Tossing a coins 3 times & noting the sequence of outcomes.

$\text{TTT TTH THT HTT HHT HTH THH HHH}$

Consider 100 tosses: therefore $2^{100}$ many sequence.

A random variable is a maping for $\Omega \rightarrow \mathbb{R}$. Consider $x$ to be a random variable such that

X$(\omega)$ = # of $\mathrm{H}$ is in the sequence.

$\text{TTT} \ldots (0\text{[this number is show as number of heads(H)]})$

$\text{TTH THT HTT} \ldots (1)$

$\text{HTH HHT THH} \ldots (2)$

$\text{ HHH } \ldots (3)$

• $x$ takes values ${0,1,2,3}$ when number of tosses $=3$
  $\therefore$ If number n of tosses is $n$ then corresponding random variable $x$ will take values: $\{0,1, \cdots n\}$

• The probabilities of $i, \ i \in \{0,1, \cdots n\}$ is not same $\ \forall i$.
  $x: 0 \quad 1 \quad 2 \quad 3$

$\quad\quad 1 \quad 3 \quad 3 \quad 1$

Suppose $p$ is the probability that the coin gives a $H$ in one toss

$P(T T T)=(1-p)(1-p)(1-p)=(1-p)^3=q^3$

$P(1 H)=P(T T H)+P(T H T)+P(H T T)$

$=q q p+q p q+p q q=3 p q^2$

Similary $P(2 H)=3 p^2 q$ we can see that $2 p(3 H)=p^3 \text {. }$

the probability can be obtained in $(q+p)^3$ & $p(x=i)$

${^3 C_2 p^i {(i-p)^{3-2}}}$

In general. If a coin has probability ’ $p$ ’ for $H$ then the random variable $x$ : giving the count of $H$ in $n$ tosses can be obtained for $(q+p)^n$.

& $\quad$P(x-i)=$\quad$ ${{^n{c_i} p^i q^{n-i}}}$

This is the probability mass function for Binomial distribution with parameters $n$ & $p$ {Binomial} $(n, p)$

Suppose we have $x$ following $\operatorname{Bin}(6,p)$ It is given that $9.P(4)=P(2)$ what is the value of $‘p’$?

Solution

$P(4)={ }^6C_4 p^4 q^2 \quad P(2)= {}^6C_2 p^2 q^4$

$\frac{P(2)}{P(4)}=9 \quad \text {i.e } \quad \frac{{ }^6 C_2 p^2 q^4}{{}^6C_4 p^4 q^2} =9$

$\text { i.e }\quad \frac{q^2}{p^2}=9 \quad \text { i.e } \quad \frac{q}{p}=3 \quad \therefore q=3 t .$

Now $p+q=1 \quad p+3p=1 \Rightarrow p=0.25 \ \ \text{or} \ \ \frac{1}{4}$

Suppose $x$ follows $\operatorname{Bin}(n, p)$ when $n=8$.

$P(1)=0.2048 \ \ \& \ \ P(2)=0.1024$

Find the value of $p$.

Solution

${}^8C_{1} p^{1} q^7=0.2048 \cdots (i)$
${}^8C_{2} p^2 q^6=0.1024 \cdots (ii)$
By dividing, we have,

$\frac{8 p q^7}{\frac{8 !}{2 ! 6 !} p^2 q^6}=\frac{0.2048}{0.1024} \quad \text { or } \quad \frac{8 p q^7}{28 p^2 q^6}=2$

$\text { or} \quad \frac{2 q}{7 p}=2 \quad \text { i.e } \quad q=7 p$

$\therefore$ value of $p=\frac{1}{8}$

An important concept of a random variable is its mean or Expectation.

The mean of a Bin(n,p) random variable $=\underline{np.}$

& its variance in $\underline{npq.}$

If the Mean of a $\operatorname{Bin}(n, x)$ random variable $n$ is 4 and its variance is half of its mean Then compute the probability that the random variable takes value $\geq 2$.

Solution

$\text { Mean }=n p, \ \ \text{ Variance }=n p q$

$[ \text { given that } \ \ n p q=\frac{1}{2} n p \Rightarrow q=\frac{1}{2} ]$

$\quad \therefore p=\frac{1}{2} .$

$\therefore \operatorname{Mean}=4 \Rightarrow n \cdot \frac{1}{2}=4 \Rightarrow n=8 .$

$\therefore \text{Bin} (8, \frac{1}{2})$

We need to compute $P\left(\operatorname{Bin}\left(8, \frac{1}{2}\right)\right) \geq 2$

i.e. $\ \ 1-P(0)-P(1)$

$= 1- ^8{C_0}\left(\frac{1}{2}\right)^0\left(\frac{1}{2}\right)^8-{ }^8 C_1\left(\frac{1}{2}\right)^1\left(\frac{1}{2}\right)^7$

$= 1-\left(\frac{1}{2}\right)^8-8 \cdot\left(\frac{1}{2}\right)^8$

$= 1-9 \cdot\left(\frac{1}{2}\right)^8$

$= 1-9 \cdot \frac{1}{256}=\frac{256-9}{256}$

$= \frac{247}{256}$

Use of Binomial theorem in counting the number of certain events. For illustration. Consider the problem that in how many ways you can have 5 different positive integers

$0

$\text {such that} \quad \sum_{i=1}^5 x_i=20$

The number of solution in 7 .

But we have done it in a mechanical way: such a solution $n$ not guaranteed to give all possible combination

The concept in “Generating Function” essentially a power series which may be finite or infinite.

