# Convert probability example to a distribution

Consider a box which contains 30 balls. 10 red, 10 white and 10 blue balls. There will be picked 4 balls randomly, without putting the balls back to the experiment.

My question is: How do I create a distribution of this senario, so I can find the probability that none of the 4 balls are red, and to calculate the mean and variance?

I am aware the math is:

$$\frac{20}{30} \times \frac{19}{29} \times \frac{18}{28} \times \frac{17}{27}=0.1768$$

But I need to convert it into a distribution and use the Probability function.

If we reformulate your question as:

Given a box with 30 balls, of which 10 are red, what is the probability that 4 balls drawn at random (without replacement) don't have any red balls in them

you can model it by the HypergeometricDistribution:

A hypergeometric distribution gives the distribution of the number of successes in $n$ draws from a population of size $n_{\mathrm{tot}}$ containing $n_{\mathrm{succ}}$ successes.

box[n_] := HypergeometricDistribution[n, 10, 30];
Probability[x == 0, x \[Distributed] box]
(* 323/1827 *)


or 0.176793 (use N or NProbability).

More generally, such problems can be modeled using theMultivariateHypergeometricDistribution. For the original question:

draw[n_] := MultivariateHypergeometricDistribution[n, {10, 10, 10}];
NProbability[r == 0, {r, b, w} \[Distributed] draw]
(* 0.176793 *)


We can also look at the probability of getting exactly 2 blue balls and at least 1 red ball:

NProbability[b == 2 && r > 0, {r, b, w} \[Distributed] draw]
(* 0.238095 *)


or the probability of getting exactly two balls, conditioned on the fact that at least one is red

NProbability[b == 2 \[Conditioned] r > 0, {r, b, w} \[Distributed] draw]
(* 0.289229 *)

• I was wondering. How would you calculate the conditional probability that there are exactly two blue balls in the sample, given that there are some red? 0.76969 Aug 7, 2013 at 13:22
• @JensJensen See update
– rm -rf
Aug 7, 2013 at 15:11