Question

A, B and C are events such that P(A) = 1/3, P(B) = ¼, and P(C) = 1/5. Find P(AUBUC) under each of the following assumptions: a) If A, B, and C are mutually exclusive. b) If A, B, and C are independent

Answer 1

(a) If $A,B,C$ are mutually exclusive, then

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

so we have

$P(A\cup B\cup C)=P(A)+P(B)+P(C)=\dfrac{47}{60}$

(b) If $A,B,C$ are mutually independent, then

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

$P(A\cap C)=P(A)P(C),$

$P(B\cap C)=P(B)P(C),$

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

so that

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

$P(A\cup B\cup C)=\dfrac{47}{60}-\left(\dfrac1{12}+\dfrac1{15}+\dfrac1{20}\right)+\dfrac1{60}$

$P(A\cup B\cup C)=\dfrac35$