three bags with black and white balls problem

There are three bag containing white and black balls. The first bag contains 2 white balls, second bag contains 2 black balls and third bag contains 1 black and 1 white balls. A bag is chosen in random and a ball is drawn from it. The drawn ball is white. The process is repeated without including the drawn ball. What is the probability that ball drawn in second round is white?

Probability of getting white ball in first attempt: (1/3)(1) + (1/3)(1/2) = 1/2

Probability of getting white ball in second attempt:

(i) First white ball was drawn from bag of 2 white balls ((1/3)(1/1) + (1/3)(1/2))

(ii) First white ball was drawn from 1 white and 1 black ball bag ((1/3)*(2/2))

Total probability of drawing white ball second time ((1/3)(1) + (1/3)(1/2)) + ((1/3)*(1))

To get them in sequence = (1/2)*(5/6) = 5/12

I dont know the answer and hence can not confirm if my attempt above is correct. Please clarify on the solution.

• Have added my attempt. Jun 17, 2019 at 9:27
• Hey, I'm just a bit confused: you're not asking any question... Do you simply want us to confirm whether this is correct? Jun 17, 2019 at 9:33
• Yes I dont know the answer and wanted clarification on the solution. I have mentioned my attempt as @MichaelRozenberg asked. Sorry about the ambiguity. Fixed the question accordingly. Jun 17, 2019 at 9:39
• Is the second ball taken out of the bag that also contained the first ball that was taken? Jun 17, 2019 at 10:28

I preassume that all bags contain exactly $$2$$ balls and that at random a bag is chosen from which the $$2$$ balls are taken one by one (if wrong then please tell me, so that I can delete the answer).

The first bag contains $$2$$ white balls and the question can be rephrased as:"If a white ball was selected at first draw then what is the probability that this ball was located in the first bag?"

There are exactly $$3$$ white balls in total and each of them has equal probability to be the ball that was selected at first draw. $$2$$ of these balls are located in a bag that contains another white ball and $$1$$ of them is located in a bag that does not contain another white ball.

So the probability that one of the $$2$$ balls located in a bag that contains another white ball was selected by first draw equals $$\frac23$$.

This event is the same as the event that the second draw will result in a white ball.

edit1:

If the above interpretation is wrong and the second ball can be chosen out each of the $$3$$ bags then the probability that the second ball is white is $$\frac25$$.

This because at the second round there are $$5$$ balls in total (all having equal probability to be chosen) of which $$2$$ are white.

edit2

If both interpretations above are wrong and by the second round each bag has the same probability to be chosen then the following calculation:

The probability that after drawing the first ball (which appeared to be white) we are in situation $$|W\mid WB\mid BB|$$ (i.e. one bag contains a white ball, one contains a white and a black ball and the third contains $$2$$ black balls) is $$\frac23$$ (i.e. the probability that the first ball was taken from the bag containing $$2$$ white balls; see first interpretation for that).

In this situation the probability that the second balls is white is $$\frac13\cdot1+\frac13\frac12+\frac13\cdot0=\frac12$$.

The probability that after drawing the first ball we are in situation $$|WW\mid B\mid BB|$$ is $$1-\frac23=\frac13$$.

In this situation the probability that the second balls is white is $$\frac13\cdot1+\frac13\cdot0+\frac13\cdot0=\frac13$$.

We conclude that the probability that the second ball is white is:$$\frac23\frac12+\frac13\frac13=\frac49$$

• The ball is chosen randomly in two rounds. I have re edited the question. Jun 17, 2019 at 11:06
• If the "process is repeated" and the second ball can be chosen out each of the $3$ bags then the probability that the second ball is white is $\frac25$. This because at the second round there are $5$ balls in total (all having equal probability to be chosen) of which $2$ are white. Jun 17, 2019 at 11:09
• I was under the impression that there is 1/3 chance of choosing a bag. Chances of choosing white ball in first attempt = (1/3)(probability of getting white ball from bag of 2 white balls) or(+) (1/3)(probability of getting white ball from bag of 1 white and 1 black ball). Jun 17, 2019 at 11:20
• That is correct for the first ball: $\frac13\cdot1+\frac13\cdot\frac12=\frac12$. Things are different for the second ball if one ball (a white one) is taken away. Jun 17, 2019 at 11:24
• Thanks for the answer. I would really appreciate if you could add comment to my question pointing the mistake in my calculation. Jun 17, 2019 at 11:35

$${Urn}_{x}$$ => 2 White Balls, $${Urn}_{y}$$ => 1 White and 1 Black Balls. $${Urn}_{z}$$ => 2 Black Balls,

P($${Urn}_{x}$$) = P($${Urn}_{y}$$) = P($${Urn}_{z}$$) = $$\frac{1}{3}$$

There can be 2 cases.

Case 1: Both the Balls are from $${Urn}_{x}$$.Both times, the Urns are selected Randomly. Take a look at Diagram 1. Diagram 1 contains another but same sort of problem, which shall make you understand this problem better. The only difference is, Here, we shall multiply with another $$\frac{1}{3}$$, as according to this problem, both the times, Urns are selected randomly. Therefore the equation for this problem will be = $$\frac{1}{3}(\frac{2}{3})$$ = $$(\frac{2}{9})$$

Case 2: 1 ball from $${Urn}_{x}$$ and 1 ball from $${Urn}_{y}$$. here the Equation will $$\frac{1}{3}W_{x}+\frac{1}{3}W_{y} = (\frac{1}{3}).(\frac{1}{3})(1 + 1) = \frac{2}{9}$$.Both times, the Urns are selected Randomly. Therefore, we have multiplied with $$\frac{1}{3}$$ both $$W_{x}$$ and $$W_{y}$$.

Now The Final Equation and answer : Case 1 + Case 2 = $$\frac{2}{9} + \frac{2}{9}$$ = $$\frac{4}{9}$$

Please take a look at this diagram also. This is a very straightforward and the easiest method to solve this problem.