Chapter 0.6, Problem 6E

To determine

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

Answer to Problem 6E

The probability of two of the same number 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:

  $P\left(\text{two of the same number}\right)$ .

The probability of two $4s$ is:

$\begin{array}{l}P\left(\text{two of the same number}\right)=P\left(2.1s\right)\times P\left(2.2s\right)\times P\left(2.3s\right)\times P\left(2.4s\right)\times P\left(2.5s\right)\times P\left(2.6s\right).\\ P\left(2.1s\right)=\frac{1}{6}\times \frac{1}{6}\text{=}\frac{1}{36}\text{ and }P\left(2.2s\right)=\frac{1}{6}\times \frac{1}{6}\text{=}\frac{1}{36}.\\ P\left(2.3s\right)=\frac{1}{6}\times \frac{1}{6}=\frac{1}{36}\text{ and }P\left(2.4s\right)=\frac{1}{6}\times \frac{1}{6}=\frac{1}{36}.\\ P\left(2.5s\right)=\frac{1}{6}\times \frac{1}{6}\text{=}\frac{1}{36}\text{ and }P\left(2.6s\right)=\frac{1}{6}\times \frac{1}{6}\text{=}\frac{1}{36}.\end{array}$

Substitute the value and simplify it as follows:

  $\begin{array}{l}P\left(\text{two of the same number}\right)=\left(\frac{1}{6}\times \frac{1}{6}\right)+\left(\frac{1}{6}\times \frac{1}{6}\right)+\left(\frac{1}{6}\times \frac{1}{6}\right)+\left(\frac{1}{6}\times \frac{1}{6}\right)+\left(\frac{1}{6}\times \frac{1}{6}\right)+\left(\frac{1}{6}\times \frac{1}{6}\right).\\ \Rightarrow \frac{1}{36}+\frac{1}{36}+\frac{1}{36}+\frac{1}{36}+\frac{1}{36}+\frac{1}{36}.\\ \Rightarrow \frac{6}{36}=\frac{1}{6}.\end{array}$

Hence, the probability of two of the same number is $\frac{1}{36}$ .