Assume that the tunnel in Exercise 3.132 is observed during ten two-minute intervals, thus giving ten independent observations Y1, Y2,..., Y10, on the Poisson random variable. Find the probability that Y > 3 during at least one of the ten two-minute intervals

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answer:

0.1745

Step-by-step explanation:

Determine  P ( Y > 3 )

Given that y follows Poisson distribution with mean

P( Y = y ) = [tex]\frac{e^{-1} }{y!}[/tex]

assuming that yi represents number of autos that will enter the tunnel and

also let A represent Y > 3 at one of the ten two-minute interval

step 1 : hence finding  P(A )

= P(Y > 3 ) = 1 - P( Y≤ 3 )

= 1 - [ [tex][\frac{e^{-1} }{0!} +\frac{e^{-1} }{1!} +\frac{e^{-1} }{2!} +\frac{e^{-1} }{3!} ][/tex]

= 0.018988

since the ten observations are independent

P(X≥ 1 ) = 1 - P( X = 0 )

            = 1 -  [tex]\left[\begin{array}{ccc}10\\0\\\end{array}\right][/tex]  * (0.018988)^0 ( 1 - 0.018988)^10-0

            = 1 - 0.8255 = 0.1745

                 

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