Independent solution

How to solve this Inclusion–Exclusion question

Setup

Setup

Let L and R be the two events. Because 0.35 is the probability of neither event, first obtain the probability of their union.

Pr⁡(L∪R)=1−0.35=0.65\Pr(L\cup R)=1-0.35=0.65

Model

Model

Use the two-event addition rule and solve it for the intersection probability.

Pr⁡(L∩R)=Pr⁡(L)+Pr⁡(R)−Pr⁡(L∪R)\Pr(L\cap R)=\Pr(L)+\Pr(R)-\Pr(L\cup R)

Compute

Compute

The union is 0.65, so substituting the two marginal probabilities gives an intersection of 0.40+0.30-0.65=0.05.

Pr⁡(L∩R)=0.40+0.30−0.65=0.05\Pr(L\cap R)=0.40+0.30-0.65=0.05

Answer

Answer

Thus the probability that both events occur is 0.05, corresponding to choice A.

0.05(A)\boxed{0.05\quad\text{(A)}}