Entrance at gymnasium Bill gave exams for the entrance at some specific gymnasium. $602$ students took part, which were classified, after the exams, in an ascending order, and the first  $108$ students will be taken, which will accept to enter. Every student that has the possibility to enter will not enter with a small possibility $p=0.02$, same for all, and independent from the rest. Bill is at the position $113$, so he will be accepted if at least $5$ students from the first $112$ will not enter at the gymnasium. I want to give an exact expression for the probability $q$ that Bill gets accepted. I also want to give an approximate expression for the probability $q$ .
Is the probability that Bill get accepted equal to
$$5 \cdot 0.02?$$
Or do we have to take also something else into consideration?
 A: This is a binomial distribution with $p=.02$, $n=112$, and five successes required. So the simple way to find the answer is simply to find a binomial calculator. For instance, https://stattrek.com/online-calculator/binomial.aspx gives 7.49%
If you want to do it by hand, you can take 
$\sum_5^{112} \binom{112}{n}(.02)^n(.98)^{112-n}=1-\sum_0^4 \binom{112}{n}(.02)^n(.98)^{112-n}$
You can also treat this as being approximated by a Poisson distribution with $\lambda = 112*.02=2.24$ and find the probability $x\geq5$, which gives 7.68%, which is close to the exact answer of 7.49%.
A: You have to take into consideration the number of ways you can choose $5$ from $112$ and the probability of remaining chosen $(.98)$. Also its the probability of at least $5$ not being chosen. This makes it more complicated as it then becomes one minus the probability of from $0$ to $4$ not chosen. The probability of Bill being chosen is therefore:
$P_B =1 - (^{112}C_0\cdot .98^{112} + ^{112}C_1\cdot .02\cdot .98^{111} + ^{112}C_2\cdot .02^2\cdot .98^{110} + ^{112}C_3\cdot .02^3\cdot .98^{109} + ^{112}C_4\cdot .02^4\cdot .98^{108})$
$P_B = 1 - 0.9251 = 0.0749$
A: easier to compute by hand (or calculator) if you re-write it as nested terms
$$
1-p^n(1+rn(1+r\frac{(n-1)}2(1+r\frac{(n-2)}3(1+r\frac{(n-3)}4))))
$$
where
$$
p=0.98  \\
n=112  \\
r=(1-p)/p  \\
$$
