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richardwander
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Hi.
I've done a search already to see if this question has been answered but haven't found anything. Hope it's okay to ask it here.
I have this brainteaser that's been bugging me a little bit. It's not homework, but is a question that I heard from someone who's already had an interview at a job I'm applying for. It was asked in the interview and goes like this:
You have three pancakes. One of them is burned on both sides, one of them is burned on one side, and one of them is perfectly cooked and burned on no side. You stack the pancakes on a plate and note that the visible side of the top pancake is burned. What is the probability that the top pancake in the stack is burned on both sides?
To me, this sounds like a Bayes' rule problem. I define the event X to be that the top pancake is burned on both sides, and the event Y to be that the top pancake is burned on at least one side. I then apply Bayes' rule to get
[tex]P(X|Y) = \frac{P(Y|X)P(X)}{P(Y)}[/tex]
I have that P(Y|X) = 1 since if the top pancake is burned on both sides, the probability that it's burned on at least one side is 1.
I also have that P(X) = 1/3 since the marginal probability that the top pancake is burned on both sides is one in three.
I calculate the marginal probability P(Y) as follows:
[tex]P(Y) = P(Y|\textrm{no sides})P(\textrm{no sides}) + P(Y|\textrm{one side})P(\textrm{one side}) + P(Y|\textrm{both sides})P(\textrm{both sides})[/tex]
where the conditionals refer to the burned sides on the pancakes. I find that
[tex]P(Y) = 0\times \frac{1}{3} + 1\times\frac{1}{3} + 1\times\frac{1}{3} = \frac{2}{3}[/tex]
This means that the probability that the top pancake is burned on both sides given that we know it's burned on at least one side is given by
[tex]P(X|Y) = \frac{1\times\frac{1}{3}}{\frac{2}{3}} = \frac{1}{2}[/tex]
However, I don't think this is correct. Intuitively, it seems that the answer should be 2/3. Can anyone point out where I'm going wrong?
I've done a search already to see if this question has been answered but haven't found anything. Hope it's okay to ask it here.
I have this brainteaser that's been bugging me a little bit. It's not homework, but is a question that I heard from someone who's already had an interview at a job I'm applying for. It was asked in the interview and goes like this:
You have three pancakes. One of them is burned on both sides, one of them is burned on one side, and one of them is perfectly cooked and burned on no side. You stack the pancakes on a plate and note that the visible side of the top pancake is burned. What is the probability that the top pancake in the stack is burned on both sides?
To me, this sounds like a Bayes' rule problem. I define the event X to be that the top pancake is burned on both sides, and the event Y to be that the top pancake is burned on at least one side. I then apply Bayes' rule to get
[tex]P(X|Y) = \frac{P(Y|X)P(X)}{P(Y)}[/tex]
I have that P(Y|X) = 1 since if the top pancake is burned on both sides, the probability that it's burned on at least one side is 1.
I also have that P(X) = 1/3 since the marginal probability that the top pancake is burned on both sides is one in three.
I calculate the marginal probability P(Y) as follows:
[tex]P(Y) = P(Y|\textrm{no sides})P(\textrm{no sides}) + P(Y|\textrm{one side})P(\textrm{one side}) + P(Y|\textrm{both sides})P(\textrm{both sides})[/tex]
where the conditionals refer to the burned sides on the pancakes. I find that
[tex]P(Y) = 0\times \frac{1}{3} + 1\times\frac{1}{3} + 1\times\frac{1}{3} = \frac{2}{3}[/tex]
This means that the probability that the top pancake is burned on both sides given that we know it's burned on at least one side is given by
[tex]P(X|Y) = \frac{1\times\frac{1}{3}}{\frac{2}{3}} = \frac{1}{2}[/tex]
However, I don't think this is correct. Intuitively, it seems that the answer should be 2/3. Can anyone point out where I'm going wrong?