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There are $x$ white marbles and $y$ black marbles in a bag of $n$ marbles where $x+y=n$ and $0 \leq x,y≤n$. I will now choose k marbles to remove from the bag, where $0 \leq k \leq n$.

  1. If there is $n-1$ white marbles and 1 black marbles in the bag, how would the probability for black marble to be removed equate to $p=\frac{k}{n}$? My argument would be: $$\frac{C_{k-1}^{n-1}\cdot C_{1}^{1}}{C_{k}^{n}}=\frac{k}{n}$$ which involves choosing k-1 white from n-1 marbles after one black marble, which contain the thinking that all white marbles are different, so I'm not sure if this is valid. $$$$

  2. If there are $x$ white and $y$ black marbles, how can I calculate the general probability for each combination of possible white and black marbles removed? (e.g. if there is 5 black and 4 white marbles, and I plan to remove 5 marbles, how do I calculate the probability that 3 white and 2 black marbles are removed, or similarly the probability that 1 white and 4 black marbles are removed? Here I am also thinking to use: $$\frac{C_{a}^{x}\cdot C_{k-a}^{y}}{C_{k}^{n}}$$ to solve the probability that $a$ white marbles is selected from $n$ marbles containing $x$ white and $y$ black marble, which I'm again not sure if it is the correct approach.

Note: If my method turns out to be valid, no answers to this question will be necessary.

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If a bag contains 5 Black marbles and 4 White marbles, and you withdraw 5 marbles at random without replacement, then then number $X$ of White marbles among the 5 withdrawn has a hypergeometric distribution with $$P(X = k) = \frac{{4 \choose k}{5 \choose 5-k}}{{9 \choose 5}},$$

for $k =0, 1, 2,3,4.$

In particular, $$P(X = 3) = \frac{{4 \choose 3}{5 \choose 2}}{9\choose 5} = \frac {4\cdot10}{126} = 0.3174603.$$

In R statistical software, where dhyper is a hypergeometric PDF, you can make a probability table for the distribution as shown below. [You can ignore row numbers in brackets.]

k = 0:5;  PDF = dhyper(k, 4,5, 5)
cbind(k, PDF)
     k         PDF
[1,] 0 0.007936508
[2,] 1 0.158730159
[3,] 2 0.476190476
[4,] 3 0.317460317   # Shown above
[5,] 4 0.039682540
[6,] 5 0.000000000   # Impossible to get 5 White

Notice that I tried to find the probability of getting 5 white marbles, which is impossible because there are only 4 white marbles in the bag. In writing the PDF you can either (i) be careful to restrict $k$ only to possible values or (ii) use the convention that the binomial coefficient ${a \choose b} = 0,$ if integer $b$ exceeds integer $a.$ If your text includes a formal statement of the hypergeometric PDF, you should notice which method is used.

Here is a plot of the specific hypergeometric distribution mentioned in your problem.

enter image description here

Computations of $\mu = E(X) = \sum_{k=0}^5 k*p(k) = 5(4/9) = 2.2222$ and $$\sigma^2 = Var(X) = \sum_{k=0}^5 (k-\mu)^2p(k)\\ = \sum_{k=0}^5 k^2p(k) - \mu^2 = 0.6173$$ are shown below. You may find formulas for these in your text.

mu = sum(k*PDF);  mu
[1] 2.222222

vr = sum((k-mu)^2*PDF);  vr
[1] 0.617284
sum(k^2*PDF) - mu^2
[1] 0.617284
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Your method is indeed valid.

For the first question, note that all marbles are equally likely to be chosen, and the probabilities of marbles being chosen must add up to $k$, so the answer is trivially $\frac{k}{n}$.

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  • $\begingroup$ Could you please elaborate on "the probabilities of marbles being chosen must add up to k?". I presume it is a typo as k may be greater than 1. But otherwise thank you for the confirmation. $\endgroup$ – LHC2012 Jun 6 '19 at 15:46
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    $\begingroup$ @LHC2012 I don’t think it’s a mistake. When you have mutually exclusive events, the sum of their probabilities cannot be greater than $1.$ But these events are not mutually exclusive. It’s like two friends estimating their chances of passing the next exam who say each has a $0.95$ chance to pass. The sum of probabilities is $1.9 > 1,$ which is OK because it’s possible they both pass. $\endgroup$ – David K Jun 7 '19 at 10:57

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