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.
Best Answer
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?