# Probability Mass function and Cumulative distribution function of Z

An (incomplete) card deck contains 36 cards, of ranks 2, 3, 4, 5, 6, 7, 8, 9, 10 (of all four suits). Two cards are picked at random without replacement. Let Z denote the random variable which is the maximal rank of the two cards picked (for example, if 6, 7 are picked then Z = 7).

Compute the probability mass function and the cumulative distribution function of Z.

PMF = P(X=x)= ????

What is the expected value (i.e. math. expectation) of Z?

I have been reading about the concept of probability mass functions for 20 minutes and I cannot really understand how it is applied to this problem... I understand that it deals with discrete random variables which is obviously relevant to this problem.

• Welcome to MSE.. please consider using either MathJax or LaTeX to format the question. – Ahmad Bazzi Sep 14 '18 at 4:43

There are $\binom{36}{2}$ ways to choose two cards.

$6$ will give $Z=2$

$6+4\cdot4=22$ will give $Z=3$

$6+4\cdot8=38$ will give $Z=4$

...

$6+4\cdot32=134$ will give $Z=10$

Therefore the answer is $Z=\frac{6*2+22*3+...+134*10}{\binom{36}2}=\frac{158}{21}$, or $7.523809$...

• So we can write the probability mass function as a function of $z$ as $f_Z(z)=\Pr(Z=z)=\frac{6+16(z-2)}{\binom{36}{2}}$ for $z\in\{2,\dots,10\}$. – molarmass Sep 14 '18 at 5:20