4 A B C are events with PA 45 PB 35 PC 5 PAB 2 PAC 15

4. A, B, C are events with P(A) = .45, P(B) = .35, P(C) = .5, P(AB) = .2, P(AC) = .15 = P(BC), and P(A U B U C) = .9, Find, P(ABC), and P(A^cB^cC^c).

Solution

We know that P(AUBUBC) = P(A) + P(B) +P(C) - P(AB) - P(BC) - P(AC) + P(ABC)

Therefore P(ABC) = P(AUBUBC) - P(A) - P(B) -P(C) + P(AB) + P(BC) + P(AC)

Substituting the given values, we get

P(ABC) = 0.9 - (0.45+0.35+0.5) + (0.2+0.15+0.15) = 0.1

a) P(ABC) = 0.1

And P(AUBUC)^c = P(A^cB^cC^c) and P(AUBUC)^c = 1-P(AUBUC)

Therefore P(A^cB^cC^c) = 1-P(AUBUC)

Putting the values we get

P(A^cB^cC^c) = 1- 0.9 = 0.1

B) P(A^cB^cC^c) = 0.1