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[[Category: Probability and Combinatorics]]
 
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[[Category: GATE2013]]
 
[[Category: Previous year GATE questions]]
 
[[Category: Previous year GATE questions]]

Revision as of 12:56, 6 January 2014

Suppose <math>p</math> is the number of cars per minute passing through a certain road junction between 5 PM and 6 PM, and <math>p</math> 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) $8/(2e^{3})$

(B) $9/(2e^{3})$

(C) $17/(2e^{3})$

(D) $26/(2e^{3})$

Solution

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

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

=$(1/e^3) + (3/e^3) + (9/2e^3)$

=$17/(2e^{3})$





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Suppose <math>p</math> is the number of cars per minute passing through a certain road junction between 5 PM and 6 PM, and <math>p</math> 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) $8/(2e^{3})$

(B) $9/(2e^{3})$

(C) $17/(2e^{3})$

(D) $26/(2e^{3})$

Solution[edit]

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

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

=$(1/e^3) + (3/e^3) + (9/2e^3)$

=$17/(2e^{3})$





blog comments powered by Disqus