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!
 A: 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.
