Question

the probability that at least one of events A and B occurs is 0.6 if A and B occur simultaneously with probability 0.2, evaluate P(A) + P(B)​

Answer 1

hey mate!

Given :

P(A∪B)=0.6,P(A∩B)=0.2P(A∪B)=0.6,P(A∩B)=0.2

∴P(A∪B)=P(A)+P(B)−P(A∩B)∴P(A∪B)=P(A)+P(B)−P(A∩B)

0.6=P(A)+P(B)−0.20.6=P(A)+P(B)−0.2

⇒0.6+0.2=P(A)+P(B)⇒0.6+0.2=P(A)+P(B)

⇒P(A)+P(B)=0.8⇒P(A)+P(B)=0.8

Step 2:

P(A¯)+P(B¯)=(1−P(A))+1−P(B))P(A¯)+P(B¯)=(1−P(A))+1−P(B))

⇒2−(P(A)+P(B))⇒2−(P(A)+P(B))

⇒2−0.8=1.2⇒2−0.8=1.2

Hence (C) is the correct answer.