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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

Respuesta :

LammettHash

(a) If [tex]A,B,C[/tex] are mutually exclusive, then

[tex]P(A\cap B)=P(A\cap C)=P(B\cap C)=P(A\cap B\cap C)=0[/tex]

so we have

[tex]P(A\cup B\cup C)=P(A)+P(B)+P(C)=\dfrac{47}{60}[/tex]

(b) If [tex]A,B,C[/tex] are mutually independent, then

[tex]P(A\cap B)=P(A)P(B),[/tex]

[tex]P(A\cap C)=P(A)P(C),[/tex]

[tex]P(B\cap C)=P(B)P(C),[/tex]

[tex]P(A\cap B\cap C)=P(A)P(B)P(C)[/tex]

so that

[tex]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)[/tex]

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

[tex]P(A\cup B\cup C)=\dfrac35[/tex]

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