# 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.

• Show please your attempts. – Michael Rozenberg Jun 17 at 9:11
• Have added my attempt. – Udhai kumar Jun 17 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? – Milloupe Jun 17 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. – Udhai kumar Jun 17 at 9:39
• Is the second ball taken out of the bag that also contained the first ball that was taken? – drhab Jun 17 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. – Udhai kumar Jun 17 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. – drhab Jun 17 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). – Udhai kumar Jun 17 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. – drhab Jun 17 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. – Udhai kumar Jun 17 at 11:35