Probability question?

[匿名]

2017-02-22 12:59 pm

Students sign up at a desk during the course of an afternoon. The arrival of each student is independent of the arrival of any other student and the number of students arriving per hour can be modelled as a Poisson distribution with a mean of λ.

The desk is open for 4 hours. If exactly 5 people arrive to sign up for the activity during that time, find the probability that exactly 3 of them arrived during first hour.

回答 (1)

Lopez

2017-02-23 8:33 pm

✔ 最佳答案

Let X, Y, T denote the number of students arrived
during first hour, next 3 hours, all 4 hours, respectively.

P( X = x ) = λ^x * e^(-λ) / x!
P( X = 3 ) = λ^3 * e^(-λ) / 3! = (1/6) * λ^3 * e^(-λ)

P( Y = y ) = (3λ)^y * e^(-3λ) / y!
P( Y = 2 ) = (3λ)^2 * e^(-3λ) / 2! = (9/2) * λ^2 * e^(-3λ)

P( T = t ) = (4λ)^t * e^(-4λ) / t!
P( T = 5 ) = (4λ)^5 * e^(-4λ) / 5! = (1024/120) * λ^5 * e^(-4λ)

P( X = 3 , Y = 2 ｜T = 5 )
= P( { X = 3 , Y = 2 } ∩ { T = 5 } ) / P( T = 5 )
= P( X = 3 , Y = 2 ) / P( T = 5 )
= (1/6) * λ^3 * e^(-λ) * (9/2) * λ^2 * e^(-3λ) / (1024/120) * λ^5 * e^(-4λ)
= (1/6)(9/2) / (1024/120)
= (3/4) / (128/15)
= 45/512 ..... Ans

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