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swmi
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Interview Question about conditional probability

June 19th, 2014, 8:11 am

HiI was recently asked this question in a phone interview and I am still not sure why the answer is what the interviewer said it was.Here is the question"You have two bags. One bag has 3 white and 1 black ball and the other bag has 1 white and 3 black balls. You select one of the bags at random (bags have equal probability of being selected) and pick a ball. This ball is black. You then replace the ball. What is the probability that if you pick a ball again it is black."To be the first and second events are independent and therefore have the same probability of 1/2. The interviewer, however, said they are not and the second event has higher probability than the first. Can someone explain why?Thanksswmi
 
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secret2
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Interview Question about conditional probability

June 19th, 2014, 9:02 am

The answer should be 5/8. Think Bayesian.
 
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swmi
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Interview Question about conditional probability

June 19th, 2014, 9:05 am

But why are they not independent? The first ball that was picked is replaced.
 
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secret2
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Interview Question about conditional probability

June 19th, 2014, 9:11 am

Because conditional on the ball being black, the probabilities of bag A and bag B being chosen are not equal
 
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swmi
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Interview Question about conditional probability

June 19th, 2014, 9:41 am

Ok, even though I am finding it hard to convince myself, am I right in thinking that these would be the steps to solve it:Let A: event of selecting first bagB: even of selecting the second bagX: first selection of black ball (with replacement)Y: second selection of black ball (with replacement)P(X) = 1/2 * 1/4 + 1/2 * 3/4 = 1/2P(A|X) = P(X|A) / P(X) = 1/4P(B|X) = P(X|B) / P(X) = 3/4P(Y|X) = P(A|X) * 1/4 + P(B|X) * 3/4 = 1/16 + 9/16 = 5/8
 
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secret2
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Interview Question about conditional probability

June 19th, 2014, 10:02 am

Looks correct other than P(A) and P(B) missing in the denominators.
 
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Vanubis1
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Interview Question about conditional probability

June 20th, 2014, 12:55 pm

It's not clearly written in the first topic that you pick the second ball in the same bag.
Last edited by Vanubis1 on June 19th, 2014, 10:00 pm, edited 1 time in total.
 
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katastrofa
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Interview Question about conditional probability

June 22nd, 2014, 6:46 pm

If it weren't the same bag in the second draw, the information about the bags (and the bags themsevles) would be redundant so it's 5/8. I'm not sure how the OP got this answer from their (incorrect) formulas above, though.a
Last edited by katastrofa on June 21st, 2014, 10:00 pm, edited 1 time in total.
 
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bluetrin
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Interview Question about conditional probability

June 23rd, 2014, 2:50 pm

edit ...
Last edited by bluetrin on June 22nd, 2014, 10:00 pm, edited 1 time in total.
 
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MattF
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Interview Question about conditional probability

June 26th, 2014, 9:05 am

The chance of the first ball being black is three times greater when the three-black bag is selected so will contribute 3/4 of the instances it occurs. The chance of the second ball being black is therefore 3/4 * (chance of drawing black when black bag selected) + 1/4 * (chance of drawing black when white bag selected) = 3/4 * 3/4 + 1/4 * 1/4 = 5/8.