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.


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marbles, and you withdraw 5 marbles at random without replacement, then then number $X$ of White marbles among the 5 withdrawn has a hypergeometric distribution with $$P(X = k) = \frac{{4 \choose k}{5 \choose 5-k}}{{9 \choose 5}},$$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

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 $\frac{k}{n}$.

  • $\begingroup$ 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
  • 1
    $\begingroup$ @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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