Question

let A and B be two event such that P(A)=0.6 P(B)=0.4 P((not A) and (not B))=0.2 Find P(A or B)

find P((not A) and (B))

find P(A/B)

Answer 1

We are given here that:
P(A) = 0.6,
P(B) = 0.4,
and P(not A and not B) = 0.2

This means that:
and P(not A and not B) = 1 - P(A or B)
Therefore, P(A or B) = 1 - P(not A and not B) = 1 - 0.2 = 0.8

Therefore P(A or B) = 0.8 is the required probability here.

Using law of addition of probability, we have here:
P(A and B) = P(A) + P(B) - P(A or B) = 0.6 + 0.4 - 0.8 = 0.2

P(not A and B) = P(B) - P(A and B) = 0.4 - 0.2 = 0.2
Therefore P(not A and B) = 0.2 is the required probability here.

P(A | B) is computed using Bayes theorem here as:
P(A | B) = P(A and B) / P(B) = 0.2 /0.4 = 0.5
Therefore 0.5 is the required probability here.