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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]

1 Answer

2 votes

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

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