Question

Let A.B and C be three disjoint events defined over the same place S. Assume AUBU C = S, P(A) = 0.4, and P(B) = 0.4. 1) Compute P(C) [The answer should be a number rounded to five decimal places, don't use symbols such as % 2) Compute the P(AUB) [The answer should be a number rounded to five decimal places, don't use symbols such as %

Answer 1

We have

P (A∪B∪C) = P(A) +P(B) +P(C) −P(AB) −P(AC) −P(BC) +P(ABC)

since A∪B∪C=S

P (A∪B∪C)=P(S)=1. By Probability Axiom

Since A , B and C are disjoint events we have P(AB) =P(AC) =P(BC) =P(ABC)=0

Hence

P (A∪B∪C) = P(A) +P(B) +P(C)

then

1) P(C) =P (A∪B∪C)- P(A) -P(B)

=1 - 0.4 - 0.4

=0.2

2) P(A∪B)=P(A)+P(B)-P(AB)

since A and B are disjoint P(AB)=0

Then P(A∪B)=P(A)+P(B)

=0.4+0.4

=0.8