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**samuel.bradley.99****Member**- Registered: 2016-03-16
- Posts: 51

An bag contains 10 black balls and some white ones. We intend to randomly pull one ball out of the bag. Before we do this, one more ball is added in the bag (presumably by someone else, so that we don't see its color). Then we pull out one ball and it is a black one. What is the probability that the extra ball was black?

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**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,322

hi samuel.bradley.99

I think you can do this using Bayes theorem but, as I can never be bothered to learn it, I prefer to make a probability tree diagram.

Working from left to right, we have the initial set up of 10 blacks and x white. Then either a black or white is added shown by the two branches.

Then we choose a ball which may be black or white. So a further two branches are shown for each of the branches already added.

The final outcomes are shown as 'its black' given 'black' and so on.

As we know the final ball is black we only have two branches to consider: P(b|b) and P(b|w). So I think the answer is

Hope that helps,

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**iasonidis****Member**- Registered: 2017-02-11
- Posts: 2

Dear Bob,

My solution is exactly the same, with the only difference that, since the extra ball can be black or white with probability Pb=Pw=1/2, in the fraction, the numerator and the two terms of the denominator are multiplied by 1/2, which in any case is simplified without affecting the final result.

Many thanks for your assistance!

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**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,322

hi iasonidis

Yes, you are correct. I got lazy as it's 1/2 on both choices so it doesn't effect the final answer.

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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