Probability with people 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, what is the probability that it will be able to accomodate everyone?
I tried to solve this problem but am not sure how to do this. I tried saying that since 20% of people never show up then 80% of them do. I also said that we can subtract it from 1 since they are trying to accomodate everyone. Here is my working.
$1-(80/100)^{51}$ $(1/4)^{52}$
Can someone please help me. Please show and explain how I can correct this working.
 A: The probabilty would be 1 since we are considering that out of 52 people who make reservations, 20% do not show up, there for 10 people will not show up, which leaves 42 people who reserved their tables being accommodated. This, of course is only true if the 20% of people not showing up is always the case
A: You’re to interpret the statement that $20$% of those who reserve tables never show up to mean that the probability that someone who reserves a table fails to show up is $\frac15$; the probability that someone who reserves a table does show up is then $1-\frac15=\frac45$.
Your idea of computing the probability that the restaurant will not be able to accommodate everyone and subtracting that from $1$ is good, but you calculated that probability incorrectly. In particular, I’ve no idea where the $\left(\frac14\right)^{52}$ came from.
There are two ways in which the restaurant can be in trouble: $51$ people show up, or all $52$ show up.


*

*The probability that all $52$ show up is $\left(\frac45\right)^{52}$.

*The probability that a particular set of $51$ people shows up and the $52$-nd person does not is $\left(\frac45\right)^{51}\left(\frac15\right)$. However, there are $\binom{52}{51}=52$ different sets of $51$ that could show up, so the probability that some set of $51$ shows up and the $52$-nd person does not is $52\left(\frac45\right)^{51}\left(\frac15\right)$.
What should you do with these two probabilities in order to complete the problem?
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 : $(1-p) = 1 - 80/100 = 20/100$
$\Pr(X \leq 50) = 1 - \Pr(X > 50) = 1 - \sum_{k = 51}^{52} \binom{52}{k}(0.8)^k(0.2)^{52 - k} = 1-[P(X = 51) + P(X = 52)]= 1- [\binom{52}{51}(0.8)^{51}(1 - 0.8)^{52 - 51} + \binom{52}{52}(0.8)^{52}(1 - 0.8)^{52 - 52} ]$
