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)
A die is tossed twice. If a 'success' is getting an even number on a toss, then find the variance of the number of successes.
Let $X$ be the random variable for a 'success' for getting an even number on a toss.
$\therefore \quad X=0,1,2, n=2, p=\frac{3}{6}=\frac{1}{2}$ and $q=\frac{1}{2}$
$$\begin{array}{ll} \text { At } X=0, & P(X=0)={ }^2 C_0\left(\frac{1}{2}\right)^0\left(\frac{1}{2}\right)^{2-0}=\frac{1}{4} \\ \text { At } X=1, & P(X=1)={ }^2 C_1\left(\frac{1}{2}\right)^1\left(\frac{1}{2}\right)^{2-1}=2 \cdot \frac{1}{2} \cdot \frac{1}{2}=\frac{1}{2} \\ \text { At } X=2, & P(X=2)={ }^2 C_2\left(\frac{1}{2}\right)^2\left(\frac{1}{2}\right)^{2-2}=\frac{1}{4} \end{array}$$
| $X$ | 0 | 1 | 2 |
|---|---|---|---|
| $P(X)$ | $\frac{1}{4}$ | $\frac{1}{2}$ | $\frac{1}{4}$ |
| $XP(X)$ | 0 | $\frac{1}{2}$ | $\frac{1}{2}$ |
| $X^2P(X)$ | 0 | $\frac{1}{2}$ | 1 |
$\therefore \quad \Sigma X P(X)=0+\frac{1}{2}+\frac{1}{2}=1\quad\text{.... (i)}$
and $$\Sigma X^2 P(X)=0+\frac{1}{2}+1=\frac{3}{2}\quad\text{.... (ii)}$$
$$\begin{aligned} \because \operatorname{Var}(X) & =E\left(X^2\right)-[E(X)]^2 \\ & =\Sigma X^2 P(X)-[\Sigma X P(X)]^2=\frac{3}{2}-(1)^2=\frac{1}{2} \quad \text { [using Eqs. (i) and (ii)] } \end{aligned}$$
There are 5 cards numbered 1 to 5, one number on one card. Two cards are drawn at random without replacement. Let $X$ denotes the sum of the numbers on two cards drawn. Find the mean and variance of $X$.
Here, $$S=\{(1,2),(2,1),(1,3),(3,1),(2,3),(3,2),(1,4),(4,1),(1,5),(5,1),(2,4),(4,2),(2,5),(5,2),(3,4),(4,3),(3,5),(5,3),(5,4),(4,5)\}$$
$$\Rightarrow n(S)=20$$
Let random variable be $X$ which denotes the sum of the numbers on two cards drawn.
$\therefore\qquad X=3,4,5,6,7,8,9$
$$\begin{aligned} & \text { At } X=3, P(X)=\frac{2}{20}=\frac{1}{10} \\ & \text { At } X=4, P(X)=\frac{2}{20}=\frac{1}{10} \\ & \text { At } X=5, P(X)=\frac{4}{20}=\frac{1}{5} \\ & \text { At } X=6, P(X)=\frac{4}{20}=\frac{1}{5} \\ & \text { At } X=7, P(X)=\frac{4}{20}=\frac{1}{5} \\ & \text { At } X=8, P(X)=\frac{2}{20}=\frac{1}{10} \\ & \text { At } X=9, P(X)=\frac{2}{20}=\frac{1}{10} \end{aligned}$$
$\therefore \quad$ Mean, $E(X)=\Sigma X P(X)=\frac{3}{10}+\frac{4}{10}+\frac{5}{5}+\frac{6}{5}+\frac{7}{5}+\frac{8}{10}+\frac{9}{10}$
$=\frac{3+4+10+12+14+8+9}{10}=6$
$$\begin{aligned} &\text { Also, }\\ &\begin{aligned} \Sigma X^2 P(X) & =\frac{9}{10}+\frac{16}{10}+\frac{25}{5}+\frac{36}{5}+\frac{49}{5}+\frac{64}{10}+\frac{81}{10} \\ & =\frac{9+16+50+72+98+64+81}{10}=39 \end{aligned} \end{aligned}$$
$$\begin{aligned} \therefore \quad \operatorname{Var}(X) & =\Sigma X^2 P(X)-[\Sigma X P(X)]^2 \\ & =39-(6)^2=39-36=3 \end{aligned}$$