# Probability problem (Urn 1 contains 3 white and 4 black balls, and Urn 2 contains 2 white and 6 black balls ...)

I'm studying probability. This is not homework. I have been studying for a graduate master's since September 2015. The textbook is Probability : An Introduction (Grimmett & Welsh).

You are presented with two urns. Urn 1 contains 3 white and 4 black balls, and Urn 2 contains 2 white and 6 black balls. (a) You pick a ball randomly from Urn 1 and place it in Urn II. Next you pick a ball randomly from Urn 2. What is the probability that the ball is black?

Prob of picking black from the Urn 1 = P(B1) = 4/7
Prob of picking white from the Urn 1 = P(W1) = 3/7
Prob of picking black from the Urn 2, after placing the picked ball in it = P(B2) = ?

B1 and W1 are a partition of Omega so I can use the partition theorem:

P(B2) = P(B2|B1)*P(B1) + P(B2|W1)*P(W1)

I intuitively know that P(B2|B1) = 7/9, because, given that I picked a black ball from the Urn 1 and placed it in the Urn II, now I have 7 possibility of success and 9 possible outcomes.

I intuitively know that P(B2|W1) = 6/9 for the same reason.

Then P(B2) = 7/9*4/7+6/9*37 = 0.73

I would like to know if my reasoning is correct and what is the formal way to show that P(B2|B1) = 7/9 and P(B2|W1) = 6/9.

• Please add the [self-study] tag & read its wiki. Feb 6, 2015 at 15:50

The formality you request uses conditional probabilities.. A probability tree is a good way to reason with them. Use it as a guide to the formal demonstration.

Let's begin at the end: what is the chance the ball is black? It depends on what happened before. That's the conditioning event. So:

• Let $$\mathcal B$$ be the event that the first ball is black and $$\mathcal B ^\prime$$ be the event it is not black. Evidently $$\Pr(\mathcal B) = 4/(3+4) = 4/7$$ and $$\Pr(\mathcal B^\prime) = 1 - 4/(3+4) = 3/7.$$

• Let $$\mathcal C$$ be the event that the ball selected from the second urn is black. When $$\mathcal B$$ occurs, the second urn has $$2$$ non-black and $$6+1$$ black balls. Otherwise, when $$\mathcal B^\prime$$ occurs, the second urn has $$2+1$$ non-black and $$6$$ black balls. Consequently

$$\Pr(\mathcal C\mid \mathcal B) = \frac{6+1}{2 + (6+1)} = \frac{7}{9}$$

and

$$\Pr(\mathcal C\mid \mathcal B^\prime) = \frac{6}{(2+1) + 6} = \frac{6}{9}.$$

Because $$\mathcal B$$ and $$\mathcal B^\prime$$ are disjoint and partition the initial outcomes, the Law of Total Probability asserts

$$\Pr(\mathcal C) = \Pr(\mathcal C\mid \mathcal B)\Pr(\mathcal B) + \Pr(\mathcal C\mid \mathcal B^\prime)\Pr(\mathcal B^\prime) = \frac{7}{9}\times \frac{4}{7} + \frac{6}{9}\times \frac{3}{7} = \frac{28 + 18}{63} = \frac{46}{63}.$$