Chapter 6.8, Problem 11E

To determine

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

Answer to Problem 11E

The probability that the spinnerA stops on 1 and then 2 is $\frac{1}{25}$ .

Explanation of Solution

Given information :

Consider, the information provided in the question,

The spinner A is divided into 5 equal parts namely 1, 2, 3, 4 and 5.

Calculation :

Here, the number of possible outcomes is 5. And to stop the spinner at 1 the number of favourable outcome will be also 1. Therefore, probability that spinner stops at 1 is

  $\begin{array}{l}P\left(\text{Stops on 1}\right)=\frac{\text{Number of favaourable outcome}}{\text{Total number of outcome}}\\ P\left(\text{Stops on 1}\right)=\frac{\text{1}}{\text{5}}\end{array}$

Similarly, the probability that spinner stops at 2 is

  $\begin{array}{l}P\left(\text{Stops on 2}\right)=\frac{\text{Number of favaourable outcome}}{\text{Total number of outcome}}\\ P\left(\text{Stops on 2}\right)=\frac{\text{1}}{\text{5}}\end{array}$

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

  $\begin{array}{l}P\left(\text{Stops on 1 and then 2}\right)=P\left(\text{Stops on 1}\right)\cdot P\left(\text{Stops on 2}\right)\\ P\left(\text{Stops on 1 and then 2}\right)=\frac{1}{5}\cdot \frac{1}{5}\\ P\left(\text{Stops on 1 and then 2}\right)=\frac{1}{25}\end{array}$

Hence,

The probability that the spinner Astops on 1 and then 2 is $\frac{1}{25}$ .