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|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. 

Answer 1

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|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. 

SOLUTIONS

(a) Find P(A ∩ B ∩ C).

Solution

$P(A\bigcap B\bigcap C)=P(C|A\bigcap B)*P(A\bigcap B)$

$=(0.20)(0.225 )$

$Answer=0.0455$

(b) Find P(B' ∩ C).

Solution

$P(B'\bigcap C)=P(C)-P(B\bigcap C)$

$=P(A'\bigcap B'\bigcap C)+P(A\bigcap B'\bigcap C)$

$=0.504+0.06$

$Answer=0.564$

(c) Find P(C).

Solution

$P(C)=P(A'\bigcap B'\bigcap C)+P(A\bigcap B' \bigcap C)+P(A'\bigcap B\bigcap C)+P(A\bigcap B\bigcap C)$

$=0.504+0.06+0.02+0.045$

$Answer=0.63$

(d) Find the probability that the river is polluted, given that fishing is permitted and the sample tested did not detect pollution.

Solution

This is the same to computing $P(A|B'\bigcap C)$

$P(A|B'\bigcap C)=\frac{P(A\bigcap B'\bigcap C)}{P(B'\bigcap C)}=\frac{0.06}{0.564}=0.1064$

$Answer=0.1064$