Mathematics-III (BT-301) Unit 5 Notes - RGPV

How unit 5 is examined

This unit covers PMF and PDF, then the Binomial, Poisson, Normal and Exponential distributions; Poisson (mean and variance derivation, numericals) and Binomial numericals carry the marks.

Probability Mass Function

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Definition. For a discrete random variable $X$, the probability mass function is $p(x_i)=P(X=x_i)$, the probability that $X$ takes exactly the value $x_i$.

Key points.

A PMF must satisfy $p(x_i)\ge 0$ for every value $x_i$.
The probabilities add up to one, $\sum_i p(x_i)=1$.
The distribution function is obtained by adding, $F(x)=P(X\le x)=\sum_{x_i\le x}p(x_i)$.
Binomial and Poisson are the standard discrete distributions, each defined by its own PMF.

==A PMF assigns a probability to each value of a discrete variable, with $p(x_i)\ge 0$ and $\sum p(x_i)=1$.==

Probability Density Function

Definition. For a continuous random variable $X$, the probability density function $f(x)$ gives probabilities as areas: $P(a<X<b)=\int_a^b f(x)\,dx$.

Key points.

A PDF satisfies $f(x)\ge 0$ everywhere.
The total area under it is one, $\int_{-\infty}^{\infty} f(x)\,dx=1$; this normalisation fixes an unknown constant.
For a continuous variable $P(X=a)=0$, so $<$ and $\le$ give the same probability.

Example. $f(x)=cx^2$, $0<x<1$. Normalise: $\int_0^1 cx^2dx=\frac{c}{3}=1$, so $c=3$. Then $P\left(\frac13<x<\frac12\right)=\int_{1/3}^{1/2}3x^2dx=\left(\frac12\right)^3-\left(\frac13\right)^3=\frac18-\frac1{27}=\frac{19}{216}$.

$c=3$ and $P=\frac{19}{216}\approx 0.088$.

==A PDF is non-negative and $\int f(x)\,dx=1$; probability is the area under it.==

Asked: [7 marks] (Dec 2025) If $f(x)=cx^2$, $0<x<1$, find the value of $c$ and determine the probability that $\frac13<x<\frac12$.

Discrete Distribution: Binomial

Definition. In $n$ independent trials, each with success probability $p$ and failure probability $q=1-p$, the number of successes $X$ follows the binomial distribution $P(X=k)=\binom{n}{k}p^kq^{n-k}$, $k=0,1,\dots,n$.

Key points.

The trials are independent, there are a fixed number $n$ of them, and each has only two outcomes, success or failure.
The probability $p$ of success is the same in every trial and $p+q=1$.
The mean is $np$ and the variance is $npq$, so the variance is always less than the mean.
The probabilities $P(0),\dots,P(n)$ are the terms of $(q+p)^n$, so they add up to 1.
"Between" limits are strict: $P(1<X<4)=P(2)+P(3)$, and "at least" is done through the complement.

Formula. $P(X=k)=\binom nk p^kq^{n-k}$.

Example (20% defective, $n=5$). $p=0.2$, $q=0.8$.

Part	Working	Value
None defective	$(0.8)^5$	0.32768
One defective	$5(0.2)(0.8)^4$	0.4096
$P(2)$	$10(0.2)^2(0.8)^3$	0.2048
$P(3)$	$10(0.2)^3(0.8)^2$	0.0512

$P(1<X<4)=P(2)+P(3)=0.2048+0.0512=$ 0.256.

Example (fair coin, 10 tosses). $n=10$, $p=q=\frac12$. $P(X=6)=\binom{10}{6}\frac1{2^{10}}=\frac{210}{1024}=\frac{105}{512}\approx0.205$. $P(X\ge6)=\frac{210+120+45+10+1}{1024}=\frac{386}{1024}=\frac{193}{512}\approx0.377$.

Example (find $p$). $n=6$, $9P(4)=P(2)$: $9\binom64p^4q^2=\binom62p^2q^4$. As $\binom64=\binom62=15$, $9p^2=q^2$, so $3p=q=1-p$, giving $p=\frac14$, $q=\frac34$. Then $P(X=1)=6\cdot\frac14\cdot\left(\frac34\right)^5=\frac{729}{2048}\approx$ 0.356.

Answer frame. Open with the definition and write $n,p,q$ from the data first; state the formula; compute each required term in a small table; for "at least" use $1-$ complement; close with the boxed values.

==$P(X=k)=\binom nk p^kq^{n-k}$, with mean $np$ and variance $npq$.==

Pitfall: Writing $P(1<X<4)$ as $P(1)+P(2)+P(3)+P(4)$; strict limits leave out both ends.

Asked: [7 marks] (Nov 2022, Jun 2023) 20% of items produced from a factory are defective. Find the probability that in a sample of 5 chosen at random (i) none is defective (ii) one is defective (iii) $P(1<x<4)$. A fair coin is tossed 10 times, what is the probability of getting exactly 6 heads and at least six heads. Asked: [7 marks] (Jun 2025) A binomial variable $X$ satisfies $9P(X=4)=P(X=2)$ when $n=6$. Find the parameter $p$ and $P(X=1)$.

