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There is a vase with $6$ red and $4$ black balls. We choose in a random way $5$ balls, and let $X$ to be the number number of red balls from those we picked. Determine the probability function of $X$, if the sampling done with repositioning.

I am not going through the whole procedure that I followed. So I am just posting my results, and I would be extremely grateful if someone could tell me if its correct and if I should re write in a better way.

So $$f(x) = \begin{cases}0.01024 &x=0 \\ 0.01536\cdot a &x=1 \\ 0.02304\cdot b &x=2 \\ 0.03456\cdot c &x=3 \\ 0.05184\cdot d &x=4 \\ 0.0776 &x=5 \\ \text{else } 0 \end{cases}$$

where $a,\ b,\ c,\ d$ is the number of times we can find these events happening.

So what do you think pals?

Thanks!

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    Does repositioning mean replacement since "sampling with replacement" is a commonly-used term in probability theory? and if not, could you explain it?2012-03-31
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    @Dilip Sarwate , Hey thanks for your responce. Anyway, by repositioning, i mean that when we pick one ball , then we put it back into the vase again and we pick another (what is this called by the way?). Thanks2012-03-31
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    @Mat I've merged your accounts, please feel free to ask for moderator help if you have future troubles logging in.2012-04-01
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    This is _almost_ sampling with replacement which calls for putting the ball back into the vase (usually called an urn) **and then shaking well** before picking another ball. Note that the ball picked on the next trial _could_ very well be the ball picked on the previous trial; it does not have to be _another_ ball in the sense of different from the first. Shaking well is often not mentioned but is very important practically; else the ball picked previously will quite likely be picked again because it will be in the top layer of balls in the vase.2012-04-01
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    It would be better to give your results in formula terms rather than decimals. For example, your result for $x=0$ is $(\frac 4{10})^5$. This makes the logic more clear and easier to assess.2012-06-03

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