Chapter 0.6, Problem 3E

To determine

To find: the probability of getting two and three when die is rolled twice.

Answer to Problem 3E

The probability of $2\text{ and }3$ 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:

  $P\left(2\text{ and }3\right)$ .

The probability of two $4s$ is:

$\begin{array}{l}P\left(2\text{ and }3\right)=P\left(2\right)\times P\left(3\right).\\ P\left(2\right)=\frac{1}{6}\text{ and }P\left(3\right)=\frac{1}{6}.\\ P\left(2\text{ and }3\right)=\frac{1}{6}\times \frac{1}{6}=\frac{1}{36}.\end{array}$

Hence, the probability of $2\text{ and }3$ is $\frac{1}{36}$ .