mathematics-class-12-unit-13-chapter-09-probability-random-variables-l-9-10-krglsh65qp4

### <Success style="font-size:25px"> Probability Random Variables L-9 </Success>
### <primary style="font-size:24px"> Definition random variable </primary>
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- Definition: A random variable $X$ is a real valued function defined on the sample space.
- $X:$

- $S \rightarrow \mathbb{R} \longrightarrow$ set of all real numbers To each $w \in S, \quad X(w)$ is a real number height of an adult male $X \in[120,240] \mathrm{cm}$ age of a person $\quad[0,125]$ yare.

- Number of attempts needed to hit a longer

- $\Rightarrow 1,2,3, \ldots$.
- If a random variable takes a finite or countably infinite number of values, then

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<footer style = "font-size: 18px">Random variables $\rightarrow$ Random variable example $\rightarrow$ <span style = "color:blue"> Definition random variable </span> $\rightarrow$ Discrete random variable $\rightarrow$ Distribution random variable </footer>

### <Success style="font-size:25px"> Probability Random Variables L-9 </Success>
### <primary style="font-size:24px"> Discrete random variable </primary>
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- It is called a discrete random variable. If the random variable takes values over an interval it is called a continuous r.u.

- Thus we can say that for a discrete random variable $X$ the set of all possible values can be discribled by E={$X_1$, $X_2$...$X_n$} or F={$X_1$, $X_2$......}.
- The probability distribution of a discrete random variable is an assignment of
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### <Success style="font-size:25px"> Probability Random Variables L-9 </Success>
### <primary style="font-size:24px"> Example-1 </primary>
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- Dist' of number of tails.

- $p_0=\frac{1}{8}, \quad p_1=\frac{3}{8}, \quad p_2=\frac{3}{8}, \quad p_3=\frac{1}{8}$

- So the mean of X (number of tails) is

-  No. of tails.

- $\quad p_0=\frac{1}{8}, \quad p_1=\frac{3}{8}, \quad p_2=\frac{3}{8}, \quad p_3=\frac{1}{8} 
\mu=E(X)=\frac{3}{2} $.

- $\mu=E(X)=\frac{3}{2}$.
- $ \operatorname{Var}(X)= E(X-\mu)^2=E\left(X-\frac{3}{2}\right)^2$

- =$\left(0-\frac{3}{2}\right)^2 \cdot \frac{1}{8}+\left(1-\frac{3}{2}\right)^2 \cdot \frac{3}{8}+\left(2-\frac{3}{2}\right)^2 \cdot \frac{3}{8} $
- $+\frac{A}{}\left(3-\frac{3}{2}\right)^2 \cdot \frac{1}{8}$

- $=  \frac{9}{4} \cdot \frac{1}{8}+\frac{1}{4} \cdot \frac{3}{8}+\frac{1}{4} \cdot \frac{3}{8}+\frac{9}{4} \cdot \frac{1}{8}$
- $= \frac{24}{32}=\frac{3}{4} .$

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

### <Success style="font-size:25px"> Probability Random Variables L-9 </Success>
### <primary style="font-size:24px"> Example-2 </primary>
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- $ p_1=P(X=1) =\frac{\binom71  \binom31}{\binom{10}2}= \frac{21}{45} = \frac{7}{15}$

- $ p_2=P(X=2)=\frac{\binom32}{\binom{10}2}= \frac{3}{45} = \frac{1}{15}$

- Number of defective bulbs

- $ p_0=\frac{7}{15}, \quad p_1=\frac{7}{15}, \quad p_2=\frac{1}{15}$ 
- $ E(X)=0 \cdot \frac{7}{15}+1 \cdot \frac{7}{15}+2 \cdot \frac{1}{15}=\frac{9}{15}=\frac{3}{5}$

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

### <Success style="font-size:25px"> Probability Random Variables L-9 </Success>
### <primary style="font-size:24px"> Example-3 </primary>
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- Suppose we want the probability that the score is at learnt 10 .

- $P(X \geqslant 10)  =P(X=10)+P(X=15)+P(X=18)$

- $ =\frac{1}{13}+\frac{3}{13}+\frac{1}{13}=\frac{5}{13}$

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

### <Success style="font-size:25px"> Probability Random Variables L-9 </Success>
### <primary style="font-size:24px"> Example-3 </primary>
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- $ \operatorname{Var}\left(X_2\right)= E\left(X_2-\mu_2\right)^2, \quad \mu_2=E\left(X_2\right)$ 
- $=E\left(X_2\right)^2=4 \cdot \frac{1}{3}+0 \cdot \frac{1}{3}+4 \cdot \frac{1}{3}$ 
- $= \frac{8}{3}$

- Thus $\operatorname{Var}\left(X_2\right)>\operatorname{Var}\left(X_1\right)$.

- Thus we can see that random variable $X_2$ has more variation than random variable $X_1$.

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