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suppose that bag contains $3$ red and $5$ black ball,each ball one by one,without replacement is selected at random,what is a probability that fourth ball is black?probability from the beginning of red ball is $3/8$ and black one is $5/8$,but after third selection,how could we determine how much black left?we only know that $5$ ball left,but what about amount of black?should we try all of combination?for example if all three is red,then probability of black on fourth selection is $5/5=1$,but if all three is black,then probability is $2/5$,please help me

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up vote 4 down vote accepted

I think what you need to do is to consider the number of arrangements of RRRBBBBB where B denotes a black ball and R denotes a red ball. The number of arrangements is $x=\frac{8!}{3!5!}$. However if we fix B at the 4th position we will have $y=\frac{7!}{3!4!}$ arrangements.

Thus the probability of getting a black ball at the 4th pick is $\frac yx$.

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Very nice answer! –  Eric Auld Aug 4 '13 at 14:16
    
so why fix?what does mean fix in this situation? –  dato datuashvili Aug 4 '13 at 14:21
    
we fix one of the B's in the 4th position so that we are guaranteed to draw a black ball at the 4th pick –  hchengz Aug 4 '13 at 14:22
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what does also $8!/3!*5!\$ express?number of selection of all ball from $8$ ball? –  dato datuashvili Aug 4 '13 at 14:22
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$\frac{8!}{5!3!}$ expresses the number of arrangements of the characters RRRBBBBB –  hchengz Aug 4 '13 at 14:24
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Since you'e picking the balls at random, without replacement, just decide before you start that you'll put the first ball you pick in the #4 slot.

So, the probability of a black in Slot #4 is $\frac{5}{3+5}=0.625$

Or, think of it this way: what are the odds that the first ball you pick is black? Clearly, it's the same $0.625.$ But then, after picking all eight and putting them down in normal 1, 2, 3, ... order, you take the last three and move them (without rearranging) to the front! Now, that black is in Slot #4...

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