mathematics-class-12-unit-13-chapter-01-probability-vanance-binomial-distribution-l-1-10-auw1-wtz6ym

### <Success style="font-size:25px"> Probability Vanance Binomial Distribution L-1 </Success>
### <primary style="font-size:24px"> Properties of expectation </primary>
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- $ \text { (ii) }$                        
- $ E(a X+b) =a E(X)+b$                          
- $ E(a X+b)=\left(a x_1+b\right) p_1  +\left(a x_2+b\right) p_2+\cdots$                         
- $ +\left(x_n+b\right) p_n$                          
- $=a\left(x_1 p_1\right.  \left.+x_2 p_2+\cdots+x_n p_n\right)$                           
- $ +b\left(p_1+p_2+\cdots+p_n\right)$                          
- $=a E(X) +b$

- So using these properties in (1), we get                    
- $ \operatorname{Var}(X)  =E\left(X^2\right)-2 \mu E(X)+\mu^2$                         
- $ =E\left(X^2\right)-2 \mu^2+\mu^2\\\\=E\left(X^2\right)-\mu^2$      
- Or, $ \operatorname{Var}(X) = E(X^2)-[E(X)]^2$

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<footer style = "font-size: 18px">Probability lecture-1 $\rightarrow$ Variance $\rightarrow$ <span style = "color:blue"> Properties of expectation </span> $\rightarrow$ Example-1 $\rightarrow$ Example-1 </footer>

### <Success style="font-size:25px"> Probability Vanance Binomial Distribution L-1 </Success>
### <primary style="font-size:24px"> Example-1 </primary>
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- **Score of the card problem**

- $ E\left(x^2\right)=\sum_{i=2}^{10} i^2 \cdot \frac{1}{13}+(15)^2 \cdot \frac{3}{13}+(18)^2 \cdot \frac{1}{13}$

- $ =\frac{1}{13}\left(\frac{10 \times 11 \times 21}{6}-1\right)+\frac{225\times3}{13}+\frac{324}{13} =\frac{1383}{13}$          
 
- $  \operatorname{Var}(X)={E(X^2)}-\\{E(X)\\}^2 =\frac{1383}{13}-(9)^2=\frac{330}{13}$        
 
- Standard deviation of a random variable $X$ is defined as         
- $\text { s.d. (X) }=\sqrt{\operatorname{Var}(X)}$

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<footer style = "font-size: 18px">Properties of expectation $\rightarrow$ Example-1 $\rightarrow$ <span style = "color:blue"> Example-1 </span> $\rightarrow$ Example-2 $\rightarrow$ Example-2 </footer>

### <Success style="font-size:25px"> Probability Vanance Binomial Distribution L-1 </Success>
### <primary style="font-size:24px"> Example-2 </primary>
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-  Two distinct fair dies are rolled once.

- Let $X$ denote the absolute difference of the numbers shown on the upper faces of the two dice.               
- Find the distribution of $X$ and Var(X).            
 - **Solution:**        
- The possible values of $X$ (absolute difference) are $0,1,2,3,4,5$.            
- The absolute difference is $0$ for                          
- $ \\{ (1,1), (2,2), (3,3),(4,4),(5,5),(6,6)\\}$

- So, $P(X=0)=\frac{6}{36}=\frac{1}{6}$                          
- The absolute difference is $1$ for                          
- $\\{(1,2),(2,1),(2,3),(3,2),(3,4),(4,3),(4,5),(5,4)(5,6),(6,5)\\}$ $ ~ $ So, $P(X=1)=\frac{10}{36}=\frac{5}{18}$

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<footer style = "font-size: 18px">Example-1 $\rightarrow$ Example-1 $\rightarrow$ <span style = "color:blue"> Example-2 </span> $\rightarrow$ Example-2 $\rightarrow$ Example-2 </footer>

