Description
Unformatted Attachment Preview
Purchase answer to see full attachment
Explanation & Answer
Please have a look at the solution.
4-a) (E1 ∩ E3) – E2
4-b) U – (E1 ∪ E2 ∪ E3) (U is the universal set)
4-c) E1 ∪ E2 ∪ E3
4-d) E1 ∩ E2 ∩ E3
4-e) (E1 ∪ E2) – (E1 ∩ E2)
4-f) (E1 ∪ E2 ∪ E3) - [ ((E1 ∩ E2) ∪ (E1 ∩ E3) ∪ (E2 ∩ E3)) – 3 (E1 ∩ E2 ∩ E3)]
5-a) We have,
Pr(A and B) = Pr(A)*Pr(B) where A and B are independent events
Pr(getting 6 in one roll) = 1/6
So, Pr(getting sixes in all n rolls) = 1/6*1/6*1/6---n times = [1/6]n
5-b) We have,
Pr(A and B) = Pr(A)*P...