Problem90: VAD

¡Puedes leer este texto en español! Problema90: VAD

Ej90. An Internet user gets an average of 9 emails per hour to his personal account. This user also has a second account for professional use which gets an average of 36 daily emails.

From previous studies it is known that the probability of getting SPAM is aproximately 0.571429 on both accounts.

The user has noticed that emails appear of homogeneous form all along hour of the day.

Calculate the probability of getting more than 10 SPAM in an hour between both accounts.


We perform a data compilation given by the problem:

· A ≡ 'Number of emails received to personal account'.
· B ≡ 'Number of emails received to professional account'.

Emails received per hour on both events (we have considered that a day has 24 hours):

· A = 9 emails/hour.
· B = 36/24 = 1.5 emails/hour.

So, the average total number of emails on both accounts are: 9 + 1.5 = 10.5 emails/hour.

· S ≡ 'The email is SPAM'.

· P(S) = 0.571429

· C ≡ 'Number of SPAM emails on either of both accounts'.

We know two things: the average number of emails received on both accounts and the probability of being SPAM. So, we can create the next event:

C = λ = 10.5·0.571429 = 6.000005

So, the C event follows a Poisson distribution: C ~ P(6.000005) SPAM/hour.

We must obtain the following probability:

P(C > 10) = 1 - P(C ≤ 10) = 1 - [P(C = 0) + ··· + P(C = 10)]

There are many elements to sort out, then we're going to use the R software for that purpose:

> ppois(c(10), 6.000005, lower.tail = F)
[1] 0.04262113

Finally, the probability getting more than 10 SPAM emails in an hour on either both accounts, is aproximately 0.042621.