GATE CS 2013 Q.2 (CS Q.2): Probability and Statistics

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?

$8/(2e^3)$

$9/(2e^3)$

$17/(2e^3)$

$26/(2e^3)$

Official answer and solution

Official answer: C
From: Official GATE 2013 answer key

Solution

Use the Poisson probability mass function with parameter lambda = 3 to sum the probabilities for k = 0, 1, 2.

Let $X$ be the Poisson random variable with parameter $\lambda = 3$ .
The probability mass function is:

P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!} = \frac{e^{-3} 3^k}{k!}

We need the probability of observing fewer than 3 cars, which is $P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)$ :

$P(X = 0) = \frac{e^{-3} 3^0}{0!} = e^{-3}$
$P(X = 1) = \frac{e^{-3} 3^1}{1!} = 3 e^{-3}$
$P(X = 2) = \frac{e^{-3} 3^2}{2!} = \frac{9}{2} e^{-3} = 4.5 e^{-3}$

Summing these probabilities:

P(X < 3) = e^{-3} \left(1 + 3 + \frac{9}{2}\right) = e^{-3} \left(\frac{17}{2}\right) = \frac{17}{2 e^3}

This matches option C.

Final answer

Option (C)

Common trap

Including $k = 3$ (fewer than 3 means strictly $k < 3$ ).

The answer is the one in the official GATE answer key. The solution is this site's own explanation.

Source

Paper: GATE 2013 Computer Science and Information Technology (CS)
Question: 2 of 65, printed as CS Q.2
Section: Computer Science and Information Technology

Report an error in this question

GATE questions are the work of the organising institutes. This site is independent and is not affiliated with them.

Question

Solution

Source