Bayes Rule Probability Problem

[Vendatte]

Homework Statement

Event B is cow has BSE
Event T is the test for BSE is positive
P(B) = 1.3*10^-5
P(T|B) = .70
probability that the test is positive, given that the cow has BSE
P(T|Bc^c) = .10
probability that the test is positive given that the cow does not have BSE
Find P(B|T) and P(B|T^c)
probability that the cow has BSE given that the test is positive, and probability that the cow has BSE given that the test is negative

Homework Equations

P(C_i|A) = P(A|C_i)/P(A) = P(A|C_i) / (P(A|C₁)P(C₁)+P(A|C₂)*P(C₂) ... + P(A|C_m)*P(C_m)

The Attempt at a Solution

The equation I use to find P(B|T) is

P(B|T) = P(T|B) / (P(T|B)*P(B)+P(T|B^C)*P(B^c)

plugging in the values, I get P(B|T) = .70/(.70*(1.3*10^-5) + .1(1-(1.3*10^-5))), however that value is close to 7, which is clearly wrong.

To find P(B|T^c I plan on using the equation P(B) = P(B|T)P(T)+P(B|T^c)*P(T^c)

Any help would be greatly appreciated.

 

[KataKoniK]

it is important to use accurate and precise calculations to determine probabilities. In this case, we are dealing with conditional probabilities which require us to consider both the likelihood of the event occurring and the prior probability of the event.

To solve for P(B|T), we can use Bayes' Theorem which states that P(B|T) = P(T|B)P(B)/P(T). Plugging in the given values, we have P(B|T) = .70*(1.3*10^-5)/(.70*(1.3*10^-5) + .1*(1-(1.3*10^-5))). This gives us a more reasonable value of approximately 0.0000902, which is the probability that the cow has BSE given that the test is positive.

To solve for P(B|Tc), we can use the equation P(B|Tc) = P(Tc|B)P(Bc)/P(Tc). Plugging in the values, we have P(B|Tc) = .10*(1-(1.3*10^-5))/(.10*(1-(1.3*10^-5)) + .70*(1.3*10^-5)). This gives us a value of approximately 0.99991, which is the probability that the cow has BSE given that the test is negative.

It is important to note that these calculations are based on the given probabilities and do not necessarily reflect the true probability of a cow having BSE or the accuracy of the test. it is important to continuously evaluate and improve upon these calculations by considering new evidence and data.