Dinner Reservations Experience shows that $20\%$ of the people reserving tables at a certain restaurant never show up. If the restaurant has 50 tables and takes $52$ reservations, then what is the probability that it will be able to accommodate everyone?
My understanding: If the reservation is $52,$ and $20\%$ people don't show up, would not the probability of accommodating every one be $1$?
 A: Given that a table that has been reserved will be taken by a party who reserved it is $80\%$, the probability that the restaurant accommodates everyone is 
$$1 - \Pr[52\ \ \mbox{parties show up}]  - \Pr[51\ \mbox{parties show up}],$$ 
where probability that $k$ out of $52$ parties show up is
$$\Pr[k\ \mbox{parties show up}] = \binom{52}{k}0.8^{k}(1-0.8)^{52-k}.$$
A: In a statement like "$20\%$ of people don't show up", you're not talking about a portion of every group of people. You're talking about the proportion of people in general. For example, when a fair coin is flipped, it comes up heads $50\%$ of the time - but that doesn't mean that every time I flip my coin twice it'll come up one heads and one tails! It just means that if I were to flip my coin over and over again, on average it would come up heads half the time.
In your situation, the "$20\%$" fact at the beginning tells you not that exactly $0.2 \cdot 52 = 10.4$ people will fail to show up (what would that even mean? Someone comes in but leaves an arm and a leg at home?) but instead that each person has a $20\%$ chance of not showing up.
The restaurant will have enough space only if at least two people don't show up. That's the same as saying they won't have enough space if only only one or zero people don't show up. The probability of zero people not showing up is $0.8^{52} \approx 0.0009\%$, because every single one of the $52$ people would have to hit the $80\%$ chance of showing up. The probability of exactly one person not showing up is $52 \cdot 0.2 \cdot 0.8^{51} \approx 0.012\%$. The probability of either one of those happening is $0.0009\% + 0.012\% \approx 0.013\%$. So the probability of neither one happening - in other words, the probability that the restaurant will have enough space - is $100\% - 0.013\% = 99.987\%$.
A: 50 tables can accommodate 50 men. But they have taken 52 reservations.
The restaurant will be able to accommodate everyone, if 50 men or fewer men show up.
Everyone gets a seat when $(X \leq 50)$.
To find this, we use the complement rule.
$1 -$ [Find the chance, Too many men show up with $X = 51$ OR $X = 52$]
Given : 20% of people don't show up. Then,  Success : Men who show up : $p = 80  $% and Failure : Men who didn't show up : $(1-p) = 1 - 80/100 = 20/100$
$\Pr(X \leq 50) = 1 - \Pr(X > 50) = 1 - \sum_{k = 51}^{52} \binom{52}{k}(0.8)^k(0.2)^{52 - k} = 1-[P(X = 51) + P(X = 52)]= 1- [\binom{52}{51}(0.8)^{51}(1 - 0.8)^{52 - 51} + \binom{52}{52}(0.8)^{52}(1 - 0.8)^{52 - 52} ]$