### <Success style="font-size:25px"> Probability Vanance Binomial Distribution L-1 </Success>
### <primary style="font-size:24px"> Example-2 </primary>
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- The absolute difference is $2$ for                          
- $ \\{(1,3),(3,1),(2,4),(4,2),(3,5),(5,3),(4,6),(6,4)\\}$                                
- $ \rightarrow 8 \text { cases }$ $ ~ $  P$(X=2)=\frac{8}{36}=\frac{2}{9}$                              
- The absolute difference is 3 for                          
- $ \\{(1, 4),(4,1),(2,5),(5,2),(3,6),(6,3)\\}  \rightarrow 6 \text { cases }$ $ ~ $  $ P(X=3)=\frac{6}{36}=\frac{1}{6}$                
- The absolute difference is 4 for                          
- $ \\{(1,5),(5,1),(2,6),(6,2)\\} \rightarrow 4 \text { cases. }$ $ ~ $  $ P(X=4)=\frac{4}{36}=\frac{1}{9}$                            
- The absolute difference is 5 for
- $ \\{(1,6),(6,1)\\} \rightarrow 2 \text { cases }$ $ ~ $  $ P(x=5)=\frac{2}{36}=\frac{1}{18}$

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<footer style = "font-size: 18px">Example-1 $\rightarrow$ Example-2 $\rightarrow$ <span style = "color:blue"> Example-2 </span> $\rightarrow$ Example-2 $\rightarrow$ Example-3 </footer>

### <Success style="font-size:25px"> Probability Vanance Binomial Distribution L-1 </Success>
### <primary style="font-size:24px"> Example-2 </primary>
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- So the probability distribution of $x$ is          
- $ p_0=\frac{1}{6}, p_1=\frac{5}{18}, p_2=\frac{2}{9}, p_3=\frac{1}{6}, p_4=\frac{1}{9}, p_5=\frac{1}{18}$

- $ E(X)=0 \cdot \frac{1}{6}+1 \cdot \frac{5}{18}+2 \cdot \frac{2}{9}+3 \cdot \frac{1}{6}+4 \cdot \frac{1}{9}+5 \cdot \frac{1}{18}$ $ ~ $ $ =\frac{35}{18}$

- $ E\left(X^2\right)=0^2 \cdot \frac{1}{6}+1^2 \cdot \frac{5}{18}+2^2 \cdot \frac{2}{9}+3^2 \cdot \frac{1}{6}+4^2 \cdot \frac{1}{9}+5^2 \cdot \frac{1}{18}$

- $ =\frac{35}{6} \text {. }$ $ ~ $ $ \operatorname{Var}(X)=E\left(X^2\right)-\[{E(X)\}]^2=\frac{35}{6}-\left(\frac{35}{18}\right)^2=\frac{665}{324} \text {. }$

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<footer style = "font-size: 18px">Example-2 $\rightarrow$ Example-2 $\rightarrow$ <span style = "color:blue"> Example-2 </span> $\rightarrow$ Example-3 $\rightarrow$ Example-3 </footer>

### <Success style="font-size:25px"> Probability Vanance Binomial Distribution L-1 </Success>
### <primary style="font-size:24px"> Example-3 </primary>
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- $\underline{\text{Example}}$: An urn contains $n$ distinct marbles. Marbles are drown successively with replacement from the urn at random until a marble is repeated.

- Let $X$ denote the number of draws needed to complete the experiment. Find the distribution of $X$.

- Solution: $X$ can take values $2,3, \cdots,(n+1)$ $P(X=2)=\frac{1}{n}$ (ie. the marble drawn on the first draw is drawn again)

- $P(X=3)=\frac{n-1}{n} \cdot \frac{2}{n}$

- (ie. the marble on second is distinct from the first and third is any of the first row)

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<footer style = "font-size: 18px">Example-2 $\rightarrow$ Example-2 $\rightarrow$ <span style = "color:blue"> Example-3 </span> $\rightarrow$ Example-3 $\rightarrow$ Bernoullian Experiments </footer>

