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The problem is the following: You have 10 balls:6 of them are black and 4 are white. 3 balls are removed from a box that have them,but their colors are not observed. What is the probability that the fourth ball removed is WHITE?all balls have the same probability to be removed. this last information made me get different answers...

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Ops,I´m so very sorry!!!!4=WHITE 6=BLACK.THANKS FOR NOTING!!!!!!!!!!!!!!! – HipsterMathematician May 27 '12 at 20:03
Now they match. – HipsterMathematician May 27 '12 at 20:14
up vote 1 down vote accepted

The probability is $\dfrac{4}{10}$.

Maybe think of it this way. Label the balls from $1$ to $10$, say $1$ to $4$ for the whites, $5$ to $10$ for the blacks, and mix thoroughly. Now remove the balls, one at a time. Since all orders of labels are equally likely, the probability that the $k$-th ball removed is (say) the ball with label $1$ is $\frac{1}{10}$. The probability that the $k$-th ball removed is the ball with label $2$ is also $\frac{1}{10}$, and so on. So the probability that the $k$-th ball removed is white is $\frac{4}{10}$.

The point is that any of the balls is as likely to be in position $k$ of a permutation as any other.

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At first ,I got 2/5...please explain! – HipsterMathematician May 27 '12 at 19:54
@MarylinM.J: You could be right. In the title $6$ are white. In the body, $4$ are white. In the solution I wrote out, I went with the numbers in the title. Have changed answer to go with numbers in the body. Now get $4/10$. Please change title or body so that they are consistent! – André Nicolas May 27 '12 at 20:06
Oh,now i got it!I was having more work to calculate it...and for some reason i had forgot about that last information and i got different answers.THANKS ! – HipsterMathematician May 27 '12 at 20:10
Just one more consideration:wouldn´t you have to considerate the fact that white balls could have been removed? – HipsterMathematician May 27 '12 at 20:16

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