A box has 3 balls numbered from 1 to 3. Two balls are selected randomly and without replacement. Let X denote the number of the first ball and Y denote the number of the second ball. Find the pmf of X and the pmf of Y.
Thoughts: the random variables look like they have hypergeometric distribution but that wouldn’t fit the experiment properly. For example P(X = 3) would have undefined binomial coefficients.
Best Answer
Since $\{X,Y\}$ is a sample of size two, their marginal pmfs are easy: $$f_X(t)=f_Y(t)=\frac13[1\le t\le 3]$$ where $[.]$ is the Iverson bracket ($1$ if the condition inside is true, $0$ otherwise). This can also be derived from a probability tree.
As to the expectation, covariance and correlation, $E(X)=E(Y)=2$ and $E(XY)=\frac{2+3+6}3=\frac{11}3$, so $\operatorname{cov}(X,Y)=-\frac13$. $\operatorname{Var}(X)=\operatorname{Var}(Y)=\frac{1+4+9}3-2^2=\frac23$, so $\operatorname{corr}(X,Y)=\frac{-1/3}{2/3}=-\frac12$.