A box has an unknown number of tickets serially numbered 1,2,...,N. Two tickets are drawn using simple random sampling without replacement A box has an unknown number of tickets serially numbered 1,2,...,N. Two tickets are drawn using simple random sampling without replacement (SRSWOR) from the box. If X and Y are the numbers on these two tickets and $Z = max(X,Y )$, show that
(a) Z is not unbiased for N
(b) $aX+bY +c$ is unbiased for N if and only if $ a+b = 2 \text{ and}\ c = −1.$
What is the the pmf of Z,  Any tips on how to proceed?
 A: Comment (continued) However, it should not be too hard for you
to find the distribution of $Z.$
Here is a simple simulation in R statistical software
for the case $N = 5.$ It seems that $Z$ takes values 2 through 5 with
respective probabilities $.1, .2, .3, .4,$ and that $E(Z) = 4.$ (Of course
simulation only provides a good approximation.)
n = 5;  box = 1:n
z = replicate(10^5, max(sample(box,2)))
hist(z, prob=T, br = (box)+.5, col="skyblue2")
mean(z)
## 4.00825  # Consistent with $E(Z) = 4.$ 


You can easily get the exact distribution of $Z$ by making a $5 \times 5$ square
array of points, representing possible values $X$ and $Y.$ What is the
probability of each point? What is the value of $Z$ for each point?
What is the exact distribution of $Z$? What is $E(Z)?$
Then maybe you can generalize the answers to larger values of $N$ without
making arrays of points.
For example, here is the histogram for $N = 10.$ It seems clear that $E(Z) < 10.$ (The red line shows the position of the mean.)

Addendum: In your last comments you turned to part (b). Consider
the special case in which $a = b = 1.$ Then let $W = X+Y-1.$ In case it helps, simulation
with $N = 10$ gives $E(W) \approx 10$ and the following histogram.

A: so i solved this question so you have to start in this way 
think what values Z can take? so z can 2,3,4,,,,N. Z can not  take the value 1 because if we take any two number them maximum value can not be one so what is the probability to take this no is given 
if z take 2 them its probability is 1/ (N c 2)     { there N c 2 is the N choose 2}
Z take   3 them its probability is  2/ (N c 2)     
{ here numerate is 2 because there is two ways to take the value 3 is {(3,2) ,(3,1)} 
and so on if now if z take N so its probabilty is  (N-1)/(N c 2) 
so in this way you got the probabiltiy distribution .
