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Official PYQ · 2013Q2Probability: Random Variables and DistributionsMCQ1 mark

GATE CS 2013 Q2 — Probability: Random Variables and Distributions

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Suppose pp is the number of cars per minute passing through a certain road junction between 55 PM and 66 PM, and pp has a Poisson distribution with mean 33. What is the probability of observing fewer than 33 cars during any given minute in this interval? 

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Answer: C

Let XX be the random variable representing the number of cars passing through the junction per minute. We are given that XX follows a Poisson distribution with mean λ=3\lambda = 3.

The probability mass function (PMF) of a Poisson random variable is given by:

P(X=k)=e−λλkk!=e−33kk!P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!} = \frac{e^{-3} 3^k}{k!}

We want to find the probability of observing fewer than 33 cars, which is P(X<3)P(X < 3):

P(X<3)=P(X=0)+P(X=1)+P(X=2)P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)

Calculating each term:

  • P(X=0)=e−3300!=e−3P(X = 0) = \frac{e^{-3} 3^0}{0!} = e^{-3}
  • P(X=1)=e−3311!=3e−3P(X = 1) = \frac{e^{-3} 3^1}{1!} = 3e^{-3}
  • P(X=2)=e−3322!=92e−3P(X = 2) = \frac{e^{-3} 3^2}{2!} = \frac{9}{2}e^{-3}

Adding these probabilities together:

P(X<3)=e−3(1+3+92)=e−3(172)=172e3P(X < 3) = e^{-3} \left(1 + 3 + \frac{9}{2}\right) = e^{-3} \left(\frac{17}{2}\right) = \frac{17}{2e^3}

Thus, the correct option is C.

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