Calculating Posterior Prob. of an Impurity's Presence

[_N3WTON_]

Homework Statement

A chemical engineer is interested in determining whether a certain trace impurity is present in a product. An experiment has a probability of 0.80 of detecting the impurity if it is present. The probability of not detecting the impurity if it is absent is 0.90. The prior probabilities of the impurity being present and being absent are 0.40 and 0.60 respectively. Three separate experiments result in only two decisions. What is the posterior probability that the impurity is present?

Homework Equations

The Attempt at a Solution

So I know that the event that an impurity is present and not present are disjoint and exhaustive, so Bayes Theorem does apply. I let $D$ denote the event that an impurity was detected in 2 of 3 tests and $I$ denotes the event that an impurity is present.
$\mathcal P{(I|D)} = \frac{\mathcal P{(D|I)}* \mathcal P{(I)}}{\mathcal P{(D|I)}* \mathcal P{(I)} + \mathcal P{(D|I')}* \mathcal P{(I')}}$
Also, I know that the probabilities for $\mathcal P{(I)} = 0.40$ and $\mathcal P{(I')} = 0.60$
However, at this point I am not sure where to go with the problem. I understand that I need to determine $\mathcal P{(D|I)}$ and $\mathcal P{(D|I')}$ but I'm not sure how to go about doing this..

 

[jedishrfu]

Can you relate the other percent values that is the 80% and 90% to them?

 

[_N3WTON_]

Can you relate the other percent values that is the 80% and 90% to them?

Would I need to use the fact that there are $\dbinom{3}{2}$ ways an impurity could be detected in 2 of 3 tests in order to calculate the needed probabilities? Then I could relate the 80% and 90% somehow, yes?

 

[Stephen Tashi]

Three separate experiments result in only two decisions.

One question is whether the experiments are done on the same sample of product. I suppose we must assume so in order to have an interesting problem.

Another question is: what is the meaning of "two decisions"? It would be clear if the problem said "two detections". I suppose we must assume that interpretation.

Number the experiments 1,2,3. Assume an impurity is present. One way to get two detections in three experiments is a detection in experiment 1, a detection in experiment 2 and a no-detection in experiment 3. Write down the expression for that probability.

Then imagine adding up the expressions for all possible ways of getting two detections in three experiments and you'll see where combinations comes into the picture.

 

[_N3WTON_]

Another question is: what is the meaning of "two decisions"? It would be clear if the problem said "two detections".

I'm really sorry about that, upon reading your comment I went back and realized that I didn't copy the problem correctly, it does indeed say "two detections". Also, I believe that we are meant to assume that the same sample is being used. Anyhow, thank you for the help.

 

[_N3WTON_]

So I think I have found the values I need, hopefully someone can confirm/deny:
$\mathcal P{(D|I)} = \dbinom{3}{2} * (0.8)^{2} * (1-0.8)^{2} = 0.384$
$\mathcal P{(D|I')} = \dbinom{3}{2} * (1-0.9)^{2} * (0.9) = 0.027$
So after substituting these values into the Bayes Theorem equation my final answer is $\mathcal P{(I|D)} = 0.91$

 

[Ray Vickson]

So I think I have found the values I need, hopefully someone can confirm/deny:
$\mathcal P{(D|I)} = \dbinom{3}{2} * (0.8)^{2} * (1-0.8)^{2} = 0.384$
$\mathcal P{(D|I')} = \dbinom{3}{2} * (1-0.9)^{2} * (0.9) = 0.027$
So after substituting these values into the Bayes Theorem equation my final answer is $\mathcal P{(I|D)} = 0.91$

The exact answer is $P(I|D) = 256/283 \doteq 0.9046$. So, yes, I guess you could be right.

 

[_N3WTON_]

The exact answer is $P(I|D) = 256/283 \doteq 0.9046$. So, yes, I guess you could be right.

May I ask how you arrived at the exact answer?

 

[_N3WTON_]

Nvm, I think I figured it out :D