Let $X$ be a discrete random variable whose probability distribution is defined as follows.
$$P(X=x)= \begin{cases}k(x+1), & \text { for } x=1,2,3,4 \\ 2 k x, & \text { for } x=5,6,7 \\ 0, & \text { otherwise }\end{cases}$$
where, $k$ is a constant. Calculate
(i) the value of $k$.
(ii) $E(X)$.
(iii) standard deviation of $X$.
$$\begin{aligned} &P(X=x)= \begin{cases}k(x+1), & \text { for } x=1,2,3,4 \\ 2 k x, & \text { for } x=5,6,7 \\ 0, & \text { otherwise }\end{cases}\\ &\text { Thus, we have following table } \end{aligned}$$
| $X$ | 1 | 2 | 3 | 4 | 5 | 6 | 7 | Otherwise |
|---|---|---|---|---|---|---|---|---|
| $P(X)$ | $2k$ | $3k$ | $4k$ | $5k$ | $10k$ | $12k$ | $14k$ | 0 |
| $XP(X)$ | $2 k$ | $6 k$ | $12 k$ | $20 k$ | $50 k$ | $72 k$ | $98k$ | 0 |
| $X^2P(X)$ | $2 k$ | $12 k$ | $36 k$ | $80k$ | $250 k$ | $432 k$ | $686k$ | 0 |
(i) Since, $\Sigma P_j=1$
$$\Rightarrow \quad k(2+3+4+5+10+12+14)=1 \Rightarrow k=\frac{1}{50}$$
$$\begin{aligned} \text{(ii)}\quad \because \quad E(X) & =\Sigma X P(X) \\ \therefore \quad E(X) & =2 k+6 k+12 k+20 k+50 k+72 k+98 k+0=260 k \\ & =260 \times \frac{1}{50}=\frac{26}{5}=5.2\quad \left[\because k=\frac{1}{50}\right] \quad\text{.... (i)} \end{aligned}$$
(iii) We know that,
$$\begin{array}{rlr} \operatorname{Var}(X) & =\left[E\left(X^2\right)\right]-[E(X)]^2=\Sigma X^2 P(X)-[\Sigma\{X P(X)\}]^2 & \\ & =[2 k+12 k+36 k+80 k+250 k+432 k+686 k+0]-[5.2]^2 & \text { [using Eq. (i) }] \\ & =[1498 k]-27.04=\left[1498 \times \frac{1}{50}\right]-27.04 & {\left[\because k=\frac{1}{50}\right]} \\ & =29.96-27.04=2.92 & \end{array}$$
We know that, standard deviation of $X=\sqrt{\operatorname{Var}(X)}=\sqrt{2.92}=1.7088=1.7$ (approx)
The probability distribution of a discrete random variable $X$ is given as under
| $X$ | 1 | 2 | 4 | 2A | 3A | 5A |
|---|---|---|---|---|---|---|
| $P(X)$ | $\frac{1}{2}$ | $\frac{1}{5}$ | $\frac{3}{25}$ | $\frac{1}{10}$ | $\frac{1}{25}$ | $\frac{1}{25}$ |
Calculate
(i) the value of $A$, if $E(X)=2.94$.
(ii) variance of $X$.
(i) We have, $\begin{aligned} \Sigma X P(X) & =\frac{1}{2}+\frac{2}{5}+\frac{12}{25}+\frac{2 A}{10}+\frac{3 A}{25}+\frac{5 A}{25} \\ & =\frac{25+20+24+10 A+6 A+10 A}{50}=\frac{69+26 A}{50}\end{aligned}$
$$\begin{array}{ll} \text { Since, } & E(X)=\sum X P(X) \\ \Rightarrow & 2.94=\frac{69+26 A}{50} \end{array}$$
$$\begin{array}{ll} \Rightarrow & 26 A=50 \times 2.94-69 \\ \Rightarrow & A=\frac{147-69}{26}=\frac{78}{26}=3 \end{array}$$
$$\begin{aligned} &\text { (ii) We know that, }\\ &\begin{aligned} \operatorname{Var}(X) & =E\left(X^2\right)-[E(X)]^2 \\ & =\Sigma X^2 P(X)-[\Sigma X P(X)]^2 \\ & =\frac{1}{2}+\frac{4}{5}+\frac{48}{25}+\frac{4 A^2}{10}+\frac{9 A^2}{25}+\frac{25 A^2}{25}-[E(X)]^2 \\ & =\frac{25+40+96+20 A^2+18 A^2+50 A^2}{50}-[E(X)]^2 \\ & =\frac{161+88 A^2}{50}-[E(X)]^2=\frac{161+88 \times(3)^2}{50}-[E(X)]^2 \quad[\because A=3] \\ & =\frac{953}{50}-[2.94]^2 \quad[\because E(X)=2.94] \\ & =19.0600-8.6436=10.4164 \end{aligned} \end{aligned}$$
The probability distribution of a random variable $x$ is given as under
$$P(X=x)=\left\{\begin{array}{l} k x^2, x=1,2,3 \\ 2 k x, x=4,5,6 \\ 0, \text { otherwise } \end{array}\right.$$
where, $k$ is a constant. Calculate
(i) $E(X)$
(ii) $E\left(3 X^2\right)$
(iii) $P(X \geq 4)$
| $X$ | 1 | 2 | 3 | 4 | 5 | 6 | Otherwise |
|---|---|---|---|---|---|---|---|
