Question

Let 𝐴 and 𝐵 be two events. Suppose that the probability that neither event occurs is 3

8. What is the probability that at least one of the events occurs?

Let 𝐶 and 𝐷 be two events. Suppose 𝑃(𝐶) = 0.5, 𝑃(𝐶 ∩ 𝐷) = 0.2 and 𝑃((𝐶 ∪ 𝐷)′) = 0.4. What

is 𝑃(𝐷)?

Answer 1

1)    The event that at least one of A and B occurs is opposite to the event that neither of A and B occurs, hence its probability is

$P=1-\frac{3}{8}=\frac{5}{8}$

2)    We have

$P\left( \left( C\cup D \right) ' \right) =0.4\Rightarrow P\left( C\cup D \right) =1-0.4=0.6$

By the inclusion-exclusion formula

$P\left( C\cup D \right) =P\left( C \right) +P\left( D \right) -P\left( C\cap D \right) \Rightarrow \\\Rightarrow P\left( D \right) =P\left( C\cup D \right) +P\left( C\cap D \right) -P\left( C \right) =0.6+0.2-0.5=0.3$