Standard Discrete & Continuous Distributions
“Binomial, Poisson approximation, Uniform, Exponential memoryless property, and Normal / Gaussian curve.”
1.The Classical Family of Parametric Distributions
§ 5.1.The GATE CS syllabus tests five standard distribution models thoroughly:
1. Binomial Distribution: Models the number of successes in n independent Bernoulli trials with constant success probability p.
2. Poisson Distribution: Models the count of rare events occurring independently at a constant average rate lambda in a fixed time/space interval. It is the limiting form of Binomial as n -> infinity, p -> 0, with np = lambda.
3. Continuous Uniform: Maximum entropy distribution on a bounded interval [a, b] when no prior information exists.
4. Exponential Distribution: Models the waiting time between Poisson arrivals. Unique continuous distribution possessing the memoryless property.
5. Normal (Gaussian) Distribution: By the Central Limit Theorem, the sum/mean of many independent random variables approaches a normal distribution regardless of their original distributions.
If a GATE question states that a discrete count variable has Mean = Variance, immediately recognize it as a Poisson distribution. If Mean > Variance, it is Binomial (np > np(1-p)).
2. Worked Problem: Exponential Memoryless Property in Server Queueing
Using full conditional probability integrals when the memoryless property gives the answer directly.
Problem [Original Practice Problem · Modeled on GATE Pattern]:
The service time of requests at a web server is exponentially distributed with a mean of 4 milliseconds.
A request has already been running for 6 milliseconds without completing.
What is the probability that its total service time exceeds 10 milliseconds?
Step 1: Identify distribution parameters:
Mean = 1 / lambda = 4 ms => lambda = 1/4 = 0.25 ms^{-1}.
Let X be the total service time. X ~ Exp(lambda = 0.25).
Step 2: Formulate the conditional probability query:
We want P(X > 10 | X > 6).
Step 3: Apply the Memoryless Property P(X > s + t | X > s) = P(X > t):
Here s = 6 and s + t = 10 => t = 4.
P(X > 10 | X > 6) = P(X > 4).
Step 4: Compute survival function P(X > t) for Exponential distribution:
P(X > t) = integral_t^infinity lambda e^{-lambda x} dx = e^{-lambda * t}.
P(X > 4) = e^{- (0.25) * 4} = e^{-1} ≈ 0.3679.
Conclusion: The probability that it takes more than 10 ms given it took > 6 ms is exactly 1/e ≈ 36.79%.
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