Question

If A, B, and C are mutually exclusive events with P(A) = 0.21, P(B) = 0.32, and P(C) = 0.43, determine the following probabilities. Round your answers to two decimal places.
​(a)​ P(A U B U C) ​
(b) P(A n B n C) ​
(c)​ P(A n B) ​
(d) P[(A U B) n C]

Answer 1

By their mutual exclusivity,

$P(A\cup B\cup C)=P(A)+P(B)+P(C)=0.96$

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

$P(A\cap B)=0$

For the last probability, first distribute the intersection:

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

Recall that for two event $X,Y$,

$P(X\cup Y)=P(X)+P(Y)-P(X\cap Y)$

so that

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

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