Answer to Question #256194 in Statistics and Probability for rk12345

Question

The owner of a small hotel with 5 rooms is considering buying TV sets to rent to room occupants.

He expects that about half of his customers would be willing to rent sets and finally he buys 3 sets.

Assuming 100% occupancy at all times

i) What fraction of evenings will there be more requests than TV sets?

ii) What is the probability that a customer who requests a TV set will receive one?

Answer 1

Let $X=$ the number of requests: $X\sim Bin (n, p)$

Given $n=5, p=0.5, q=1-p=0.5$

i)

$$P(X>3)=P(X=4)+P(X=5)$$

$$=\dbinom{5}{4}(0.5)^4(0.5)^{5-4}+\dbinom{5}{5}(0.5)^5(0.5)^{5-5}=0.1875$$

ii)

$$P(X\leq3)=1-P(X=4)-P(X=5)$$

$$=1-\dbinom{5}{4}(0.5)^4(0.5)^{5-4}-\dbinom{5}{5}(0.5)^5(0.5)^{5-5}$$

$$=0.8125$$