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,

$\mathbb P(D)=\mathbb P(A\cup B\cup C)$
$\mathbb P(D)=\mathbb P(A)+\mathbb P(B)+\mathbb P(C)-\bigg(\mathbb P(A\cap B)+\mathbb P(A\cap C)+\mathbb P(B\cap C)\bigg)+\mathbb P(A\cap B\cap C)$
$\mathbb P(D)=\dfrac14+\dfrac16+\dfrac14-\dfrac39+\dfrac1{13}$
$\mathbb P(D)=\dfrac{16}{39}\neq1$

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