GATE CSE 2013 | Question: 2

Question

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. $\dfrac{26}{(2e^{3})}$

Comments

In case you are wondering why λ is taken as mean

In poison distribution, λ is the expected value, which is mean.

https://stats.stackexchange.com/questions/30365/why-is-expectation-the-same-as-the-arithmetic-mean#:~:text=The%20expectation%20is%20the%20average,variable%20not%20a%20probability%20distribution.&text=The%20sample%20mean%20(or%20sample,arithmetic%20average%20of%20the%20data.

Answer 1

Answer is (C)

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

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

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

$=\dfrac{17}{\left(2e^{3}\right)}$

Comments

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.

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Yes, probability mass function is right.

Answer 2

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 / 2e³ 

Comments

perfect explanation.