### <Success style="font-size:25px"> Probability Vanance Binomial Distribution L-1 </Success>
### <primary style="font-size:24px"> Example-3 </primary>
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- $P(X=4)=\frac{n-1}{n} \cdot \frac{n-2}{n} \cdot \frac{3}{n}$     
- $ \vdots$     
- $ P(X=i)=\frac{n-1}{n} \cdot \frac{n-2}{n} \cdots \frac{n-i+2}{n} \cdot \frac{i-1}{n}$     
- $ \vdots$      
- $ P(X=n+1)=\frac{n-1}{n} \cdot \frac{n-2}{n} \cdots \frac{1}{n+1} \cdot \frac{n}{n}$

- We can verify that $\sum_{i=2}^{n+1} P(X=i)=1$ (stat from the end and use term by term sum).

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<footer style = "font-size: 18px">Example-2 $\rightarrow$ Example-3 $\rightarrow$ <span style = "color:blue"> Example-3 </span> $\rightarrow$ Bernoullian Experiments $\rightarrow$ Bernoullian Experiments </footer>

### <Success style="font-size:25px"> Probability Vanance Binomial Distribution L-1 </Success>
### <primary style="font-size:24px"> Bernoullian Experiments </primary>
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- In many real life experiments, one is interested in only two possible outcomes.

- For example: hitting a target   
- $\quad \quad \downarrow$    
- (success or failure)

- $\rightarrow$ treating a patient  
- $\quad \quad \downarrow$  
- (cured or not cured)

- $\rightarrow$ appearing in exam   
- $\quad \quad \downarrow$   
- (qualifying not qualifying)

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<footer style = "font-size: 18px">Example-3 $\rightarrow$ Example-3 $\rightarrow$ <span style = "color:blue"> Bernoullian Experiments </span> $\rightarrow$ Bernoullian Experiments $\rightarrow$ Binomial distribution </footer>

### <Success style="font-size:25px"> Probability Vanance Binomial Distribution L-1 </Success>
### <primary style="font-size:24px"> Binomial distribution </primary>
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- Let $X \rightarrow$ the number of successes in $n$ trials.          
- Then $X$ is a discrete $r . v$. and it can take values $0,1,2, \ldots, n$          
- $p_k=p(x=k)\\\\= \binom{n}{k} p^k q^{n-k}, \quad k=0,1,2, \ldots, n$                      
- This is called binomial distribution.              
- We first show that this is a valid probability distribution.              
- $ \text { i.e. }  \sum_ {k=0}^n p_ k  ~  =\sum_{k=0}^n\binom{n}{k} p^k  q^{n-k}$

- $ =   \binom{n}{0} q^n+ \binom{n}{1}  p q^{n-1}+\binom{n}{2} p^2 q^{n-2}+\cdots+\binom{n}{n} p^n$ $ =  (q+p)^n=(1)^n =1 $

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<footer style = "font-size: 18px">Bernoullian Experiments $\rightarrow$ Bernoullian Experiments $\rightarrow$ <span style = "color:blue"> Binomial distribution </span> $\rightarrow$ Binomial distribution E(x) and var(x) $\rightarrow$ Binomial distribution E(x) and var(x) </footer>

### <Success style="font-size:25px"> Probability Vanance Binomial Distribution L-1 </Success>
### <primary style="font-size:24px"> Binomial distribution E(x) and var(x) </primary>
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- We now calculate $E(X)$ and $\operatorname{Var}(X)$             
- $ \mu=E(X)  \\\\=\sum_{k=0}^n k p_k\\\\=\sum_{k=0}^n k \cdot\binom{n}{k} p^k q^{n-k}$                   
- $ =\sum_{k=1}^n k\binom{n}{k} p^k q^{n-k}$ $ =\sum_{k=1}^n k \cdot \frac{n !}{k !(n-k) !} p^k q^{n-k}$           
- $ =\sum_{k=1}^n \frac{n ! p^{k} q^{n-k}}{(k-1) !(n-1-\overline{k-1}) !}\quad \quad k-1=m$              
- $ =n p \sum_{m=0}^{n-1} \frac{(n-1) ! p^{k-1} q^{n-1-m}}{m !(n-1-m) !} $                   
- $ =n p \sum_{m=0}^{n-1} \binom{n-1}{m} p^{m} q^{n-1-m}$ $ =np(q+p)^{n-1}=n p $