In how many ways we can have two variables

$x \ \& \ y \ \Rightarrow \ 0 \leq x \leq 2 \quad$ & $\quad 1 \leq y \leq 2$

& ${x+y}=3 \text { here } X \text { and } Y \text { are integer variables }$

Consider the following two polynomial & compute their product.

$\left(z^0+z^1+z^2\right)\left(z^1+z^2\right)$

$= \left(1+z^2+ z^2\right)\left(z+z^2\right)$

$= z+z^2+z^3+z^2+z^3+z^4$

$= z+2 z^2+2 z^3+z^4$

Suppose we want to find out $x+y+z=10$ when $0 \leq x \leq 4$

$y>0, \ \ z \geq 0$ here $x, y, z$ are integer variables

$\left(x^0+x^1+x^2+x^3+x^4\right) \cdot\left(x+x^2+\cdots\right)\left(1+x+x^2+\cdots\right)$

Obtain the coefficient of $x^{10}$ that will give the number of possible solution to the problem. $\left(1+x+x^2+x^3+x^4\right) x \cdot\left(1+x+x^2+\cdots\right)^2$

Now we know that

$\left(1+x+a^2-\cdots\right)^n =\frac{1}{(1-x)^n}=(1-x)^{-n}$

$=1+\sum_{r=1}^{\infty}{ }^{n-1+r} C_r x^r$

$\therefore$ we need to find out coefficient of $x^9, x^8, \cdots x^5$

Let us consider $x^5$.

Here, $n=2, r=5$
$\therefore \quad \text{ coeff of } x^5= \quad {^{2+5-1}c_5} ={ }^6 c_5=6 .$

Let us compute for $x^9$. Here $r=9 ,n =2$

$\ \therefore \text { coeff. } x^9= {}^{9+2-1}c_9 ={ }^{10}c_9=10$

You understand that the total number of possible solution $=10+9+8+7+6=40$

$\therefore$ The number possible solutions is 40 .

• Note that we are forming the power series corresponding to each variable by considering the value that it may take.

For example:
In how many ways we can get $x+y+z=50$ such that $x$ is a multiple of 2, $y$ is a multiple of 3, $z$ is positive & multiple of 5.

$\therefore$ The no of solution for $x+y+z=50$ with the above restriction will be given by coeff. of $x^{50}$ in:

$\left(1+x^2+x^4+\cdots\right) \times(1+x^3+x^6+x^9+\cdots)$

$\times\left(x^5+x^{10}+x^{15}+\cdots\right)$

coeff. $\underline{x}^{50}$

$0

Note that all the 5 variables are integers. Let

$m_2=n_2-n_1$

$m_3=n_3-n_2$

$m_4=n_4-n_3$

$m_5=n_5-n_4$

$\therefore n_1$ & $m_2, m_3, m_4, m_5$ are all $>0$

More over

$n_1+n_2+n_3+n_n+n_5=20$

$5 n_1+4\left(n_2-n_1\right)+3\left(n_3-n_2\right)+2\left(n_4-n_3\right)+\left(n_5-n_4\right)=20$

or

$5 n_1+4 m_2+3 m_3+2 m_4+m_5=20$

when,

$n_1$ & all $m_i>0$

Let us now make the following subtitution.

$x_1=n_1-1$

$x_2=m_2-1$

$x_3=m_3-1$

$x_4=m_4-1$

$x_5=m_5-1$

$\therefore \quad \text { Each } x_i \geq 0$

$\therefore 5 x_1+4 x_2+3 x_3+2 x_4+x_5$

$=5\left(x_1-1\right)+4\left(m_2-1\right)$

$+3\left(m_3-1\right)$

$+2\left(m_4-1\right)$

$+\left(m_5-1\right)$

$=20-(1+2+3+4+5)$

$\underline{=20-15=5}$

when $x_i \geq 0$

Now $5 x_1, \quad \text{taken value} \quad \{0,5,10, \cdots\}$

$4 x_2 \quad \text{taken value} \quad \{0,4,8,12 \cdots\}$

$3 x_3 \quad \text{taken value} \quad \{0,3,6, \cdots\}$

$2 x_4 \quad \text{taken value} \quad \{0,2,4,6, \cdots\}$

$x_5 \quad \text{taken value} \quad \{0,1,2,3 \ldots\}$

$\therefore$ The power series for us in

$\left(1+x^5+x^{10} \cdots\right) \times\left(1+x^4+x^8+\cdots\right.) \times \left(1+x^3+x^6+\cdots\right) \times$

$\left(1+x^2+x^4+\cdots\right) \times \left(1+x+x^2+x^3-\cdots\right)$

We need to compute the coeff. of $x^5$.

$\therefore$ The problem is to compute cofficent of $x^5$ in

$(1+x^5)(1+x^4) (1+x^3)(1+x^2+x^4) (1+x+x^2+x^3+x^4+x^5)$

$(1+x^4+x^5)(1+x^2+x^4+x^3+x^5+x^7)(1+x+x^2+x^3+x^4+x^5)$

$(1+x^4+x^5)(1+x^2+x^3+x^4+x^5)(1+x+x^2+x^3+x^4+x^5)$

$(1+x^2+x^3+x^4+x^5+x^4+x^5)(1+x+\cdots+x^5)$

$=(1+x^2+x^3+2 x^4+2 x^5)(1+x+\cdots+x^5)$

coefficent of $x^5 = 7$

number of possible solution = 7

$0

such that $\sum_{i=1}^n n_i=16$ Note that all the 4 variables are integers

& we found that the number of solution $=9$