Let S be a sample space and E and F be events associated with S. Suppose that

Pr left parenthesis Upper E right parenthesis equals 0.6Pr(E)=0.6,

Pr left parenthesis Upper F right parenthesis equals 0.2Pr(F)=0.2

and

Pr left parenthesis Upper E intersect Upper F right parenthesis equals 0.1Pr(E∩F)=0.1.

Calculate the following probabilities.

(a) Pr left parenthesis E|F right parenthesisPr(E|F)

(b) Pr left parenthesis F|E right parenthesisPr(F|E)

(c) Pr left parenthesis E| Upper F prime right parenthesisPrE|F′

(d) Pr left parenthesis Upper E prime | Upper F prime right parenthesisPrE′|F′

a) P(E|F) = 0.5

b) P(F|E) = 0.167

c) P(E|F') = 0.625

d) P(E′|F′) = 0.375

Step-by-step explanation:

P(E) = 0.6

P(F) = 0.2

P(E n F) = 0.1

a) P(E|F) = Probability of E occurring, given F has already occurred. It is given mathematically as

P(E|F) = [P(E n F)]/P(F) = 0.1/0.2 = 0.5

b) P(F|E) = Probability of F occurring, given E has already occurred. It is given mathematically as

P(F|E) = [P(E n F)]/P(E) = 0.1/0.6 = 0.167

c) P(E|F′) = Probability of E occurring, given F did not occur. It is given mathematically as

P(E|F') = [P(E n F')]/P(F')

But P(F') = 1 - P(F) = 1 - 0.2 = 0.8

P(E n F') = P(E) - P(E n F) = 0.6 - 0.1 = 0.5

P(E|F') = 0.5/0.8 = 0.625

d) P(E′|F′) = [P(E' n F')]/P(F')

P(F') = 0.8, P(Universal set) = P(U) = 1

P(E' n F') = P(U) - [P(E n F') + P(E' n F) + P(E n F) = 1 - (0.5 + 0.1 + 0.1) = 0.3

P(E′|F′) = 0.3/0.8 = 0.375