Question

A machine in a laundromat breaks down an average of 3 times per month. Using the Poisson distribution, find:

a) The probability that there are exactly four breakdowns in a month.

b)The probability that the number of breakdowns in a month is no more than 3.

Answer #1

Solution :

Given that ,

mean = = 3

Using poisson probability formula,

P(X = x) = (e^{-}
*
^{x} ) / x!

a)

P(X = 4) = (e^{-3} * 3^{4)} / 4! = 0.1680

The probability that there are exactly four breakdowns in a month is 0.1680

b)

P(X 3) =P(X = 0 ) + P(X = 1) + P(X = 2) + P(X = 3)

= (e^{-3} * 3^{0)} / 0!
+ (e^{-3} * 3^{1)} / 1!
+ (e^{-3} * 3^{2)} / 2!
+ (e^{-3} * 3^{3)} / 3!

= 0.0498 + 0.1494 + 0.2240 + 0.2240

= 0.6472

The probability that the number of breakdowns in a month is no more than 3 is 0.6472

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