Question

Consider three events A, B and C with following properties.

P (A) =1/4, P (B) = 1/6, P (C) = 1/4
P (AՈB) = P (BՈC) = P (AՈC) = 1/9
P (AՈBՈC) = 1/13
Let D = P (AUBUC) c, find P (D). Do the events A, B and C constitute the sample space? Briefly Justify your answer.

Answer 1

By the inclusion/exclusion principle,


`\mathbb P(D)=\mathbb P(A\cup B\cup C)`

`\mathbb P(D)=\mathbb P(A)+\mathbb P(B)+\mathbb P(C)-\bigg(\mathbb P(A\cap B)+\mathbb P(A\cap C)+\mathbb P(B\cap C)\bigg)+\mathbb P(A\cap B\cap C)`

`\mathbb P(D)=\frac14+\frac16+\frac14-\frac39+\frac1{13}`

`\mathbb P(D)=(16)/(39)\\eq1`

So the union of A, B, and C does not constitute the entire sample space.