Chapter 4.1, Problem 26E

To determine

To find:

The classical probability.

Answer to Problem 26E

Solution:

In rolling a die twice, the probability that the second number is at least as large as the first is 0.5833.

Explanation of Solution

Given:

Number of times a six-sided die is rolled = 2

Formula Used:

Let E denote the event that the second number is at least as large as the first and S denote the sample space. Then, $P\left(E\right)$ denotes the probability that E occurs. The probability is given by,

$P\left(E\right)=\frac{n\left(E\right)}{n\left(S\right)}$

Where, $n\left(E\right)$ denotes the number of elements in E and $n\left(S\right)$ denotes the number of elements in S.

Calculation:

The elements of the sample space are given as,

$\begin{array}{rll}\text{S }& =& \left\{\begin{array}{l}\left(1,1\right),\left(1,2\right),\left(1,3\right),\left(1,4\right),\left(1,5\right),\left(1,6\right),\left(2,1\right),\left(2,2\right),\left(2,3\right),\left(2,4\right),\left(2,5\right),\left(2,6\right),\left(3,1\right),\\ \left(3,2\right),\left(3,3\right),\left(3,4\right),\left(3,5\right),\left(3,6\right),\left(4,1\right),\left(4,2\right),\left(4,3\right),\left(4,4\right),\left(4,5\right),\left(4,6\right),\left(5,1\right),\left(5,2\right),\\ \left(5,3\right),\left(5,4\right),\left(5,5\right),\left(5,6\right),\left(6,1\right),\left(6,2\right),\left(6,3\right),\left(6,4\right),\left(6,5\right),\left(6,6\right)\end{array}\right\}\\ n(S)& =& 36\end{array}$ The elements of the event E are,

$\begin{array}{rll}E& =& \left\{\begin{array}{l}\left(1,1\right),\left(1,2\right),\left(1,3\right),\left(1,4\right),\left(1,5\right),\left(1,6\right),\left(2,2\right),\left(2,3\right),\left(2,4\right),\left(2,5\right),\left(2,6\right),\left(3,3\right),\left(3,4\right),\\ \left(3,5\right),\left(3,6\right),\left(4,4\right),\left(4,5\right),\left(4,6\right),\left(5,5\right),\left(5,6\right),\left(6,6\right)\end{array}\right\}\\ n\left(E\right)& =& 21\end{array}$ Thus, the probability that the second number is at least as large as the first is given by,

$\begin{array}{rll}P\left(E\right)& =& \frac{n\left(E\right)}{n\left(S\right)}\\ & =& \frac{21}{36}\\ & \approx & 0.5833\end{array}$