Chapter CSB, Problem 4.6E

To determine

To find: the probability of two of the same number when die is rolled twice.

Answer to Problem 4.6E

The probability of two $4\text{'}s$ is $\frac{1}{36}$ .

Explanation of Solution

Given information:

As mentioned in the problem the die is rolled twice.

Formula:

  $\text{Number of probability}=\frac{\text{No}.\text{ of outcome}}{\text{Total no}.\text{ of outcome}}.$

Calculation:

Consider the statement that a die is rolled twice.

Then total number of favorable outcome.

  $\begin{array}{l}\left(1,1\right),\left(1,2\right),\left(1,3\right),\left(1,4\right),\left(1,5\right),\left(1,6\right).\\ \left(2,1\right),\left(2,2\right),\left(2,3\right),\left(2,4\right),\left(2,5\right),\left(2,6\right).\\ \left(3,1\right),\left(3,2\right),\left(3,3\right),\left(3,4\right),\left(3,5\right),\left(3,6\right).\\ \left(4,1\right),\left(4,2\right),\left(4,3\right),\left(4,4\right),\left(4,5\right),\left(4,6\right).\\ \left(5,1\right),\left(5,2\right),\left(5,3\right),\left(5,4\right),\left(5,5\right),\left(5,6\right).\\ \left(6,1\right),\left(6,2\right),\left(6,3\right),\left(6,4\right),\left(6,5\right),\left(6,6\right).\end{array}$

The total number of outcome for two $4\text{'}s$ willbe $\left(4,4\right)=1$ .

Recall that $\text{number of probability}=\frac{\text{No}.\text{ of outcome}}{\text{Total no}.\text{ of outcome}}.$

Therefore $\text{number of probability of two 4's}=\frac{1}{36}.$