# Probability of picking various numbers of red balls from a vase of red and black balls

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? – Dilip Sarwate Mar 31 '12 at 22:48
@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?). Thanks – Mat Sullivan Mar 31 '12 at 23:28
@Mat I've merged your accounts, please feel free to ask for moderator help if you have future troubles logging in. – Zev Chonoles Apr 1 '12 at 2:03
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. – Dilip Sarwate Apr 1 '12 at 2:10
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. – Ross Millikan Jun 3 '12 at 5:45

With replacement, this is a Binomial distribution (let's call it X) with 5 trials (n=5) and probability 0.6 (p=0.6) We might say X~Bin(5,0.6)

A binomial distribution occurs when we do a fixed number of independent "trials" with identical probability, the trials have a success or failure result (success = get a red ball), and we count the number of results.

If you didn't replace the balls, it wouldn't be a binomial distribution, because the probabilities wouldn't be identical and the trials wouldn't be independent.

Some of your probabilities don't really match the correct result, although it's clear you have partially the correct numbers.

You should calculate them like this:

$$P(X=2) = 5C2 * (0.6)^2 * (0.4)^3 = 0.2304$$

Where 5C2 is the binomial coefficient, which your calculator should be able to deal with.

It doesn't seem to make sense to have a, b, c and d in your answers. But, otherwise, your answer is nicely displayed.

There is a nice check for your answer: once you've worked out all the probabilities correctly, the probabilities should add up to 1 (since you've covered every situation).

This Wolfram Alpha search might offer you some help, as well.

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