Finding the joint p.f.

Question

I have two questions here that, I believe, revolves around the same topic.

1) Suppose there are 10 red balls, 9 yellow balls, and 6 blue balls. Five balls are selected at random without replacement.

Let $X$ be the number of red balls selected and $Y$ be the number of yellow balls selected. What is the joint p.f. of $X$ and $Y$?

I am not sure how to start this question. I know it's hypergeometric distribution. but the question require to find the p.f. of the "number of red/yellow balls" throws me off.

2) The Hardy-Weinberg law of genetics states that, under certain conditions, the relative frequencies with which three genotypes $AA$, $Aa$ and $aa$ occur in the population will be $θ^2$, $2θ(1 − θ)$ and $(1 − θ)^2$ respectively where $0 < θ < 1$. Suppose $n$ members of the population are selected at random. Let $X$ be the number of $AA$ types selected and let $Y$ be the number of $Aa$ types selected.

I feel the two questions are similar in nature and our variables are defined as the "number of ..." throws me off and I am not sure how to tackle these.

Answer 1

The questions are different. The first is more difficult (because you have to keep track of how many balls remain after each selection), so let's address it.

Generally, suppose there are $\rho$ red balls, $\upsilon$ yellow balls, and $\beta$ blue balls (as is conventional, I use lower case Greek letters to describe the properties of the distribution) and you select $k$ balls without replacement. Let $(x,y)$ be any possible ordered pair of values of $(X,Y):$ that is, $x$ and $y$ are integers between $0$ and $k$ and $x+y\le k.$

The chance that $(X,Y)=(x,y)$ is the chance that the section included exactly $x$ red balls, exactly $y$ yellow balls, and (therefore) exactly $k-x-y$ red balls. Equivalently -- here is the key step in the reasoning -- your selection consists of an $x$-subset of the $\rho$ red balls (of which there are $\binom{\rho}{x}$), a $y$-subset of the $\upsilon$ yellow balls, and a $k-x-y$-subset of the $\beta$ blue balls. Because all possible combinations of these subsets can occur, the total number of possible samples is the product of these three Binomial coefficients:

The number of possible samples is $\binom{\rho}{x}\binom{\upsilon}{y}\binom{\beta}{k-x-y}.$

The uniform random selection means every possible sample has the same chance of occurring. Since there are $\binom{\rho+\upsilon+\beta}{k}$ such subsets, the chance we seek must be the proportion of such samples among all possible samples,

$$f(x,y) = \Pr((X,Y)=(x,y)) = \frac{\binom{\rho}{x}\binom{\upsilon}{y}\binom{\beta}{k-x-y}}{\binom{\rho+\upsilon+\beta}{k}}.$$

The joint probability function is $f,$ by definition. One way to look at it is to display it as a matrix indexed by $x$ and $y:$

   y
x        0      1      2      3      4      5
  0 0.0001 0.0025 0.0136 0.0237 0.0142 0.0024
  1 0.0028 0.0339 0.1016 0.0949 0.0237     
  2 0.0169 0.1143 0.1829 0.0711     
  3 0.0339 0.1220 0.0813     
  4 0.0237 0.0356     
  5 0.0047

(I have rounded the values to four decimal digits and show only the nonzero probabilities.)