Raine is a birdwatcher and counts the birds in her garden, which arrive according to a Poisson process of rate A. If an hour passes with no birds, she gets bored and goes to make tea. (a) What is the probability that she doesn't see any birds? (b) How much time on average does she spend birdwatching? Hint: Condition on the time of the ?rst observed bird. (0) What is the distribution of the number of birds seen?

Answer (a) :

P(she doesn't see any bird for next t hours  |  she doesn't see any bird for 1^st one hour) = e^-λτ

Answer (b) :

Average time she spends in birdwatching  = 

t  =  λ1?  

Answer (c) :

The distribution of the number of birds seen is the uniform distribution.

Step-by-step explanation

And :- Rate of arriving of birds = 1 birds / hour. => Number of average birds she see in 1 hour = 1 27 Number of average birds she see in t hours = it for poisson process p ( X z x ) = le " where I = average x 1 no. of events and x is no. of next first (a ) P ( No birds for't hours No birds for one hour , required occurances - = P ( No birds for ( Itt ) hours .. P (A / B ) = P ( Aand B ) P ( No birds for ist ne hour) P ( B = P ( X = 0 ; x = > ( 1+1 ) P ( x 2 0 ; 1 = > () ] = e- . . 0 ! = 1 and = elt ( + at ) = 1 2 1 - 7 P ( she doesn't see any birds for next t hours ) = e-it - 7 P ( she doesn't see any bird for next I hour ) = e

Ans 6 let she spends time t to watch a bird, then ; time to watch * number of birds = It = 1 hour given : - birds arrive at rate 1 7 1 number of birds in 1 hour = > X t = 1 = > t= 1/ = Average time to spend birdwatching Average interarrival time . And C Number of birds seen in a unit time = 1 ( given ) 7 Number of birds seen in time t units = 1+ -> Nbirds a t time t spent watchy I 2 3 4 5 t Number of birds seen 2 1 3 x 57

Hence, we can see that the number of birds seen are in the uniform distribution.