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songoku
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Homework Statement
A dealer has a stock of 6 similar television sets which he rents out to customers on a monthly basis. It is known from past experience of the dealer that the monthly demand for the television sets have a Poisson distribution with mean 3.56
(i) Find the probability that in any month at least two sets are not rented out
(ii) Calculate the probability that in any month the demand is not fully met
(iii) Find the probability that exactly one set not rented out in exactly two out of four months.
(iv) If one of the television sets is temporarily withdrawn for repairs, find the expected number of television sets rented out in a month.
Homework Equations
Probability = e-λ . λx / x!
Binomial distribution
The Attempt at a Solution
λ = 3.56 / month
(i) P (x ≤ 4) = P (x = 0) + P (x = 1) + P(x = 2) + P(x = 3) + P (x = 4)
(ii) P (x ≥ 7) = 1 - P (x ≤ 6)
(iii) I am not sure about the mean. Should we use mean per month or per 2 months or per 4 months?
I tried using per month so probability exactly one set not rented = P( x = 5) = p, then using binomial distribution the probability exactly two out of four months = 4C2 . p2 . q2
(iv) expected value for Poisson distribution = λ
If no. of television = 6, mean = 3.56 / month
If no. of television = 5, mean = ... / month ---> what the question asks
How to find this one? Can simple ratio be applied for this question?Additional question: how to construct probability distribution table for Poisson distribution? I tried to find the probability from x = 0 until x = 6 (because there are 6 TV) but it doesn't add up to 1
Thanks