Pollution of the rivers in the United States has been a problem for many years. Consider the following events: a: the river is polluted, b: a sample of water tested detects pollution, c: fishing is permitted. Assume P(A) = 0.3, P(B|A) = 0.75, P(B|A) = 0.20, P(C|A∩B) = 0.20, P(C|A∩B) = 0.15, P(C|A∩B) = 0.80, and P(C|A∩B) = 0.90. A) Find P(A∩B∩C).
B) Find P(B∩C).
C) Find P(C).
D) Find the probability that the river is polluted, given that fishing is permitted and the sample tested did not detect pollution.

The answer is below

Step-by-step explanation:

P(A') = 1 - P(A) = 1 - 0.3 = 0.7

P(B'/A) = 1 - P(B/A) = 1 - 0.75 - 0.25

P(B/A') = 0.2, P(B'/A') = 1 - 0.2 = 0/8

P(C|A∩B) = 0.20,

P(C|A'∩B) = 0.15,

P(C|A∩B') = 0.80, and

P(C|A'∩B') = 0.90.

A) From conditional probability;

P(B/A) = P(B ∩ A) / P(A)

P(B∩A) = P(B/A) × P(A) = 0.75 × 0.3 = 0.225

P(C/A∩B) = P(A∩B∩C)/P(A∩B)

P(A∩B∩C) =  P(C/A∩B) × P(A∩B)

P(A∩B∩C) = 0.2 ×0.225 = 0.045

B) P(A∩B∩C) = P(A) × P(B/A) × P(C/A∩B)

P(B'∩C) = P(A∩(B'∩C)) + P((A'∩B')∩C)

P(B'∩C) = P(A) × P(B'/A) × P(C/A∩B') + P(A') × P(B'/A') × P(C/A'∩B')

P(B'∩C) = 0.3 × 0.25 × 0.8 + (0.7 × 0.8 × 0.9) = 0.06 + 0.504 = 0.564

C) P(C) =  P(A∩B∩C) +  P(A'∩B∩C)  +  P(A∩B'∩C) +  P(A'∩B'∩C)

P(A'∩B∩C) = P(A') × P(B/A') × P(C/A'∩B) = 0.7 × 0.2 × 0.15 = 0.021

P(A∩B'∩C) = P(A) × P(B'/A) × P(C/A∩B') =  0.3 × 0.25 × 0.8 = 0.06

P(A'∩B'∩C) = P(A') × P(B'/A') × P(C/A'∩B') = 0.7 × 0.8 × 0.9 = 0.504

P(C) =  P(A∩B∩C) +  P(A'∩B∩C)  +  P(A∩B'∩C) +  P(A'∩B'∩C) = 0.045 + 0.021 + 0.06 + 0.504 = 0.63

D) P(A/B'∩C) = P(A∩B'∩C)/P(B'∩C)

P(A/B'∩C) = 0.06 / 0.564 = 0.106

if you mean

then it's c. 8, -7/3

but if you mean

then it is

step-by-step explanation:

the multiplicative inverse of a fraction a/b is b/a. for the multiplicative inverse of a real number, divide 1 by the number. for example, the reciprocal of 5 is one fifth 1/5 or 0.2, and the reciprocal of 0.25 is 1 divided by 0.25, or 4.