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+11 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?

(A) $\dfrac{8}{(2e^{3})}$

(B) $\dfrac{9}{(2e^{3})}$

(C) $\dfrac{17}{(2e^{3})}$

(D) $\dfrac{26}{(2e^{3})}$
asked in Probability by Boss (18k points)
edited by | 1.8k views

2 Answers

+24 votes
Best answer
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)

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

answered by Veteran (342k points)
edited by
+9 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

answered by Loyal (6.8k points)
edited by

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