Question

the visit of a customer at a restaurant follows a Poisson process with a rate of 3 arrivals per week. A day with no visits to the restaurant is called a risky day.

a. Find the expected number of risky days in a week

b.  Given that a risky day was observed on a Sunday, what is the probability that the next risky day will appear on the following Wednesday?

c. Find the probability that the 4th day, which is Thursday, of the week is the second risky day.

Answer 1

events/time = arrival/day = 3/7

P(risky day) = P(0 arrival in 1 day) = e^(-(3/7)*1) * ((3/7)*1)^0 / (0!) = 0.65

P(risky day) = 0.65

a.

expected number of risky days in a week = (no. of days in week)*(p(risky day))

= 7*0.65

= 4.55 days

expected number of risky days in a week = 4.55 days

b.

P(next risky day on wednesday)

= P(non risky day on monday)*P(non risky day on tuesday)*P(risky day on wednesday)

= (1 - P(risky day))*(1 - P(risky day))*P(risky day)

= (1 - 0.65)*(1 - 0.65)*0.65

= 0.079625

c.

probability that the 4th day, which is Thursday, of the week is the second risky day.

= P(risky day on monday)*P(non risky day on tuesday)*P(non risky day on wednesday)*P(risky day on thursday) + P(non risky day on monday)*P(risky day on tuesday)*P(non risky day on wednesday)*P(risky day on thursday) + P(non risky day on monday)*P(non risky day on tuesday)*P(risky day on wednesday)*P(risky day on thursday)

= P(risky day)*(1 - P(risky day))*(1 - P(risky day))*P(risky day) + (1 - P(risky day))*P(risky day)*(1 - P(risky day))*P(risky day) + (1 - P(risky day))*(1 - P(risky day))*P(risky day)*P(risky day)

= 0.65*(1 - 0.65)*(1 - 0.65)*0.65 + (1 - 0.65)*0.65*(1 - 0.65)*0.65 + (1 - 0.65)*(1 - 0.65)*0.65*0.65

= 0.1553

probability that the 4th day, which is Thursday, of the week is the second risky day = 0.1553

(please UPVOTE)