1Suppose PA 9 PB 8 and PAB C 0 Show that PC 3 2 Prove If A

1.Suppose P(A) .9, P(B) .8, and P(AB C) = 0. Show that P(C) .3.

2. Prove: If A B C = , then P((A B) (B C) (C A)) = P(A B) + P(B C) + P(C A).

Solution

We have P(A\') <0.1 and P(B\')<0.2

As A B C are the only 3 possible outcomes, C should definitely be contained in A\'UB\'

Or P(C)<=0.1+0.2 = 0.3

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2) P((A B) (B C) (C A))

As AUB and BUC will have common as all B,
This is equivalent to

P{(B C) UBAUAB)

= P(AB)+P(BC)+P(CA)