Bayes Theorem for two urn draws

Question

I am struggling with the following exercise:

We have two urns. Urn 1 contains 9 black balls and 1 white ball. Urn 2 contains 7 white and 1 black ball. We randomly pick two balls from urn 1 and put them in urn 2 without looking at them. Now we draw a ball from urn 2 at random. Given that it is black calculate the probability that the 2 balls we have taken from urn 1 were both black.

After trying around a bit I strongly suppose that we need Bayes' Theorem here, but I do not get any further. Could you give me a hint?

Answer 1

I call $O$ the (stochastic) variable which refers to the final observed ball and $T=(T_1,T_2)$ the first two taken balls from the first urn and put in the second. $b$ and $w$ are the values black and white that these variables can take.

We want:

$P(T=(b,b)|O=b)=P(O=b|T=(b,b))*P(T=(b,b))/P(O=b) [1]$

It is easy to estimate probabilities for $O$ given $T$:

$P(O=b|T=(b,b))=3/10$

$P(O=b|T=(b,w))=2/10$

$P(O=b|T=(w,b))=2/10$

$P(O=b|T=(w,w))=1/10$

Further:

$P(T=(b,b))=8/10$

$P(T=(b,w))=1/10$

$P(T=(w,b))=1/10$

$P(T=(w,w))=0$

Combining these results:

$P(O=b)=P(O=b|T=(b,b))*P(T=(b,b))+P(O=b|T=(b,w))*P(T=(b,w))+P(O=b|T=(w,b))*P(T=(w,b))+P(O=b|T=(w,w))*P(T=(w,w))=28/100$

So we have all elements to insert in [1].

If I am not mistaken we have:

$P(T=(b,b)|O=b)=3/10*8/10/(28/100)=6/7$

Answer 2

$ P(B)=P(B|B1\cap B2)P(B2|B1)P(B1)+P(B|B1\cap W2)P(W2|B1)P(B1)+P(B|W1\cap B2)P(B2|W1)P(W1)$

Then the answer you are seeking is $\cfrac{P(B|B1\cap B2)P(B2|B1)P(B1)}{P(B)}$

with Bi=black ball taken from urn 1 in turn i; Wi=white ball taken from urn 1 in turn i.

B is the event that the ball taken from urn 2 is black

Answer =$\cfrac{6}{7}$