Urn of unknown composition with black and white balls exercise

E

ele

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Consider an urn of unknown composition containing 5 balls that can be white or black, but not all of the same color. Calculate the probability that the urn contains 3 white balls, assumed that in 3 draws without return you get 3 white balls.




My thoughts:




There are 4 different ways the urn can be:
- 4 White 1 Black
- 3 White 2 Black
- 2 White 3 Black
- 1 White 4 Black




Let the random variable \(\displaystyle A = \text{"The urn contains 3 white balls"}\) and \(\displaystyle B = \text{"3 white balls are drawn with 3 draws without return"}\), we need to find \(\displaystyle P(A|B)\).




Using Bayes' rule we need to find \(\displaystyle P(A|B) = \frac{P(B|A)P(A)}{P(B)} = \frac{{\frac{1}{{4}}{\left(\frac{3}{{5}}\cdot\frac{2}{{4}}\cdot\frac{1}{{3}}\right)}}}{{\frac{1}{{4}}{\left(\frac{4}{{5}}\cdot\frac{3}{{4}}\cdot\frac{2}{{3}}\right)}+\frac{1}{{4}}{\left(\frac{3}{{5}}\cdot\frac{2}{{4}}\cdot\frac{1}{{3}}\right)}}} = \frac{1}{5}\)

Is my method correct?

-Elle
 
Consider an urn of unknown composition containing 5 balls that can be white or black, but not all of the same color. Calculate the probability that the urn contains 3 white balls, assumed that in 3 draws without return you get 3 white balls.




My thoughts:




There are 4 different ways the urn can be:
- 4 White 1 Black
- 3 White 2 Black
- 2 White 3 Black
- 1 White 4 Black




Let the random variable \(\displaystyle A = \text{"The urn contains 3 white balls"}\) and \(\displaystyle B = \text{"3 white balls are drawn with 3 draws without return"}\), we need to find \(\displaystyle P(A|B)\).




Using Bayes' rule we need to find \(\displaystyle P(A|B) = \frac{P(B|A)P(A)}{P(B)} = \frac{{\frac{1}{{4}}{\left(\frac{3}{{5}}\cdot\frac{2}{{4}}\cdot\frac{1}{{3}}\right)}}}{{\frac{1}{{4}}{\left(\frac{4}{{5}}\cdot\frac{3}{{4}}\cdot\frac{2}{{3}}\right)}+\frac{1}{{4}}{\left(\frac{3}{{5}}\cdot\frac{2}{{4}}\cdot\frac{1}{{3}}\right)}}} = \frac{1}{5}\)

Is my method correct?

-Elle
So we know that the urn contains at least 3 white balls out of 5 balls and want to know if the urn contains exactly 3 white balls. I would restate this problem as: an urn has 2 balls (5-3=2) which are white or black but not both white. What is the probability that the urn has 0 white balls. This 'new' problem would be easier to solve don't you think? What answer do you get?
 
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I think we can have one of the following:
-BB
-WB or BW

So P(0 white) = P(2 black) = 1/2 ? :confused:
 
Oh ok I thought WB and BW were the same... at this point yes, P(0 whites) should be 1/3
 
Jomo, you think it is correct, for the original exercise, \(\displaystyle P(B|A)=(\frac{3}{5}\cdot\frac{2}{4}\cdot\frac{1}{3})=\frac{1}{10}\) ? It's the only way I can "view" this piece of Bayes' rule.

For \(\displaystyle P(A)\) and \(\displaystyle P(B)\) I'm not too sure, but I think \(\displaystyle P(A)=\frac{1}{4}\) since there is only \(\displaystyle 1\) urn out of \(\displaystyle 4\) with exactly \(\displaystyle 3\) white balls, and the probability of picking an urn is equiprobable.
For \(\displaystyle P(B)\), instead, I would rewrite it using the law of total probability: there are 2 urns which have at least 3 white balls, and picking one has \(\displaystyle \frac{1}{4}\) probability. Probability of picking 3 white balls from the urn with 3 white balls and 2 black balls is \(\displaystyle \frac{3}{5}\cdot\frac{2}{4}\cdot\frac{1}{3}=\frac{1}{10}\), whilst probability of picking 3 white balls from the urn with 4 white balls and 2 black balls is \(\displaystyle \frac{4}{5}\cdot\frac{3}{4}\cdot\frac{2}{3}=\frac{2}{5}\). But you need to weigh the probabilities with the probability of picking that urn so \(\displaystyle P(B)=\frac{1}{4}\cdot\frac{1}{10}+\frac{1}{4}\cdot\frac{2}{5}=\frac{1}{8}\). So the probability asked should be \(\displaystyle \frac{1}{5}\). What do you think?
 
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My professor said it's \(\displaystyle \frac{1}{5}\) is she wrong?
 
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