Calculating conditional probability of discrete uniform r.v.

Question

X is a discrete uniform random variable on $\{a, a+1, a+2, ... , b\}$ with mean 7 and variance 4.

Find $Pr[X \leq 6| X > 4]$

I'm not familiar with the discrete uniform distribution. I was tempted to do the following:

$$Pr[X \leq 6| X > 4] = \frac{Pr[4 < X \leq 6]}{Pr[X > 4]} = \frac{Pr[(4-7)/2 < Z \leq (6-7)/2]}{Pr[Z > (4-7)/2]} = \frac{Pr[-1.5 < Z \leq -.5]}{Pr[Z > -1.5]} = (.3085-.0668)/(1-.0668)$$

I'm getting $.259$, but I believe this answer is incorrect.

Any help would be much appreciated.

Answer 1

The first thing you must do is find $a,b$ from the given mean ($7$) and variance ($4$).   Only then can you try to find $\mathsf P(a\leq X\leq 6\mid 4\leq X\leq b)$

The mean and variance of a uniform discrete distribution, $X\sim\mathcal U\{a..b\}$ are: $$\mathsf E(X) = \frac{a+b}{2} = 7\\\mathsf{Var}(X)= \frac{(a-b+1)^2-1}{12} = 4$$