If we have $n$ identical balls which are to be kept in $k$ boxes at no box will remain empty then the number of possible arrangement is

$^{n-1}C_{k-1}$

Ex: If we have 3 identical balls to be kept in 2 boxes 7 ne box in empty then there are two passible way of doing it : $(1,2)$ or $(2,1)$

$\quad$ n identical balls to be kept in $k$ boxes at some boxes may remain empty then the number of possible arrangements in

$^{n+k-1}C_{x-h}$

Ex 3 balls 2 boxes possible arrangements are.

$(0,3) \quad(1,2) \quad(2,1) \quad(3,0)=4$

& we have $^{3+2-1} C_{2-1}={ } ^4 C_1= 4$

In how many ways you can choose 5 numbers $a, b, c, d, e$ there-exist

each one is $>0$ & $a+b+c+d+e=20$

then the number of possible

Solution: in $\quad{ }^{20-1} c_{5-1}={ }^{19} c_4$

However, if $a, b, c, d, e \geq 0$ then the number of possible

sol's is $^{20+5-1}C_{5-1}$ $={ }^{24} c_4$

In how many ways you can choose 5 numbers

$n_1, n_2, n_3, n_n n_5 \ni n_6>0 \forall i=1,5$

& $~ {n_1<n_2<n_3<n_4<n_5}$

& $~ {\sum_{i=1}^5 n_i=20 .}$

$\therefore$ Note that $:(1,2,3,4,10)$ is a possible sol's.

point $(1,2,4,4,9)$ in not a sol's. since 4 in repeated.

Note that smallest possible value for $n_1=1$

for ${n_2}=2 \ldots$ for $n_5=5$.

$\therefore$ Let us define 5 new variables

$x_1 x_2, x_3, x_n$ & $x_5$ as follow.

$x_1=n_1-1$ $\qquad x_2=n_2-2$

$x_3=n_3-3$ $\qquad x_4=n_4-4$

$x_5=n_5-5$

$\therefore\text { Each } x_i \leq n \geq 0$
& $x_1 \leq x_2 \leq x_3 \leq x_n \leq x_5$

Hence the problem becomes: to choose 5 numbers

$x_1 \quad x_2 \quad x_3 \quad x_4 \quad x_5 \quad \rightarrow$ all are $\geq 0$

$x_1+x_2+x_3+x_4+x_5$

$=n_1-1+n_2-2+n_3-3+n_4-4 + n_5-5$

$=\sum_{i=1}^5 n_i-(1+2+3+4+5)$

$=20-15$

$=5$

In how many ways you choose 4 numbers

${n_1<n_2<n_3<n_4} \quad n_1>0 \quad \forall$ i=1,2,3,4

${2 \sum_{i=1}^n n_1=16},$

we define:

$x_1=n_1-1$ $x_2=n_2-2$

$x_3=n_3-3$ $x_n=n_4-4$

$\rightarrow \text { each } x_i \geq 0 \quad x_1 \leqslant x_2 \leqslant x_3 \leqslant x_4$

& $\quad \sum x_i=16-10=6$

We talk about probability when the underlying experiment in random.

the sample space $\Omega$ in known. And we need to compute probability of what ??

Example

Coin tossing 5 time.

a)

what in the probability of two H's.

what in the probability of the number of $T^{\prime} s$ is prime.

Event: an event is a subset of the sample space $\Omega$.

consider : we toss a coin 3 time. we want to compute that number of H's is odd.

HTT THT TTH & HHH

$\therefore$ We are looking at the probabilitiy of the subsect: ${H T T, T H T, T T H, H H H}$.

The subsets of cardinality 1 are called elementary events.

i.e.

the outcome of one individual trial in called an elementary event.

If we throw a die, then no of elementary events in 6 .

$\text { {1}, }{2},{3}, \cdots{6} .$

def.) An event comprising of more that one elementary event in called a "compound" event.

Suppose the $\Omega$ in ${1,2,3,4,5,6,7,8,9}.$

How many compound events are possible?

Solution:

Since $|\Omega|=9 \quad \therefore$ there are $2^9$ possible subset.

of which 1 in $\phi$ & sa re elementary events.

$\therefore ~$ No. of compound events in

$2^9-(9+1)=512-10=502$

Def):

Two events $E_1$ & $E_2$ are said to be disjoint if $E_1 \cap E_2=\varnothing$

Eg: Throwing a die.

$\left.\begin{array}{l}E_1: \text { getting a number } \leq 3 \\ E_2 \text { : getting a number }>4\end{array}\right] disjoint$

Def.)

