Question

If A, B, and C are mutually exclusive events with P(A) = 0.2, P(B) = 0.3, and P(C) = 0.4, determine the following probabilities: a. P(A ∪ B ∪ C) b. P(A ∩ B ∩ C) c. P(A ∩ B) d. P[(A ∪ B) ∩ C] e. P(A′ ∩ B′ ∩ C′)

Answer 1

a. The events are mutually exclusive, so

$P(A\cup B\cup C)=P(A)+P(B)+P(C)=\boxed{0.9}$

b. For the same reason,

$P(A\cap B\cap C)=\boxed0$

Another way to see this is to use the result from part (a), and the inclusion/exclusion principle:

$P(A\cap B\cap C)=P(A)+P(B)+P(C)-P(A\cup B\cup C)$

and the right side reduces to 0.

c. Same as (b),

$P(A\cap B)=\boxed0$

d. Same as (b),

$P((A\cup B)\cap C)=\boxed0$

We can also use the distributivity rule for unions and intersections to write

$P((A\cup B)\cap C)=P((A\cap C)\cup(B\cap C))=P(A\cap C)+P(B\cap C)=0$

e. If the $A'$ is the complement of $A$, then by DeMorgan's law,

$P(A'\cap B'\cap C')=P(A\cup B\cup C)'=1-P(A\cup B\cup C)=\boxed{0.1}$