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Sour Milk Problem

OhMyMarkov

Member
Mar 5, 2012
83
Hello everyone!

I'm confusing conditional and joint probability again, here is the problem:

Given 4 bottles of milk, one of them is sour. We need to find out which of the bottles is sour by tasting the least number of bottles. By inspection, we can roughly tell the probability of every bottle being sour. Therefore we rank the bottles in non-ascending order of probability that the bottle is sour. Assume that these probabilities are $p_1, p_2, p_3,$ and $p_4$.

We also define an RV $X$ that is equal to the number of bottles we taste before determining what bottle is sour: $X\in\{1,2,3\}$.

Here is where I'm confused:

$P\{X=1\}=P\{b_1 \text{bad}\}=p_1$

Now, which of the following is correct?

(1) $P\{X=2\}=P\{b_2 \text{bad} \; | \; b_1 \text{good}\}$, $P\{X=3\}=P\{b_3 \text{bad} \;|\; b_1, b_2 \text{good}\}$
(2) $P\{X=2\}=P\{b_2 \text{bad and} \; b_1 \text{good}\}$, $P\{X=3\}=P\{b_3 \text{bad} \; \text{and} \; b_1 \text{good and} \; b_2 \text{good}\}$

Thanks!:eek:
 
Last edited:

Jameson

Administrator
Staff member
Jan 26, 2012
4,043
Hi there,

Can you clean up your notation please? It's hard to follow and I see you tried to use a Latex command but it's not rendering. The way I see this written most often is in the following way \(\displaystyle P \left[ X=x \right]\). To see how I wrote that you can quote this post and copy the Latex code. I think if you make a couple quick edits the question will be much more likely to be answered.

Jameson
 

Jameson

Administrator
Staff member
Jan 26, 2012
4,043
I've been thinking over this for a bit and I'm not certain I have a full solution but hopefully some helpful comments.

$P[B_n]$ is the probability that the sour milk is located in the nth container. From the OP the order of the bottles is always $(B_1, B_2, B_3,B_4)$ with corresponding probabilities of $P[B_1] < P[B_2] < P[B_3]<P[B_4]$. The sum of these four probabilities is 1.

$P[X=1]=P[B_1]$ if the first bottle is sour.

What does $P[X=2]$ mean? It means that the first bottle was not sour and the second one was. You could write this as $P[B_2|(B_1)']$ but this isn't necessary because there can be just one success in the experiment before it's finished. All of the probabilities correspond to the three other bottles not being sour by the conditions of the situation so because of that I don't think finding $P[X=2,3,4]$ will involve conditional formulas.

When X=2 the only way that happens is bottle 1 to not be sour and then bottle two is so it seems that \(\displaystyle P[X=2]=P[(B_1)'] \cap P[B_2]\) or the second choice you posted.

This seems correct but I wouldn't be shocked if I have not accounted for something. Forcing the order that we choose the bottles is a restriction so it makes sense that the probabilities where failures are counted before the success should be smaller than the chance of it just being sour.

Hopefully there is something useful here for you and sorry in advance if there turn out to be errors.