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<footer style = "font-size: 18px">Bernoullian Experiments $\rightarrow$ Binomial distribution $\rightarrow$ <span style = "color:blue"> Binomial distribution E(x) and var(x) </span> $\rightarrow$ Binomial distribution E(x) and var(x) $\rightarrow$ Binomial distribution E(x) and var(x) </footer>

### <Success style="font-size:25px"> Probability Vanance Binomial Distribution L-1 </Success>
### <primary style="font-size:24px"> Binomial distribution E(x) and var(x) </primary>
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- $ \operatorname{Var}(X)  =E\left(X^2\right)-\\{E(X)\\}^2$                     
- $ E\left(X^2\right)  =E\\{X(X-1)+X\\}$                     
- $ =E\\{X(X-1)\\}+E(X)$                    
- $ E\\{X(X-1)\\} \\\\ =\sum_{k=0}^n k(k-1) \cdot \binom{n}{k} p^k q^{n-k}$                     
- $ =\sum_{k=2}^n k(k-1) \cdot \binom{n}{k} p^k q^{n-k}$                      
- $ =\sum_{k=2}^n k(k-1) \cdot \frac{n ! p^k q^{n-k}}{k !(n-k) !}$                    
- $=\sum_{k=2}^n \frac{n ! p^k q^{n-k}}{(k-2) !(n-k) !}$                             
- $=n(n-1)p^2 \sum_{k=2}^n \frac{(n-2) ! p^{k-2} q^{n-2-(k-2)}}{(k-2) !(n-2-k-2) !}$

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<footer style = "font-size: 18px">Binomial distribution $\rightarrow$ Binomial distribution E(x) and var(x) $\rightarrow$ <span style = "color:blue"> Binomial distribution E(x) and var(x) </span> $\rightarrow$ Binomial distribution E(x) and var(x) $\rightarrow$ Binomial distribution E(x) and var(x) </footer>

### <Success style="font-size:25px"> Probability Vanance Binomial Distribution L-1 </Success>
### <primary style="font-size:24px"> Binomial distribution E(x) and var(x) </primary>
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- So  
- $\operatorname{Var}(X)=E\left(X^2\right)-[E(X)\]^2$          
- $=n(n-1) p^2+n p-n^2 p^2$          
- $=n p-n p^2$       
- $ =n p(1-p)=n p q$

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<footer style = "font-size: 18px">Binomial distribution E(x) and var(x) $\rightarrow$ Binomial distribution E(x) and var(x) $\rightarrow$ <span style = "color:blue"> Binomial distribution E(x) and var(x) </span> $\rightarrow$ Remark $\rightarrow$ Special case </footer>

### <Success style="font-size:25px"> Probability Vanance Binomial Distribution L-1 </Success>
### <primary style="font-size:24px"> Special case </primary>
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- <ins>In this case,</ins>            
- $ P(X=n-k)  = \binom{n}{n-k} \frac{1}{2^n}$                  
- $ = \binom{n}{n-k} \cdot \frac{1}{2^n}=P(X=k)$

- For example    
- $p_0=p_n=\frac{1}{2^n}$       
- $p_1=p_{n-1}=\frac{n}{2^n} \ldots$

- In this case,     
- $E(X)=\frac{n}{2},\\\\V(X)=\frac{n}{4}$   
- $\text { s.d. }(X)=\frac{\sqrt{n}}{2} \text {. }$

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<footer style = "font-size: 18px">Binomial distribution E(x) and var(x) $\rightarrow$ Remark $\rightarrow$ <span style = "color:blue"> Special case </span> $\rightarrow$ Thank you $\rightarrow$  </footer>