Chapter 3 - Probability - Exercises 3.1 - 3.34 - Learning the Mechanics - Page 119: 3.9b

P($E_{3}$) = 0.25

Given: P($E_{1}$)+P($E_{2}$)+P($E_{3}$)+P($E_{4}$)+P($E_{5}$) =1 P($E_{3}$)=P($E_{1}$) Therefore, substituting the given values in the above equation, P($E_{3}$)+0.1+P($E_{3}$)+0.2+0.1=1 2$\times$P($E_{3}$)= 1- 0.5 P($E_{3}$) = 0.5 $\div$ 2 P($E_{3}$) = 0.25