Question

A website for bed and breakfast inns gets approximately seven visitors per minute.

Suppose the number of website visitors per minute.

a. Compute the probability of no website visitors in a one-minute period. b. Compute the probability of two or more website visitors in a one-minute period. c. Compute the probability of one or more website visitors in a 30-second period. d. Compute the probability of five or more website visitors in a one-minute period.

Answer 1

Solution:

$\lambda=7$ per min$=7/ 60$ per sec

(a) $P(X=0)=e^{-7} \frac{7^0}{0!}=e^{-7}=0.00091$

(b) $P(X\ge2)=1-P(X=0)-P(X=1)-P(X=2)$

$=1-e^{-7} \frac{7^0}{0!}-e^{-7} \frac{7^1}{1!}-e^{-7} \frac{7^2}{2!} \\=0.97036$

(c) $\lambda$ for 30 sec period$=7(30/60)=3.5$

$P(X\ge1)=1-P(X=0)-P(X=1) \\=1-e^{-3.5} \frac{3.5^0}{0!}-e^{-3.5} \frac{3.5^1}{1!} \\=0.8641$

(d) $P(X\ge5)=1-P(X=0)-P(X=1)-P(X=2)-P(X=3)-P(X=4)-P(X=5)$

$=1-e^{-7} \frac{7^0}{0!}-e^{-7} \frac{7^1}{1!}-e^{-7} \frac{7^2}{2!}-e^{-7} \frac{7^3}{3!}-e^{-7} \frac{7^4}{4!}-e^{-7} \frac{7^5}{5!}$

$=0.69929$