Respuesta :
Answer:
a) 98.1388%
b) 11.8556%
Step-by-step explanation:
Let A, B and C be the following events
A=The sample is contaminated with organic pollutants AND is detected by the test.
B=The sample is contaminated with volatile solvents
AND is detected by the test.
C=The sample is contaminated with chlorinated compounds
AND is detected by the test.
a)
A, B and C are disjoints sets and the probability that the test will signal is
P(A∪B∪C) = P(A) + P(B) + P(C)
On the other hand, since the probability of taking a sample does not depend on previous choices, the events are independent, so
P(A) = P(The sample is contaminated with organic pollutants)*P(The sample is detected by the test) = 0.6*0.992 = 0.5952
Similarly, we find that
P(B) = 0.27*0.9994 = 0.269838
P(C) = 0.13*0.895 = 0.11635
Hence, the likelihood the test will signal is
0.5952 + 0.269838 + 0.11635 = 0.981388 or 98.1388%
In order not to make heavy the notation, let's all
P(Chl) = Probability the sample is contaminated with chlorinated compounds = 0.13
P(V) = Probability the sample is contaminated with volatile solvents = 0.27
P(O) = Probability the sample is contaminated with organic pollutants = 0.6
P(S | Chl) = the probability that the sample signals GIVEN THAT is contaminated with chlorinated compounds = 0.895
P(S | V) = the probability that the sample signals GIVEN THAT is contaminated with volatile solvents = 0.9994
P(S | O) = the probability that the sample signals GIVEN THAT is contaminated with organic pollutants = 0.992
b)
If the test signals, what is the probability that the chlorinated compounds are present?
We want P(Chl | S)
By the BAYE'S THEOREM
[tex]\large P(Chl|S)=\frac{P(Chl)P(S|Chl)}{P(Chl)P(S|Chl)+P(V)P(S|V)+P(O)P(S|O)}=\\=\frac{0.13*0.895}{0.13*0.895+0.27*0.9994+0.6*0.992}=0.1185556575\approx 11.8556\%[/tex]
So, the probability wanted is 11.8556%