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I don't know why I can't solve this problem using Polya's method .$$$$ There are two boxes,one of which contains 5 black and 7 white marbles and the other have 2 black and 2 white marbles .randomly choosing two marbles form the first mentioned box (without seeing the colors ) and putting them in the second box.now we choose two marbles out of the second box,find the probability that both of them are white. $$$$ probability that the both chosen marbles form the first box are white

p(2W1)=$ \frac{\binom 72}{\binom {12}2}$
probability that the both chosen marbles form the first box are black
p(2B1)=$ \frac{\binom 52}{\binom {12}2}$
probability that the chosen marbles form the first box are white and black

p(W1B1)=$ \frac{\binom 71\binom 51}{\binom {12}2}$

and by solving this : p(2W2)=p(2W2|2W1)p(2W1)+p(2W2|2B1)p(2B1)+p(2W2|W1B1)p(W1B1)

P(2W2)=$\frac{31}{165}$

I get different answer when solving the problem using Polya's method that's
$ \frac{\binom 22}{\binom42}$

mahyar p
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1 Answers1

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This is wrong:

p(W1B1)=p(2W1)=$ \frac{\binom 71\binom 51}{\binom {12}2}$

The probability that you choose one black and one white marble (from the first box) is the sum of the the probability that you choose a black marble first and then a white one and the probability that you choose a white marble first and then a black one:

$$\frac 5 {12}\times \frac 7{11}+\frac 7{12}\times \frac5{11}.$$

You can calculate the rest.

zoli
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  • the method used for this kind of of problem ( choosing marbles from the first box without knowing the color ) . why? because that's simpler . – mahyar p Apr 13 '19 at 10:05
  • $\dfrac{\binom 71\binom 51}{\binom {12}2}=\dfrac 5 {12}\times \dfrac 7{11}+\dfrac 7{12}\times \dfrac5{11}$ – Henry Apr 13 '19 at 14:14
  • thanks but that was a typo. I'm sure about my final answer from the first method that's 31/165 . The problem is I can't get the same answer using Polya's method – mahyar p Apr 14 '19 at 09:30
  • You will be very angry, but I still don't know whhat Pólya's method is... – zoli Apr 14 '19 at 19:05
  • No I appreciate your helps. I'm referring to the method that says choosing the marbles from the first box with close eyes and adding them to the second box wont affect the probability of choosing marbles from the second box https://math.stackexchange.com/questions/378810/a-problem-on-polyas-urn-scheme – mahyar p Apr 16 '19 at 07:21