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

TTT, TTH, THT, HTT, HHT, HTH, THA, 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

$X(\omega)$= Number of H's in the sequence.

TTT $\rightarrow 0$

TTH THT HTT $\rightarrow 1$

HTH HHT THH $\rightarrow 2$

HHH $\rightarrow 3$

Notice that x takes values $\{0,1,2,3\}$ when no. of tosses $=3$

$\therefore$ If numbers of tosses is $n$ then corresponding r.v. $X$ will take values: $\{0,1, \cdots n\}$

Note that: The probability of $i, \quad i \in\{0, 1, \ldots n\}$ 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.

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

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

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

Similarly, $P(2 H)=3 p^2 q$

And, $P(3 H)= p^3$

We can see that the probabilities can be obtained by

$(q+p)^3 \quad \&$

$P(X=i) ={ }^3 C_i p^i(1-p)^{3-i}$

In general if a coin has probability 'p' for $H$ then the r.v X :

giving the can not of $H$ in $n$ tosses can be obtained from $(q+p)^n$ and,

P(x=i)=${n_ {C_i}} p^i q^{n-i}$

Probability mass function for Binomial distribution with parameter n & p Bin (n, p)

Suppose we have $X$ following Bin(6,p)

It is given that 9 P(4) = P(2)

what in the value of 'p'?

Solution:

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

$\therefore \frac{P(2)}{P(4)}=9$

i.e $\frac{{ }^6 C_ 2 p^2 q^4}{6_ {C_4} p^4 q^2}=9$

i.e $\frac{q^2}{p^2} = 9$

i.e $\frac{q}{p}=3 \therefore q= 3 p$

Suppose $X$ follows Bin(n, p) when n = 8.

P(1)=0.2048

P(2)=0.1024

Find the value of p.

Solution:

$8_ {C_1} p^1 q^7 =0.2048$

$8_ {C_2}^{p^2 q^6} =0.1024$

On dividing,

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

or = $\frac{2 q}{7 p} =2 \Rightarrow q=7 p$

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

An important concept of a r.v is its mean or expectation.

The mean of a Bin(n, p) r.v =n p, $~$ and its variance is $~ n p q$

If the mean of a Bin(n, p) r.v is 4 and its variance in half of its mean. Then compute the probability that the r.v. takes value $\geq 2$.

Solution:

Mean =n p and variance = n p r & given that

n p q= $\frac{1}{2} n p \Rightarrow q=\frac{1}{2}$

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

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

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

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

ie $1-P(0)-P(1)$

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

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

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

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

= $\frac{247}{256}$

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

$0<n_1<n_2<n_3<n_4<n_5$

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

The number of solution in 7 .

But we have done it in a mechanical way:

such a solution is not guaranteed to give all possible combination

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 polynomials & compute their product.

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

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

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

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

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.

Consider, $\left(x^0+x^1+x^2+x^3+x^4\right)$

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

Obtain the coefficient of $x^{0}$ that will give the number of possible solution to the problem.

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

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

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

Now we know that,

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

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

we need to find out coefficient of $x^9, x^8, x^7. . . x^5$

Let us consider $x^5$

Here n=2, r=5

$\therefore$ coefficient of $x^5=^{2+5-1}C_5$

$r=5 = {}^6 C_5 =6$

Let us compute for $x^9$.

Here, $r=9 \therefore$ coeff $x^9= ^{9+2-1} C_5$

${ }^{10} C_9=10$

You understand that the total no. of

possible solution = 10+9+8+7+6 = 40

$\therefore$ The number of possible solution is 40 .

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

For example: In how many ways we can get $x+y+z=50$

$\rightarrow x$ in 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 restiction will be given by coeff. of $x^{50}$ in:

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

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

$0<n_1<n_2<n_3<n_4<n_5 \quad \sum_{i=1}^5 n_i=20$

Note that all the 5 variables are integers.

Solution:

let : $m_2=n_2-n_1$

$m_3=n_3-n_2$

$m_4=n_4-n_3$

$m_5=n_5-n_4$

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

More over

$n_1+n_2+n_3+n_4+n_5=20$

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 $n_1 2$ & all $m_i>0$

Let us now make the following subsitution.

$x_1=m_1-1$

$x_2=m_2-1$

$x_3=m_3-1$

$x_4=m_4-1$

$x_5=m_5-1$

$\therefore$ Each $x_i \geq 0$

$\therefore {5 x_ 1+4 x_ 2+3 x_ 3+2 x_ 4+x_ 5}={5\left(x_ 1-1\right)+4\left(x_2-1\right)}$

$+3\left(m_3-1\right)$ $+2\left(m_n-1\right)$ $\text { and }\left(m_5-1\right)$

$=20-(1+2+3+4+5)$ $=20-15=5$ when $x_i \geq 0$

Now $5 x_1$, take values {0, 5 , 10.........}

$4 x_2$ take value{0,4,8,12........... }

$3 x_3$ take value{0,3,6,........... }

$2 x_4$ take value{0,2,4,6........... }

$x_5$ take value{0,1,2,3........... }

$\therefore$ The power series for us is

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

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

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

$\therefore$ The problem in to compute coeff of $x^5$ in.

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

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

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

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

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

$\therefore$ coeff. $x^5=7$

No of possible solution =7

$0<n_1<n_2<n_3<n_4$

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

The no. of solution = 9