Poisson's Distribution

Definition. The Poisson distribution is the limit of the binomial when $n\to\infty$, $p\to0$ and $np=\lambda$ stays finite; it gives the probability of $r$ rare events: $P(X=r)=\dfrac{e^{-\lambda}\lambda^r}{r!}$, $r=0,1,2,\dots$

Key points.

It models rare events in a large number of trials, such as defectives in a sample, accidents or telephone calls.
Its only parameter is $\lambda=np$, the average number of events.
The mean and the variance are both equal to $\lambda$.
The probabilities add up to one because $\sum \lambda^r/r!=e^{\lambda}$.
"At least one" is found as $1-P(0)=1-e^{-\lambda}$.
In a numerical, first find $\lambda$ from the given condition or from $np$.

Derivation (mean and variance).

$$E[X]=\sum_{r=0}^{\infty} r\frac{e^{-\lambda}\lambda^r}{r!}=\lambda e^{-\lambda}\sum_{r=1}^{\infty}\frac{\lambda^{r-1}}{(r-1)!}=\lambda e^{-\lambda}e^{\lambda}=\lambda$$

Write $r^2=r(r-1)+r$:

$$E[X(X-1)]=\sum_{r=2}^{\infty} r(r-1)\frac{e^{-\lambda}\lambda^r}{r!}=\lambda^2e^{-\lambda}\sum_{r=2}^{\infty}\frac{\lambda^{r-2}}{(r-2)!}=\lambda^2$$

So $E[X^2]=\lambda^2+\lambda$, and

$$\mathrm{Var}(X)=E[X^2]-(E[X])^2=\lambda^2+\lambda-\lambda^2=\lambda.$$

Example ($P(1)=P(2)$). $\lambda e^{-\lambda}=\frac{\lambda^2e^{-\lambda}}{2}$ gives $\lambda=2$, so the mean is 2.

$P(x\ge1)=1-P(0)=1-e^{-2}\approx$ 0.8647.
$P(1<x<4)=P(2)+P(3)=e^{-2}\left(2+\frac43\right)=$ $\frac{10}{3}e^{-2}\approx0.4511$.

Example (samples). $m=np=2$, so $P(r)=e^{-2}2^r/r!$. $P(0)=e^{-2}=0.1353$. $P(r\ge3)=1-e^{-2}(1+2+2)=1-5e^{-2}=0.3233$. Out of 1000 samples: at least 3 defectives: about 323; none defective: about 135.

Answer frame. Derivation: open with the PMF and $E[X]=\sum rP(r)$; take out $\lambda e^{-\lambda}$, use $\sum\lambda^{r-1}/(r-1)!=e^\lambda$; then find $E[X(X-1)]$, then $E[X^2]$; close with variance $=\lambda$, mean equals variance. Numerical: find $\lambda$ first, write the PMF, compute each term, close with the values.

<mark>For the Poisson distribution the mean and the variance are both $\lambda$.</mark>

Pitfall: Forgetting to change $r=0$ to $r=1$ in the sum for the mean, which breaks the factorial cancellation.

Asked: [7 marks] (Nov 2022) If a random variable has a Poisson distribution such that $P(1)=P(2)$, find (i) mean (ii) $P(x\ge1)$ (iii) $P(1<x<4)$. Asked: [7 marks] (Jun 2023, Dec 2025) Find the mean and variance of the Poisson distribution. Asked: [7 marks] (Dec 2025) In sampling a large number of parts manufactured by a machine, the mean number of defectives in a sample of 20 is 2; out of 1000 such samples, how many would be expected to contain (i) at least 3 defective parts (ii) none defective?

Continuous Distribution: Normal Distribution

Definition. A continuous variable $X$ is normal with mean $\mu$ and standard deviation $\sigma$ if its PDF is $f(x)=\dfrac{1}{\sigma\sqrt{2\pi}}e^{-\frac{(x-\mu)^2}{2\sigma^2}}$, $-\infty<x<\infty$.

Key points.

The curve is bell-shaped and symmetric about $x=\mu$, so mean, median and mode are equal.
Total area under the curve is 1, and half of it lies on each side of $\mu$.
The standard normal variate is $Z=\frac{X-\mu}{\sigma}$, with mean 0 and variance 1.
Probabilities are read from the table of areas: $P(0\le Z\le z)$.
About 68.26% of the area lies within $\mu\pm\sigma$, 95.44% within $\mu\pm2\sigma$ and 99.73% within $\mu\pm3\sigma$.

Derivation (mean). $E[X]=\int_{-\infty}^{\infty}xf(x)\,dx$. Put $z=\frac{x-\mu}{\sigma}$, so $x=\mu+\sigma z$, $dx=\sigma\,dz$:

$$E[X]=\mu\int_{-\infty}^{\infty}\frac{e^{-z^2/2}}{\sqrt{2\pi}}dz+\frac{\sigma}{\sqrt{2\pi}}\int_{-\infty}^{\infty}ze^{-z^2/2}dz$$

The first integral is the total area, 1. The second integrand is odd, so its integral is 0. Hence $E[X]=\mu$.