| $P(X)$ | k | 4k | 9k | 8k | 10k | 12k | 0 |
$$\begin{aligned} &\text { We know that, } \Sigma P_i=1\\ &\Rightarrow \quad 44 k=1 \Rightarrow k=\frac{1}{44} \end{aligned}$$
$$\begin{aligned} \therefore \quad \Sigma X P(X) & =k+8 k+27 k+32 k+50 k+72 k+0 \\ & =190 k=190 \times \frac{1}{44}=\frac{95}{22} \end{aligned}$$
(i) So, $E(X)=\Sigma X P(X)=\frac{95}{22}=4.32$
(ii) Also, $\quad E\left(X^2\right)=\Sigma X^2 P(X)=k+16 k+81 k+128 k+250 k+432 k$
$$\begin{array}{ll} =908 k=908 \times \frac{1}{44} & {\left[\because k=\frac{1}{44}\right]} \\ =20.636=20.64 \text { (approx) } \end{array}$$
$$ \begin{aligned} &\therefore \quad E\left(3 X^2\right)=3 E\left(X^2\right)=3 \times 20.64=61.92=61.9\\ &\begin{aligned} \text { (iii) }\quad P(X \geq 4) & =P(X=4)+P(X=5)+P(X=6) \\ & =8 k+10 k+12 k=30 k=30 \cdot \frac{1}{44}=\frac{15}{22} \end{aligned} \end{aligned}$$
A bag contains $(2 n+1)$ coins. It is known that $n$ of these coins have a head on both sides whereas the rest of the coins are fair. A coin is picked up at random from the bag and is tossed. If the probability that the toss results in a head is $\frac{31}{42}$, then determine the value of $n$.
Given, $n$ coins have head on both sides and $(n+1)$ coins are fair coins.
Let $E_1=$ Event that an unfair coin is selected
$E_2=$ Event that a fair coin is selected
$E=$ Event that the toss results in a head
$$\begin{array}{ll} \therefore & P\left(E_1\right)=\frac{n}{2 n+1} \text { and } P\left(E_2\right)=\frac{n+1}{2 n+1} \\ \text { Also, } & P\left(\frac{E}{E_1}\right)=1 \text { and } P\left(\frac{E}{E_2}\right)=\frac{1}{2} \end{array}$$
$$\begin{array}{ll} \therefore & P(E)=P\left(E_1\right) \cdot P\left(\frac{E}{E_1}\right)+P\left(E_2\right) \cdot P\left(\frac{E}{E_2}\right)=\frac{n}{2 n+1} \cdot 1+\frac{n+1}{2 n+1} \cdot \frac{1}{2} \\ \Rightarrow & \frac{31}{42}=\frac{2 n+n+1}{2(2 n+1)} \Rightarrow \frac{31}{42}=\frac{3 n+1}{4 n+2} \\ \Rightarrow & 124 n+62=126 n+42 \\ \Rightarrow & 2 n=20 \Rightarrow n=10 \end{array}$$
Two cards are drawn successively without replacement from a well shuffled deck of cards. Find the mean and standard variation of the random variable $X$, where $X$ is the number of aces.
Let $X$ denotes a random variable of number of aces.
$$\begin{array}{l} \therefore & X =0,1,2 \\ \text { Now, } & P(X =0)=\frac{48}{52} \cdot \frac{47}{51}=\frac{2256}{2652} \\ & P(X=1)=\frac{48}{52} \cdot \frac{4}{51}+\frac{4}{52} \cdot \frac{48}{51}=\frac{384}{2652} \\ & P(X=2)=\frac{4}{52} \cdot \frac{3}{51}=\frac{12}{2652} \end{array}$$
| $X$ | 0 | 1 | 2 |
|---|---|---|---|
| $P(X)$ | $\frac{2256}{2652}$ | $\frac{384}{2652}$ | $\frac{12}{2652}$ |
| $XP(X)$ | 0 | $\frac{384}{2652}$ | $\frac{24}{2652}$ |
| $X^2P(X)$ | 0 | $\frac{384}{2652}$ | $\frac{48}{2652}$ |
$$\begin{aligned} &\text { We know that, } \operatorname{Mean}(\mu)=E(X)=\Sigma X P(X)\\ &\begin{aligned} & =0+\frac{384}{2652}+\frac{24}{2652} \\ & =\frac{408}{2652}=\frac{2}{13} \end{aligned} \end{aligned}$$
Also,
$$\begin{array}{rlr} \operatorname{Var}(X) & =E\left(X^2\right)-[E(X)]^2=\Sigma X^2 P(X)-[E(X)]^2 & \\ & =\left[0+\frac{384}{2652}+\frac{48}{2652}\right]-\left(\frac{2}{13}\right)^2 & {\left[\because E(X)=\frac{2}{13}\right]} \\ & =\frac{432}{2652}-\frac{4}{169}=0.1628-0.0236=0.1391 & \end{array}$$
$\therefore \quad$ Standard deviation $=\sqrt{\operatorname{Var}(X)}=\sqrt{0.139}=0.373$ (approx)