Chapter 8.6, Problem 49E

To determine

To find: the probability of the event.

Answer to Problem 49E

The probability of selected all three numbers less than or equal to 3 is $\frac{27}{1000}$ .

Explanation of Solution

Given information:

A random number generator selects three numbers from 1 through 10.

All three numbers are less than or equal to 3.

Calculation:

Consider the 1 to 10 numbers. Three numbers randomly selected from 1 to 10 numbers.

There are total 3 numbers less than or equal to 3, so probability of selecting one number less than equal to 3 is,

  $\begin{array}{c}P\left(E\right)=\frac{n\left(E\right)}{n\left(S\right)}\\ =\frac{3}{10}\end{array}$

Since, the selection of all three numbers is independent events.

So, the probability that the selected three numbers less than or equal to 3 product of individual probability of selecting one number,

  $\begin{array}{c}P\left(E\right)=\frac{3}{10}\cdot \frac{3}{10}\cdot \frac{3}{10}\\ =\frac{27}{1000}\end{array}$

Hence, the probability of selected all three numbers less than or equal to 3 is $\frac{27}{1000}$ .