Consider two boxes: there are 15 white and 12 black balls in the first box, and 14 white and 18 black balls in the second box. Anna puts her hand in the first box, takes at once two balls and places them in the second box. Then, she takes one ball wihout looking from the second box.

Knowing that she took a black ball from the second box, what is the probability that she transferred two balls of different colors from the first box to the second box?

I tried using Bayes Theorem:

X:=transfer 2 colours Y:= pick black from the 2nd box






Does this work like this?

  • $\begingroup$ I have $P(X)=\frac{\binom{15}{1}\cdot \binom{12}{1}}{\binom{27}{2}}=\frac{20}{39}$ $\endgroup$ – callculus Mar 22 '19 at 18:53
  • $\begingroup$ @callculus Oh, I somehow read 'from the same color' instead of 'different color'. And the rest is okay? $\endgroup$ – user655181 Mar 22 '19 at 18:59
  • $\begingroup$ The rest cannot be ok if something is wrong. Please make an edit. I think you are on the right track. But I or someone else can confirm your solution only if you make an edit. $\endgroup$ – callculus Mar 22 '19 at 19:02
  • $\begingroup$ And for $P(Y)$ you have to apply the total law of probability. $\endgroup$ – callculus Mar 22 '19 at 19:06

To summarize callculus' suggestion:

X1: 1b(black), 1w(white) from the 1st box. X2: 2b from 1st. X3: 2w from 1st.

Y: 1b from the 2nd box.

Then P(X1|Y)=$\frac{P(Y|X1)P(X1)}{P(Y)}$, where:

P(Y|X1)= $\frac{19}{34}$

P(Y) = $\sum_{i=1}^3 P(Y|Xi)P(Xi)$

P(X1)= $\frac{C_1^{15}C_1^{12}}{C_2^{27}}$

P(X2)= $\frac{C_2^{15}}{C_2^{27}}$

P(X3)= $\frac{C_2^{12}}{C_2^{27}}$

Hope I didn't type something wrong. You should be able to figure out the computations.

BTW, when you find any probability greater than 1, something is definitely wrong.

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You can use simple conditional probability i.e.:

$$P(X|Y)=\frac{P(X\cap Y)}{P(Y)}$$

Where in this case it would be:

$$P(X\cap Y) = P(\text{transfer diff colours})*P(\text{picking black|transfered diff colours})$$ $$P(X\cap Y) = \left( 2. \frac{15}{27}\frac{12}{26}\right)\frac{19}{34} = \frac{190}{663}$$

for $P(Y)$ we can use total probability:

$$P(Y) = \left( \frac{12}{27}\frac{11}{26}\right) \frac{20}{34}+ \left( \frac{15}{27}\frac{14}{26}\right)\frac{18}{34} + \left(2. \frac{15}{27}\frac{12}{26}\right)\frac{19}{34} = \frac{5}{9}$$

Result is therefore: $P(X|Y)=\frac{114}{221}\approx0.5158..$

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I would say

$P(X) = \frac{20}{39}$ (see colleagues 'calculus' comment)


Knowing that she took a black ball from the second box: $\Rightarrow P(Y)=1$

$ \Rightarrow P(X|Y)=\frac{P(Y|X)P(X)}{P(Y)}=\frac{P(Y|X)P(X)}{1}=\cdots$

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