Chapter 0.6, Problem 4E

To determine

To find: the probability of getting two $4s$ when die is rolled twice.

Answer to Problem 4E

The probability of $4\text{ and }4$ is $\frac{1}{36}$ .

Explanation of Solution

Given:

A die is rolled twice.

Concept used:

The number of ways of arranging $n$ item in order $=n!$ .

The probability formula is used to compute the probability of an event to occur as:

  $P\left(A\right)=\frac{\text{Number of favorable outcome}}{\text{Total Number of favorable outcomes}}$ .

Calculation:

Consider a die is rolled twice. According to the given the following probability is:

  $P\left(\text{two }4s\right)$ .

The probability of two $4s$ is:

$\begin{array}{l}P\left(\text{two }4s\right)=P\left(4\text{ and }4\right).\\ \Rightarrow P\left(4\right)\times P\left(4\right).\end{array}$

Here $P\left(4\right)=\frac{1}{6}$ .

Substituting the value and simplify it as follows:

  $\begin{array}{l}P\left(\text{two }4s\right)=P\left(4\text{ and }4\right).\\ \Rightarrow P\left(4\right)\times P\left(4\right).\\ \Rightarrow \frac{1}{6}\times \frac{1}{6}=\frac{1}{36}.\end{array}$

Hence, the probability of $4\text{ and }4$ is $\frac{1}{36}$ .