2003 Paper 1 Q14

Year: 2003
Paper: 1
Question Number: 14

Course: UFM Statistics
Section: Poisson Distribution

Difficulty: 1500.0 Banger: 1475.2

probability Poisson distribution expectation truncation conditional expectation independent events

Problem

Jane goes out with any of her friends who call, except that she never goes out with more than two friends in a day. The number of her friends who call on a given day follows a Poisson distribution with parameter \(2\). Show that the average number of friends she sees in a day is~\(2-4\e^{-2}\,\). Now Jane has a new friend who calls on any given day with probability \(p\). Her old friends call as before, independently of the new friend. She never goes out with more than two friends in a day. Find the average number of friends she now sees in a day.

No solution available for this problem.

Rating Information

Difficulty Rating: 1500.0

Difficulty Comparisons: 0

Banger Rating: 1475.2

Banger Comparisons: 2

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Problem source

Jane goes out with any of her friends 
who call, except that she never  goes out with more
than two friends in a day. The number of her friends who call
on a given day follows  a Poisson distribution with parameter $2$.
Show that the average number of friends she sees in a day is~$2-4\e^{-2}\,$.
Now Jane has a new friend who calls on any given day with 
probability $p$. Her old friends call as before, independently
of the new friend.  She never  goes out with more
than two friends in a day. Find the average  number of friends she now
sees in a day.

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