Assume that A, B, and C are independent events and let P(A) = 1/3, P(B) = 1/4, a

Question

Assume that A, B, and C are independent events and let P(A) = 1/3, P(B) = 1/4, and P(C) = 1/6. Then, calculate the probability that: (a) At least one of the events A, B, C occurs; (b) All three events A, B, C occur; (c) None of the events A, B, C occurs; (d) Events A and B occur, but not C; (e) Either event A occurs or, if not, then B also does not occur; (f) Exactly one of the events A, B, C occurs; (g) At most one of the events A, B, C occurs.

I got the relations but I'm not sure how to incorporate the math.

a) A U B U C

b) A B C

c) A' B' C'

d) A B C'

e) A U (A' B')

f) [A (B' C' )] U [B (C' A' )] U [C (A' B' )]

g) [(A B) U (B C) (C A)]'

Explanation / Answer

a)

P(A U B U C) = P(A) + P(B) + P(C) - P(A n B) - P(B n C) - P(A n C) + P(A n B n C)

As they are independent, the probabilities of intersections are just products of probabilities. Hence,

P(A U B U C) = P(A) + P(B) + P(C) - P(A) P(B) - P(B)P(C) - P(A)P(C) + P(A)P(B)P(C)

Plugging in the given,

P(A U B U C) = 7/12 [answer]

****************

b)

P(A n B n C) = P(A) P(B) P(C) = 1/72 [answer]

*************

c)

P(A' n B' n C') = (1-P(A))(1-P(B))(1-P(C)) = 5/12 [answer]

***************

d)

P(A n B n C') = P(A) P(B) (1 - P(C)) = 5/72 [answer]

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