What is the probability that the restaurant can provide tables for all the people who made reservations to come to dinner? Experience shows that $20\%$ of people who reserve a table in a certain restaurant don't come to dinner. If the restaurant has $50$ tables
and they only accept $52$ reservations, what is the probability that they can give a table to all the people who come to dinner?
My try: I believe that this could be solve by conditional probability.
Let $B =$ the event which $80\%$ of the people arrives.
Let $A =$ given B, the event witch all people who came have a place.
So I need to find $P(A \mid B) = \dfrac{P(A \cap B)}{P(B)}$
My trouble is that I been thinking about how to calculate $P(A \cap B)$.
 A: We are told that a person who reserves a table at the restaurant fails to come to dinner $20\%$ of the time.  The restaurant will have enough tables if at least $2$ of the $52$ people who make reservations fail to show.  The assumption here is that the probability that any particular person who makes a reservation fails to show is $0.2$.  
The probability that exactly $k$ people fail to show can be determined by using the binomial distribution, which states that the probability of exactly $k$ successes will occur in $n$ trials, each of which has probability $p$ of success is 
$$\Pr(X = k) = \binom{n}{k}p^k(1 - p)^{n - k}$$
where $p^k$ is the probability of $k$ successes, $(1 - p)^{n - k}$ is the probability of $n - k$ failures, and $\binom{n}{k}$ is the number of ways exactly $k$ successes could occur in $n$ trials.
If we define a success as a person who makes a reservation failing to show, then the probability that at least two people fail to show is 
$$\Pr(X \geq 2) = \sum_{k = 2}^{52} \binom{52}{k}(0.2)^k(0.8)^{52 - k}$$
That said, it is easier to calculate this probability by subtracting the probability that fewer than two people who make a reservation fail to show from $1$.
$$\Pr(X \geq 2) = 1 - \Pr(X < 2) = 1 - \sum_{k = 0}^{1} \binom{52}{k}(0.2)^{k - 1}(0.8)^{52 - k}$$
If you prefer to think of a success as a person who makes a reservation shows up for dinner, then $p = 1 - 0.2  = 0.8$, so the probability that at most $50$ people show up for dinner is
$$\Pr(X \leq 50) = \sum_{k = 0}^{50} \binom{50}{k}(0.8)^k(0.2)^{52 - k}$$
which is more easily calculated by subtracting the probability that more than $50$ people show up from $1$.
$$\Pr(X \leq 50) = 1 - \Pr(X > 50) = 1 - \sum_{k = 51}^{52} \binom{52}{k}(0.8)^k(0.2)^{52 - k}$$
