The Gateway to Computer Science Excellence
+18 votes

Suppose $p$ is the number of cars per minute passing through a certain road junction between $5$ PM and $6$ PM, and $p$ has a Poisson distribution with mean $3$. What is the probability of observing fewer than $3$ cars during any given minute in this interval?

  1. $\dfrac{8}{(2e^{3})}$
  2. $\dfrac{9}{(2e^{3})}$
  3. $\dfrac{17}{(2e^{3})}$
  4. $\dfrac{26}{(2e^{3})}$
in Probability by
edited by | 3.6k views

2 Answers

+39 votes
Best answer

Answer is (C)

Poisson Probability Density Function (with mean $\lambda$) =$\dfrac{ \lambda^{k}} { \left(e^{\lambda}k!\right)}$,

We have to sum the probability density function for $k = 0,1$ and $2$ and $\lambda = 3 $(thus finding the cumulative mass function)

=$\left(\dfrac{1}{e^{3}}\right) + \left(\dfrac{3}{e^{3}}\right) + \left(\dfrac{9}{2e^{3}}\right)$


edited by

As poisson distribution is for discrete random variables, isn't it supposed to be called as "probability mass function" instead of "probability density function"? check this.

Yes, probability mass function is right.
+14 votes

Let X be a random variable which indicates number of cars per minute.

Given E(X) (i.e) expected number of cars per minute (i.e) average number of cars per minute                                                         (i.e) mean number of cars per minute = 3 

probability of number of cars being 0 or 1 or 2 per minute

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

                                   = [e-3 (3)0 / 0!] + [e-3 (3)1 / 1!] + [e-3 (3)2 / 2!] 

                                   = 17 / 2e

edited by

Related questions

Quick search syntax
tags tag:apple
author user:martin
title title:apple
content content:apple
exclude -tag:apple
force match +apple
views views:100
score score:10
answers answers:2
is accepted isaccepted:true
is closed isclosed:true
52,345 questions
60,511 answers
95,354 users