Question

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|Ac ) = 0.2, P C|A∩B = 0.2, P C|Ac∩B = 0.15, P C|A∩Bc = 0.8 and P C|Ac ∩ Bc = 0.9, then find (1) P(A ∩ B ∩ C) (2) P(Bc ∩ C) (3) P(C) (4) Find the probability that the river is polluted, given that fishing is permitted and the sample tested did not detect pollution. Hint: de-Morgan’s law (A ∩ B) c = Ac ∪ B

Answer 1

Answer:

(1) 0.045 (2) 0.564 (3) 0.63 (4) 0.1064

Step-by-step explanation:

We use the multiplication rule several times

(1) $P(A\cap B\cap C) = P(C|A\cap B)P(B|A)P(A)$ = (0.2)(0.75)(0.3) = 0.045

(2) $P(B^{c}\cap C) = P[B^{c}\cap C\cap (A\cup A^{c})] = P[B^{c}\cap C\cap A] + P[B^{c}\cap C\cap A^{c}] = P(C|A\cap B^{c})P(B^{c}|A)P(A) + P(C|A^{c}\cap B^{c})P(B^{c}|A^{c})P(A^{c})$ =

(0.8)(0.25)(0.3) + (0.9)(0.8)(0.7) =  0.564

(3) $P(C\cap B) = P[C\cap B\cap (A\cup A^{c})] = P(A\cap B\cap C) + P(C\cap B\cap A^{c})$ =

0.045 + $P(C|B\cap A^{c})P(B|A^{c})P(A^{c})$ = 0.045 + (0.15)(0.2)(0.7) = 0.066

$P(C) = P(C\cap B) + P(C\cap B^{c})$ = 0.066 + 0.564 = 0.63

(4) $P(A | C\cap B^{c}) = P(A\cap C\cap B^{c})/P(C\cap B^{c}) = P(A\cap C\cap B^{c})/0.564=$

$P(C|A\cap B^{c})P(B^{c}|A)P(A)/0.564$ = (0.8)(0.25)(0.3)/0.564 = 0.1064