Answer
a)Binomial distribution with $n=10^{9}$, $p=\frac{10000}{36^6}\approx 4.59\times 10^{-6}$
b) $p(X=0)\approx 0$
c) $E(x)\approx 4595$
$V(x)\approx 4595$
Work Step by Step
Let X be the random variable of the number of users' passwords hits. then X is distributed as binomial distribution with $n=10^{9}$, and we can determine p as follows,
number of available passwords $={(10+26)}^{6}=36^6$
$p=\frac{10000}{36^6}\approx 4.59\times 10^{-6}$
this means that
$p(X=x)=\frac{10^{9}}{x!(10^{9}-x)!}(\frac{10000}{36^6})^x(1-\frac{10000}{36^6})^{10^{9}-x},x=0,1,...,10^{9}$
b) $p(X=0)=(1-\frac{10000}{36^6})^{10^{9}-x}\approx 0$
c) $E(x)=np=10^{9}(\frac{10000}{36^6})\approx 4595$
$V(x)=np(1-p)=10^{9}(\frac{10000}{36^6})\times(1-\frac{10000}{36^6})\approx 4595$
