Question

A household receives an average of 3 pieces of junk mail per day. What is the probability that this household will not receive any junk mail on a certain day? will receive exactly 10 junk mails in 4 days? will receive at least 4 junk mails in two days?

Answer 1

X ~ Poisson ()

Where = 3 pieces per day.

Poisson probability distribution is

P(X) = e^-X / X!

a)

We have to calculate, for a given day = 3 pieces per day.

P( X = 0) = e^-3 3⁰ / 0!

= 0.0498

b)

For 4 days, = 3 * 4 = 12 pieces per 4 days ,

So,

P(X = 10) = e^-12 12¹⁰ / 10!

= 0.1048

c)

For 2 days, = 2 * 3 = 6 pieces per 2 days

P( X >= 4) = 1 - P( X <= 3)

= 1 - [ P( X = 0) + P( X = 1) + P( X = 2) + P( X = 3) ]

= 1 - [ e^-6 6⁰ / 0! +  e^-6 6¹ / 1! + e^-6 6² / 2! + e^-6 6³ / 3! ]

= 1 - 0.1512

= 0.8488