A sequence of events $E_1, E_2 \cdots E_K$ in said to be mutnally exclusive if $E_i \cap E_j=\phi \quad \forall i, j \in{1,2, \cdots k}$ $i \neq j$.

Def.)

Two events $A$ and $B$ are said to be independent if $P(A \cap B)=P(A) * P(B)$.

What in the probability of an event?

probability in a mapping from the power set of $\Omega$ to $[0,1]$

i.e.

if $A \subseteq \Omega$ then $P(A)=p$

When $P$ satisfies the following:

a) $P(A) \geq 0 \quad \forall A \subseteq \Omega$.

b) $P(\Omega)=1$

c) If $A_1, A_2 \cdots A_k$ are matually exclusive then $P\left(A_1 \cup A_2 \cdots \cup A_k\right)=\sum_{i=1}^k P\left(A_i\right)$.

Given a set $A \leq \Omega \quad P(A)$ in computed as $\frac{ \text{number of elements in A}}{|\Omega|}$

when the outcomes are equally likely

Ex.

Throwing a die and probability of getting an even no. is $\frac{\# \{2,4,6\}}{|\Omega|}$

$=\frac{3}{6}=\frac{1}{2}$

Suppose probability of getting a $H$ is $p$. when $0<p<1$.

Then what is the probability of getting $2 \mathrm{H}$'s in 3 tosses .

Show that $P\left(A^c\right)=1-P(A)$.

since $A \cup A^C=\Omega$

$\therefore \quad 1= P(\Omega)=P\left(A \cup A^c\right)=P(A)+P\left(A^c\right)$

$\therefore P\left(A^c\right)=1-P(A) .$

Now consider getting $2 \mathrm{H}$ 's in 3 tosses

P(HHT)+ P(HTH)+ P(THH)

$P(HHT) = p \times p\times (1-p) , \quad P(HTH)=p^2(1-p),$

$P(THH)=p^2(1-p)]$

$\text { Total probability }=3 p^2(1-p) \text {. }$

$\text { in } P(A \cup B)=P(A)+P(B)-P(A \cap B)$

$P(A \cup B)= P\left(B \cap A^C\right)+$

$P(B \cap A)+$

$P\left(A \cap B^C\right)$

These three are disjoint.

$\because \quad$ $P\left(B \cap A^C\right)$ $=P(B)-P(A \cap B)$

$\ P(B \cap A)+P\left(A \cap B^C\right)=P(A) \text {. }$

$\therefore$ So Together we have $P(A \cup B)=P(A)+P(B)-P(A \cap B)$

Let A and B be two events $\exists$

$0<P(A)<1$ & $0<P(B)<1 \text {. }$

Then which of the following statements are True?

(A) $A$ and $A^c$ are mutually exclusive.

(B) $A$ and $A^c$ are independent.

(C) $A$ and $B$ are independent $\Rightarrow$ $A$ and $B^c$ are independent.

(D) $A$ and $B$ are independent $\Rightarrow$ $A^c$ and $B^c$ are independent.

(A) It is obvious that A & $A^c$ are mutually exclusive.

$\because {\exists} \omega \geqslant \omega \in A \omega \in A^c$.

(B) $A$ and $A^c$ are independent Now we know that $A$ and $B$ are independent if

$P(A \cap B)=P(A) P(B)$

Now, $P\left(A \cap A^c\right)=0$

$[\because \quad A \cap A^c$ in $\varnothing]$

However, $P(A) * P\left(A^C\right) \neq 0$

$[ \therefore P\left(A \cap A^c\right)]\qquad \because O<P(A)<1 =\frac{ \phi}{ \Omega} =0$

(C) $A$ & $B$ are independent $\Rightarrow A ~ \& ~ B^C$ are independent

$P\left(A \cap B^C\right) =P(A)-P(A \cap B)$

$=P(A)-P(A) P(B) \quad \because A ~ \& ~ B$ are independent

$=P(A)(1-P(B))$

$=P(A) P\left(B^C\right)^A$

$\therefore A ~ \& ~ B^C$ are independent

(D) $A$ and $B$ are independent $\Rightarrow$ $A^c$ and $B^c$ are independent

$P\left(A^c \cap B^c\right) =P(\Omega)-P(A \cup B)$

$=1-(P(A)+P(B)-P(A \cap B))$

$=1-P(A)-P(B)+P(A \cap B)$

$=(1-P(A))-P(B)(1-P(A))$

$\because P(A \cap B)=P(A).P(B)$

$=(1-P(A))(1-P(B))$

$=P\left(A^c\right) P\left(B^c\right)$

$\therefore A^c B^c$ are independent