Answer
P($E_{3}$) = 0.6
Work Step by Step
We know that probability of any outcome in a sample space is 1.
Therefore,
P($E_{1}$)+P($E_{2}$)+P($E_{3}$)+P($E_{4}$)+P($E_{5}$) =1
Substituting the given values,
0.1+0.1+P($E_{3}$)+0.1+0.1=1
P($E_{3}$)=1-0.4
P($E_{3}$)=0.6
