Let $E_1$ and $E_2$ be two independent events such that $P\left(E_1\right)=P_1$ and $P\left(E_2\right)=P_2$. Describe in words of the events whose probabilities are
(i) $P_1 P_2$ (ii) $\left(1-P_1\right) P_2$ (iii) $1-\left(1-P_1\right)\left(1-P_2\right)$ (iv) $P_1+P_2-2 P_1 P_2$
$$P\left(E_1\right)=P_1 \text { and } P\left(E_2\right)=P_2$$
(i) $P_1 P_2 \Rightarrow P\left(E_1\right) \cdot P\left(E_2\right)=P\left(E_1 \cap E_2\right)$
So, $E_1$ and $E_2$ occur.
(ii) $\left(1-P_1\right) P_2=P\left(E_1\right)^{\prime} \cdot P\left(E_2\right)=P\left(E_1^{\prime} \cap E_2\right)$
So, $E_1$ does not occur but $E_2$ occurs.
(iii) $$\begin{aligned} 1-\left(1-P_1\right)\left(1-P_2\right) & =1-P\left(E_1\right)^{\prime} P\left(E_2\right)^{\prime}=1-P\left(E_1^{\prime} \cap E_2^{\prime}\right) \\ & =1-\left[1-P\left(E_1 \cup E_2\right)\right]=P\left(E_1 \cup E_2\right) \end{aligned}$$
So, either $E_1$ or $E_2$ or both $E_1$ and $E_2$ occurs.
(iv) $$\begin{aligned} P_1+P_2-2 P_1 P_2 & =P\left(E_1\right)+P\left(E_2\right)-2 P\left(E_1\right) \cdot P\left(E_2\right) \\ & =P\left(E_1\right)+P\left(E_2\right)-2 P\left(E_1 \cap E_2\right) \\ & =P\left(E_1 \cup E_2\right)-P\left(E_1 \cap E_2\right) \end{aligned}$$
So, either $E_1$ or $E_2$ occurs but not both.
A discrete random variable X has the probability distribution as given below
| X | 0.5 | 1 | 1.5 | 2 |
|---|---|---|---|---|
| P(X) | $k$ | $k^2$ | $2k^2$ | $k$ |
(i) Find the value of $k$.
(ii) Determine the mean of the distribution.
We have,
| X | 0.5 | 1 | 1.5 | 2 |
|---|---|---|---|---|
| P(X) | $k$ | $k^2$ | $2k^2$ | $k$ |
$$\begin{aligned} &\text { (i) We know that, } \quad \sum_{i=1}^n P_i=1 \text {, where } P_i \geq 0\\ &\begin{array}{lr} \Rightarrow & P_1+P_2+P_3+P_4=1 \\ \Rightarrow & k+k^2+2 k^2+k=1 \\ \Rightarrow & 3 k^2+2 k-1=0 \\ \Rightarrow & 3 k^2+3 k-k-1=0 \\ \Rightarrow & 3 k(k+1)-1(k+1)=0 \\ \Rightarrow & (3 k-1)(k+1)=0 \\ \Rightarrow & k=1 / 3 \Rightarrow k=-1 \\ \text { Since, } & k \text { is } \geq 0 \Rightarrow k=1 / 3 \end{array} \end{aligned}$$
(ii) Mean of the distribution $(\mu)=E(X)=\sum_{i=1}^n x_i P_i$
$$ \begin{aligned} & =0.5(k)+1\left(k^2\right)+1.5\left(2 k^2\right)+2(k)=4 k^2+2.5 k \\ & =4 \cdot \frac{1}{9}+2.5 \cdot \frac{1}{3} \quad \left[\because k=\frac{1}{3}\right]\\ & =\frac{4+7.5}{9}=\frac{23}{18} \end{aligned}$$
Prove that
(i) $P(A)=P(A \cap B)+P(A \cap \bar{B})$
(ii) $P(A \cup B)=P(A \cap B)+P(A \cap \bar{B})+P(\bar{A} \cap B)$
(i) $\because \quad P(A)=P(A \cap B)+P(A \cap \bar{B})$
$$\begin{aligned} \therefore \quad \mathrm{RHS} & =P(A \cap B)+P(A \cap \bar{B}) \\ & =P(A) \cdot P(B)+P(A) \cdot P(\bar{B}) \\ & =P(A)[P(B)+P(\bar{B})] \\ & =P(A)[P(B)+1-P(B)] \quad [\because P(\bar{B})=1-P(B)]\\ & =P(A)=\mathrm{LHS}\quad\text{Hence proved.} \end{aligned}$$
$$\begin{aligned} &\begin{array}{lrl} \text { (ii) } \because P(A \cup B) =P(A \cap B)+P(A \cap \bar{B})+P(\bar{A} \cap B) \\ \therefore \quad R H S =P(A) \cdot P(B)+P(A) \cdot P(\bar{B})+P(\bar{A}) \cdot P(B) \end{array} \end{aligned}$$
$$ \begin{aligned} &\begin{aligned} & =P(A) \cdot P(B)+P(A) \cdot[1-P(B)]+[1-P(A)] P(B) \\ & =P(A) \cdot P(B)+P(A)-P(A) \cdot P(B)+P(B)-P(A) \cdot P(B) \\ & =P(A)+P(B)-P(A) \cdot P(B) \\ & =P(A)+P(B)-P(A \cap B) \\ & =P(A \cup B)=\mathrm{LHS} \quad\text { Hence proved. } \end{aligned}\\ \end{aligned}$$
If $X$ is the number of tails in three tosses of a coin, then determine the standard deviation of $X$.
