Question

A certain restaurant always overbooks. If possible. At 7:00 pm the restaurant can seat 50 parties, but takes reservations for 53. If the probability of a party not showing up is 0.04, assuming independence, what is the probability that the restaurant can coommodate all the customers who do show up? You must show all of your work including the correct formulas for calculation to receive full credit of this answer

Answer 1

Answer:

The probability that the restaurant can accommodate all the customers who do show up is 0.3564.

Step-by-step explanation:

The information provided are:

• At 7:00 pm the restaurant can seat 50 parties, but takes reservations for 53.
• If the probability of a party not showing up is 0.04.
• Assuming independence.

Let X denote the number of parties that showed up.

The random variable X follows a Binomial distribution with parameters n = 53 and p = 0.96.

As there are only 50 sets available, the restaurant can accommodate all the customers who do show up if and only if 50 or less customers showed up.

Compute the probability that the restaurant can accommodate all the customers who do show up as follows:

$P(X\leq 50)=1-P(X>50)\\=1-P(X=51)-P(X=52)-P(X=53)\\=1-[{53\choose 51}(0.96)^{51}(0.04)^{53-51}]-[{53\choose 52}(0.96)^{52}(0.04)^{53-52}]\\-[{53\choose 53}(0.96)^{53}(0.04)^{53-53}]\\=1-0.27492-0.25377-0.11491\\=0.3564$

Thus, the probability that the restaurant can accommodate all the customers who do show up is 0.3564.