(A\right)\cdot P\left(BafterA\right)$

Similarly, $P\left(oddtheneven\right)=P\left(odd\right)\cdot P\left(evenafterodd\right)$

  $P\left(event\right)=\frac{numberoffavorableoutcomes}{numberofpossibleoutcomes}$

Here, a cube is rolled twice and a cube contains number $1to6$ , so the total number of possible outcomes are $12$ .

Out of $12$ total outcomes $6$ may odd and six may even two times as cube is rolling two times.

The number of favorable outcomes $6$

So, $P\left(event\right)=\frac{numberoffavorableoutcomes}{numberofpossibleoutcomes}$

  $P\left(odd\right)=\frac{6}{12}=\frac{1}{2}$

Now $P\left(evenafterodd\right)$ ,will be,

  $11$ are number of possible outcomes in which the even number may occur six times when cube is rolled twice.

Number of favorable outcomes are two,

  $P\left(evenafterodd\right)=\frac{6}{11}$

Now calculate $P\left(oddtheneven\right)=P\left(odd\right)\cdot P\left(evenafterodd\right)$

  $\begin{array}{l}P\left(oddtheneven\right)=\frac{6}{11}\times \frac{1}{2}\\ =\frac{3}{11}\end{array}$

Hence, $P\left(oddtheneven\right)=\frac{3}{11}$ .