Chapter 13, Problem 38SGA

To determine

To find: The probability of the given event.

Answer to Problem 38SGA

The probability of getting a multiple of two or three is $\frac{9}{14}$ .

Explanation of Solution

Given:

The given events: One paper is drawn at random from a box contains slips of paper numbered from one to fourteen and selecting a multiple of two or a multiple of three.

Calculation:

The set of all possible outcomes are as:

  $\left\{1,2,3,4,5,6,7,8,9,10,11,12,13,14\right\}$ .

The probability of getting a multiple of two is calculated as:

  $\begin{array}{c}P\left(\text{getting a multiple of 2}\right)=\frac{\text{Possible outcomes}}{\text{Total outcomes}}\\ =\frac{7}{14}\end{array}$

The probability of getting a multiple of three is calculated as:

  $\begin{array}{c}P\left(\text{getting a multiple of 3}\right)=\frac{\text{Possible outcomes}}{\text{Total outcomes}}\\ =\frac{4}{14}\end{array}$

The probability of getting a multiple of both two and three is calculated as:

  $\begin{array}{c}P\left(\text{getting a multiple of both 2\& 3}\right)=\frac{\text{Possible outcomes}}{\text{Total outcomes}}\\ =\frac{2}{14}\end{array}$

Therefore the probability of getting a multiple of two r or a multiple of three is calculated as:

  $\begin{array}{c}P\left(\text{multiple of}2\text{or multiple of}3\right)=P\left(\text{multiple of}2\right)+P\left(\text{multiple of}3\right)-P\left(\text{multiple of}2\text{and multiple of}3\right)\\ =\frac{7}{14}+\frac{4}{14}-\frac{2}{14}\\ =\frac{9}{14}\end{array}$ Therefore, the probability is $\frac{9}{14}$ .