1. If we have $n$ identical balls which are to be kept in $k$ boxes such that 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 such that no box is empty then there are two possible way of doing it : $(1,2)$ or $(2,1)$.

2. $n$ identical balls to be kept in $k$ boxes such that some boxes may remain empty then the number of possible arrangements is
   ${}^{n+k-1}{C_{k-1}}.$

Ex. 3 balls 2 boxes

$\therefore$ possible arrangements are.

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

and we have ${}^{3+2-1}C_{2-1}={ }^4 c_1=4$

In how many ways you can choose 5 numbers $a, b, c, d, e$ such that each one is $>0$ and

$a+b+c+d+e=20$

then the number of possible solution, is

${ }^{20-1}c_{5-1}={ }^{19} c_4$

However, if $a, b, c, d, e \geq 0$ then the number of possible solution is ${}^{20+5-1}C_{5-1}$ $={ }^{24} C_4$

In how many ways you can choose 5 numbers $n_1, n_2, n_3, n_4$ & $n_5$ such that $\quad n_i>0 \quad \forall i=1,5$ and $\quad{n_1

$$\sum_{i=1}^5 n_i=20$$ $\therefore$ Note that $:(1,2,3,4,10)$ is a possible solution.

But $(1,2,4,4,9)$ in not a solution since 4 in repeated.

Note that smallest possible value for $n_1=1$ for $n_2=2$ & for $n_5=5$

$\therefore$ Let us define 5 new variables $x_1 x_2, x_3, x_n$ & $x_5$ as follows:

$x_1=n_1-1$

$x_2=n_2-2$

$x_3=n_3-3$

$x_4=n_4-4$

$x_5=n_5-5$

$\therefore \quad \text { Each } \ x_i \ \text{is} \geq 0 \quad x_1 \leq x_2 \leq x_3 \leq x_4 \leq x_5$

Hence the problem becomes: to choose 5 numbers,

$x_1$ & $x_2$ & $x_3$ & $x_n$ & $x_5 \$ such that $\text { all are } \geq 0$

$x_1+x_2+x_3+x_4+x_5$

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

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

$=20-15 =5 .$

$\therefore \ \text{number of possible solution is} \ 7$

In how many ways you choose 4 numbers

${n_1

${\sum_{i=1}^n n_i=16}, \quad n_1>0 \quad \forall i=1,2,3,4$

we define: $x_1=n_1-1 \\ x_2=n_2-2 \\ x_3=n_3-3 \\ x_n=n_4-4$

such that each $\quad$ $x_i \geq 0$

$x_1 \leqslant x_2 \leqslant x_3 \leqslant x_4$

$\sum x_i=16-10=6$

$9$ Possible solution

We talk about probability when the underlying experiment is random.

$\therefore$ the sample space $\Omega$ is known.

And we need to compute probability of what??

• Coin tossing 5 times.

  • what is the probability of two heads?
  • what is the probability of the number of tales is prime?

• Passenger with bags:

  • What is the probability that number of bags is even number of passenger is odd?

An event is a subset of the sample space $\Omega$.

consider : we toss a coin 3 times.

we want to compute that number of heads is odd. $\mathrm{HHH}, \mathrm{HTT}, \mathrm{THT}, \mathrm{TTH}$.

We are looking at the probability of the subset: $\{H T T, TH T, T T H, H H H\}$.

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 number of elementary events is 6 $$\{1\}, \{2\}, \{3\}, \cdots \{6\}.$$

An event comprising of more that one elementary event is called a compound event.

Suppose the $\Omega$ is $\{1,2,3,4,5,6,7,8,9\}.$
How many compound events are possible?

Solution:

Since $|\Omega|=9$

$\therefore$ there are $2^9$ possible subsets of which 1 is $\phi$ & 9 are elementary events.

Number of compound events is

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

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

  • Eg: Throwing a die
    $E_1$ : getting a number $\leq 3$.
    $E_2$ : getting a number $> 4$.
  • $E_1 \ \& \ E _2$ are disjoint.

• A sequence of events $E_1, E_2 \cdots E_K$ is said to be mutually exclusive if $E_i \cap E_j=\phi \ \ \text{for all} \ i,j \in \{1,2, \cdots k\} \ \ i\neq j$.

Two events $A$ and $B$ are said to be independent if

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

Probability is a mapping from the power set of $\Omega$ to $[0,1]$
i.e. if $A \subseteq \Omega$ then $P(A)=p | 0\leqslant p \leqslant 1$

Where $P$ 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 exculsive then
$\quad P\left(A_1 \cup A_2 \cdots \cup A_K\right)=\sum_{i=1}^k P\left(A_i\right)$

Given a set $A \subseteq\Omega$

$P(A)$ is computed as, $\ \frac{\text {Number of elements in A}}{|\Omega|}$

when the outcomes are equally likely.

1) Throwing a die and probability of getting an even number is

$\frac{\# \{2,4,6\}}{\mid\Omega\mid}$

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

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

Proof:

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

$1= P(\Omega)=P\left(A \cup A^c\right)=P(A)+P\left(A^c\right) \\ P\left(A^c\right)=1-P(A) .$

When we are tossing a coin,

if $A=\{H\}$ then $A^c=\{T\}$

$\therefore$ if $P(H)=P$ then

$P(T)=1-P \text {. }$

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?

Solution:

Now consider getting $2 \mathrm{H}$ ’s in 3 tosses, $P(H H T)+P(H T H)+P(T H H)$

• $P(H H T)$
  • $1^{st} H \rightarrow p$
  • $2^{nd} H \rightarrow p$
  • $3^{rd} T \rightarrow (1-p)$
• $P(H T H) \ \rightarrow p^2(1-p)$
• $P(T H H) \ \rightarrow p^2(1-p)$

Total probability $\ = 3p^2(1-p)$

Another important property is

$P(A \cup B)=P(A)+P(B)-P(A \cap B) \text {. }$
$P(A \cup B)= P\left(B \cap A^c\right)+P(B \cap A)+P(A \cap B^C)$
$\because$ These three are disjoint.

Now $P(B \cap A^c)=P(B)- P(A \cap B)$
$\& \ P(B \cap A)+P(A \cap B C)=P(A)$

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

Suppose $E$ is a random experiment. Let $A$ and $B$ be two events such that

$0

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 matually exclusive.

$\because\quad \nexists \ \omega \ \ \omega\in A \quad$& $\quad\omega \in A^C$

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

$\because \quad A \cap A^c$ is $\phi$

$P(A \cap A^c) = \frac{\# \phi}{\# \Omega}=0$

However, $P(A).P\left(A^C\right) \neq 0$
$\because 0 < P(A) < 1.$
$\therefore$ B is false.

(C) $A$ & $B$ are independent $\Rightarrow A < B^c$ are independent.
$P\left(A \cap B^C\right) =P(A)-P(A \cap B)$

$=P(A)-P(A) P(B)$
$\{\because \ A \ \& \ B \ \text{are independent.}\}$

$P\left(A \cap B^C\right) =P(A)(1-P(B))$

$=P(A) P(B^C)$

$\therefore A\ \& \ B^C \text { are independent }$
$\therefore$ C is true.

(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))$

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

$=(1-P(A))-P(B)(1-P(A))$
$\{ \because \ P(A \cap B)= P(A).P(B)\}$
$=(1-P(A))(1-P(B))$
$=P\left(A^c\right) \cdot P\left(B^c\right)$
$\therefore (A^c)$ & $(B^c)$ are independent.