$$\begin{aligned} &\text { Given that, random variable } X \text { is the number of tails in three tosses of a coin. }\\ &\begin{array}{l} \text { So, } & X=0,1,2,3 . \\ \Rightarrow & P(X=x)={ }^n C_x(p)^x q^{n-x}, \\ \text { where } & n=3, p=1 / 2, q=1 / 2 \text { and } x=0,1,2,3 \end{array} \end{aligned}$$
| X | 0 | 1 | 2 | 3 |
|---|---|---|---|---|
| P(X) | $\frac{1}{8}$ |
$\frac{3}{8}$ | $\frac{3}{8}$ | $\frac{1}{8}$ |
| XP(X) | 0 | $\frac{3}{8}$ | $\frac{3}{4}$ | $\frac{3}{8}$ |
| X$^2$P(X) | 0 | $\frac{3}{8}$ | $\frac{3}{2}$ | $\frac{9}{8}$ |
We know that, $\operatorname{Var}(X)=E\left(X^2\right)-[E(X)]^2\quad\text{.... (i)}$
where, $E\left(X^2\right)=\sum_{i=1}^n x_i^2 P\left(x_i\right)$ and $E(X)=\sum_{i=1}^n x_i P\left(x_i\right)$
$$\begin{aligned} &\begin{array}{ll} \therefore & E\left(X^2\right)=\sum_{i=1}^n x_i^2 P\left(X_i\right)=0+\frac{3}{8}+\frac{3}{2}+\frac{9}{8}=\frac{24}{8}=3 \\ \text { and } & {[E(X)]^2=\left[\sum_{i=0}^n x_i^2 P\left(x_i\right)\right]^2=\left[0+\frac{3}{8}+\frac{3}{4}+\frac{3}{8}\right]^2=\left[\frac{12}{8}\right]^2=\frac{9}{4}} \\ \therefore & \operatorname{Var}(X)=3-\frac{9}{4}=\frac{3}{4}\quad\text{[using Eq. (i)]} \end{array}\\ &\text { and standard deviation of } X=\sqrt{\operatorname{Var}(X)}=\sqrt{\frac{3}{4}}=\frac{\sqrt{3}}{2} \end{aligned}$$
In a dice game, a player pays a stake of ₹ 1 for each throw of a die. She receives ₹ 5 , if the die shows a 3 , ₹ 2 , if the die shows a 1 or 6 and nothing otherwise, then what is the player's expected profit per throw over a long series of throws?
Let X is the random variable of profit per throw.
| X | $-1$ | 1 | 4 |
|---|---|---|---|
| P(X) | $\frac{1}{2}$ |
$\frac{1}{3}$ | $\frac{1}{6}$ |
Since, she loss ₹ 1 on getting any of 2,4 or 5.
So, at $X=-1\quad,$ $$ P(X)=\frac{1}{6}+\frac{1}{6}+\frac{1}{6}=\frac{3}{6}=\frac{1}{2}$$
Similarly, at $X=1\quad$, $P(X)=\frac{1}{6}+\frac{1}{6}=\frac{1}{3}\quad$ [if die shows of either 1 or 6 ]
and at $X=4, \quad P(X)=\frac{1}{6}$ [if die shows a 3]
$\therefore$ Player's expected profit $=E(X)=\Sigma X P(X)$
$=-1 \times \frac{1}{2}+1 \times \frac{1}{3}+4 \times \frac{1}{6}$
$=\frac{-3+2+4}{6}=\frac{3}{6}=\frac{1}{2}=$ ₹ 0.50