On Halloween trickortreaters arrive at a house from 530 pm u

On Halloween, trick-or-treaters arrive at a house from 5:30 pm until 9:00 pm according to a Poisson process with an average of 14 trick-or-treaters per hour.

(a) What is the probability that between 4 and 6 trick-or-treaters (inclusive) arrive in the first 30minutes?

(b) What is the probability that the first trick-or-treaters arrives to the house between 5:33 pm and 5:35 pm?

(c) If the time period from 5:30 pm until 9:00 pm is split into 7 disjoint 30-minute intervals, find the probability that at most two of them have between 4 and 6 trick-or-treaters (inclusive).

Solution

(a) What is the probability that between 4 and 6 trick-or-treaters (inclusive) arrive in the first 30minutes?

Given X follows Poisson distribution with mean=14/2=7 in 30 minutes

P(X=x)=(7^x)*exp(-7)/x! for x=0,1,2,...

So the probability is

P(4 <=X<= 6) = P(X=4)+P(X=5)+P(X=6)

=(7^4)*exp(-7)/4!+...+(7^6)*exp(-7)/6!

=0.3679456

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(b) What is the probability that the first trick-or-treaters arrives to the house between 5:33 pm and 5:35 pm?

Given X follows Poisson distribution with mean=2*14/60 =0.4666667 in 2 minutes

P(X=x)=(0.4666667^x)*exp(-0.4666667)/x!

So P(X=1)=(0.4666667^1)*exp(-0.4666667)/1=0.2926416

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(c) If the time period from 5:30 pm until 9:00 pm is split into 7 disjoint 30-minute intervals, find the probability that at most two of them have between 4 and 6 trick-or-treaters (inclusive).

Given X follows Binomial distribution with n=7 and p=0.3679456

P(X=x)=7Cx*(0.3679456^x)*((1-0.3679456)^(7-x)) for x=0,1,2,...,7

So the probability is

P(X<=2) = P(X=0)+P(X=1)+P(X=2)

=7C0*(0.3679456^0)*((1-0.3679456)^(7-0))+...+7C2*(0.3679456^2)*((1-0.3679456)^(7-2))

=0.4912977