Example. $\mu=100$, $\sigma=2$; find $P(98<X<102)$. $z_1=\frac{98-100}{2}=-1$, $z_2=\frac{102-100}{2}=1$. $P(-1<Z<1)=2P(0<Z<1)=2(0.3413)=0.6826$. 68.26% of the resistors have resistance between 98 and 102 ohm.

Answer frame. Derivation: open with the PDF and the definition of $E[X]$; substitute $z$; split into two integrals; close with $E[X]=\mu$. Numerical: convert to $z$ values, draw the bell curve with the shaded area, use symmetry, close with the percentage.

==The normal curve is symmetric about $\mu$ and $Z=(X-\mu)/\sigma$ converts any normal variable to the standard one.==

Asked: [7 marks] (Nov 2022) Derive Mean of the Normal Distribution. Asked: [7 marks] (Jun 2025) A manufacturer knows that the resistance of resistors he produces has mean $\mu=100\,\Omega$ and s.d. $\sigma=2\,\Omega$. What percentage of resistors will have resistance between $98\,\Omega$ and $102\,\Omega$?

Exponential Distribution

Definition. A continuous variable has the exponential distribution with parameter $\lambda>0$ if its PDF is $f(x)=\lambda e^{-\lambda x}$ for $x>0$ and $0$ otherwise.

Key points.

The mean is $\frac1\lambda$ and the variance is $\frac1{\lambda^2}$.
The distribution function is $F(x)=1-e^{-\lambda x}$, so $P(X>x)=e^{-\lambda x}$.
It has the memoryless property: $P(X>s+t\mid X>s)=P(X>t)$.
It is used for waiting times between Poisson events, life of electronic components and service times.

==The exponential PDF is $f(x)=\lambda e^{-\lambda x}$, $x>0$, with mean $1/\lambda$ and the memoryless property.==

Asked: [7 marks] (Dec 2025) Write short note on Exponential distribution.

Last-minute revision

PMF: $p(x_i)\ge0$, $\sum p(x_i)=1$; PDF: $f(x)\ge0$, $\int f\,dx=1$.
For a continuous variable $P(a<X<b)=\int_a^b f(x)\,dx$ and $P(X=a)=0$.
$f=cx^2$ on $(0,1)$: $c=3$, $P(\frac13<x<\frac12)=\frac{19}{216}$.
Binomial: $P(k)=\binom nk p^kq^{n-k}$, mean $np$, variance $npq$.
20% defective, $n=5$: $P(0)=0.32768$, $P(1)=0.4096$, $P(1<X<4)=0.256$.
$9P(4)=P(2)$, $n=6$: $p=\frac14$, $P(1)=\frac{729}{2048}$.
Poisson: $P(r)=e^{-\lambda}\lambda^r/r!$, mean $=$ variance $=\lambda=np$.
$P(1)=P(2)$ gives $\lambda=2$; $P(x\ge1)=1-e^{-2}$; $P(1<x<4)=\frac{10}{3}e^{-2}$.
$m=2$ in 1000 samples: about 323 with at least 3 defectives, about 135 with none.
Normal: $Z=\frac{X-\mu}{\sigma}$; within $\pm1\sigma$ is 68.26%.
Exponential: mean $1/\lambda$, variance $1/\lambda^2$, memoryless.

Memory hooks

Binomial has a $q$ in the variance: $npq$; Poisson needs no $q$, so variance is just $\lambda$.
Poisson mean derivation: pull out $\lambda e^{-\lambda}$ and the rest is $e^{\lambda}$, so they cancel.
Normal: odd integrand over a symmetric range is zero, so the mean is $\mu$.
Exponential: "no memory", the past waiting time does not change the future waiting time.
Strict limits "between" drop both ends, "at least" is one minus the complement.

Coverage checklist

Probability Mass Function: definition and properties (no past questions).
Probability Density Function: $f=cx^2$, find $c$ and $P(\frac13<x<\frac12)$ (Dec 2025).
Discrete Distribution: Binomial: 20% defective sample of 5 and coin tossed 10 times (Nov 2022, Jun 2023); $9P(4)=P(2)$, $n=6$ (Jun 2025).
Poisson's: $P(1)=P(2)$ (Nov 2022); mean and variance (Jun 2023, Dec 2025); defectives in 1000 samples (Dec 2025).
Continuous Distribution: Normal Distribution: derive the mean (Nov 2022); resistors 98 to 102 ohm (Jun 2025).
Exponential Distribution: short note (Dec 2025).

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Mathematics-III (BT-301) - Unit 5 Short Notes

How unit 5 is examined

Probability Mass Function

Probability Density Function

Discrete Distribution: Binomial

Poisson's Distribution

Continuous Distribution: Normal Distribution

Exponential Distribution

Last-minute revision

Memory hooks

Coverage checklist