Chapter 6.8, Problem 12E

To determine

To find : The probability that the spinner B stops in 1, then 2 if spinner B spins two times.

Answer to Problem 12E

The probability that the spinner B stops on 1 and then 2 is $\frac{1}{16}$

Explanation of Solution

Given information :

Consider, the information provided in the question,

The spinner B is divided into 4 equal parts namely 1, 2, 3, and 4.

Calculation :

Here, the number of possible outcomes is 4. And to stop the spinner at 1 the number of favourable outcome will be also 1. Therefore, probability that spinner stops at 1 is

  $\begin{array}{l}P\left(\text{Stops on 1}\right)=\frac{\text{Number of favaourable outcome}}{\text{Total number of outcome}}\\ P\left(\text{Stops on 1}\right)=\frac{\text{1}}{4}\end{array}$

Similarly, the probability that spinner stops at 2 is

  $\begin{array}{l}P\left(\text{Stops on 2}\right)=\frac{\text{Number of favaourable outcome}}{\text{Total number of outcome}}\\ P\left(\text{Stops on 2}\right)=\frac{\text{1}}{\text{4}}\end{array}$

By counting principle the probability that the spinner stops on 1 and then 2 is

  $\begin{array}{l}P\left(\text{Stops on 1 and then 2}\right)=P\left(\text{Stops on 1}\right)\cdot P\left(\text{Stops on 2}\right)\\ P\left(\text{Stops on 1 and then 2}\right)=\frac{1}{4}\cdot \frac{1}{4}\\ P\left(\text{Stops on 1 and then 2}\right)=\frac{1}{16}\end{array}$

Hence,

The probability that the spinner B stops on 1 and then 2 is $\frac{1}{16}$ .