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I'm sure there must be a problem like this posted already, apologize for the duplicate- all the names are so similar I can't find one sifting through. The problem goes like this:

Urn 1 contains 3 white balls and one red ball. Urn 2 contains 2 white balls and 3 red balls.

a) An urn is chosen at random, and a ball is drawn from it at random. If the ball drawn is R, what is the probability the urn now contains no red balls?

answer: .25, which I'm pretty confident is right. Feel free to correct me.

This is the part where I'm stuck.

b) An urn is is chosen at random which will give a ball, and the other one will receive that ball. Then, a ball is drawn at random from the urn the received a ball. Given that this ball drawn is red, what is the probability that the donating urn now contains no red balls?

It seems like this should also just be $\frac13$, since there are only 3 balls that can be drawn initially resulting in in an urn with 0 reds; the 2 whites drawn from urn 2, or the red drawn from urn 1. Is this correct?

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  • $\begingroup$ A quick calculation gives me a different number than yours on the first problem. Note that in the current post, there are $2$ white and $3$ red in the second urn. We can use informal reasoning, or a formal conditional probability calculation. $\endgroup$ – André Nicolas Dec 12 '13 at 22:43
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Assuming that choosing an urn at random means choosing each with probability $\frac12$, the probability of drawing a red ball from Urn $1$ is $\frac12\cdot\frac14=\frac18$. The probability of drawing a red ball from Urn $2$ is $\frac12\cdot\frac35=\frac3{10}$. Given that you’ve drawn a red ball, the probability that it came from Urn $1$ is therefore $$\frac{\frac18}{\frac18+\frac3{10}}=\frac{\frac18}{\frac{17}{40}}=\frac5{17}\;,$$ not $\frac14$.

For (b), look at the possible sequences of events.

  • You draw the first ball from Urn $1$, it’s red, and then you draw a red ball from Urn $2$: the probability of this is $\frac12\cdot\frac14t\cdot\frac23$. (Why?)
  • You draw the first ball from Urn $1$, it’s white, and then you draw a red ball from Urn $2$: the probability of this is $\frac12\cdot\frac34\cdot\frac12$. (Why?)
  • You draw the first ball from Urn $2$, it’s red, and then you draw a red ball from Urn $1$; what’s the probability of this?
  • You draw the first ball from Urn $2$, it’s white, and then you draw a red ball from Urn $1$; what’s the probability of this?

The sum of these four probabilities is the probability of drawing a red ball the second time. What fraction of this probability comes from cases that leave the donating urn without red balls?

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  • $\begingroup$ for part 1, I thought that was how it worked, but I was specifically corrected to the way I did it earlier in the year. Are you sure? $\endgroup$ – Nick Dec 12 '13 at 22:51
  • $\begingroup$ @Nick: Yes, assuming that my interpretation of the problem is correct. What calculations/reasoning did you use to get $\frac14$? $\endgroup$ – Brian M. Scott Dec 12 '13 at 22:56
  • $\begingroup$ There are 4 red balls in total. The selection of an urn at random is inconsequential; it's just 1 of the 4 red balls. that's how I got to 1/4. $\endgroup$ – Nick Dec 12 '13 at 23:08
  • $\begingroup$ @Nick: But the $4$ red balls aren’t equally likely to be chosen. The probability of choosing any specific ball in Urn $1$ is the probability of choosing Urn $1$ and then of choosing that ball, or $\frac12\cdot\frac14=\frac18$. The probability of choosing any specific ball in Urn $2$ is similarly $\frac12\cdot\frac15=\frac1{10}$. $\endgroup$ – Brian M. Scott Dec 12 '13 at 23:13
  • $\begingroup$ so if the jars had an equal number of balls in them (which was the case in the other problem), then that would be how you do it? $\endgroup$ – Nick Dec 13 '13 at 0:09

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