Commutator load

The problem

Probability of client connection to commutator during 1 hour is 0.01. Commutator handle 300 clients.

What the probability of:

• a) 4 clients will connect during hour
• b) less then 4 clients will connect during hour

Solutions

The problem can be solved by Possion formula

$$\\ P_{n}(k)=\frac{\lambda ^{k}}{k!} e^{-\lambda} \\ \lambda = np$$

a)

$$\\ p = 0.01 \\ n = 300 \\ \lambda = p*n = 3 \\ P_{300}(4)=\frac{3 ^{4}}{4!} e^{-3} = 0.1680314$$

b)

$$\\ p = 0.01 \\ n = 300 \\ \lambda = p*n = 3 \\ P_{300}(0) + P_{300}(1) + P_{300}(2) + P_{300}(3) = \\ = \frac{3 ^{0}}{0!} e^{-3} + \frac{3 ^{1}}{1!} e^{-3} + \frac{3 ^{2}}{2!} e^{-3} + \frac{3 ^{3}}{3!} e^{-3} = \\ = e^{-3}(1+3+\frac{9}{2}+\frac{27}{6})=\frac{13}{e^{3}} = 0.6472319$$