A lot of 100 watches is known to have 10 defective watches. If 8 watches are selected (one by one with replacement) at random, then what is the probability that there will be atleast one defective watch?
Probability of defective watch from a lot of 100 watches $=\frac{10}{100}=\frac{1}{10}$
$$\begin{aligned} \therefore \quad p=1 / 10, q & =\frac{9}{10}, n=8 \text { and } r \geq 1 \\ \therefore \quad P(r \geq 1) & =1-P(r=0)=1-{ }^8 C_0\left(\frac{1}{10}\right)^0\left(\frac{9}{10}\right)^{8-0} \\ & =1-\frac{8!}{0!8!} \cdot\left(\frac{9}{10}\right)^8=1-\left(\frac{9}{10}\right)^8 \end{aligned}$$
Consider the probability distribution of a random variable $X$.
| X | 0 | 1 | 2 | 3 | 4 |
|---|---|---|---|---|---|
| P(X) | 0.1 | 0.25 | 0.3 | 0.2 | 0.15 |
Calculate
(i) $V\left(\frac{X}{2}\right)$. (ii) Variance of $X$.
We have,
| X | 0 | 1 | 2 | 3 | 4 |
|---|---|---|---|---|---|
| P(X) | 0.1 | 0.25 | 0.3 | 0.2 | 0.15 |
| XP(X) | 0 | 0.25 | 0.6 | 0.6 | 0.60 |
| X$^2$P(X) | 0 | 0.25 | 1.2 | 1.8 | 2.40 |
$$\begin{array}{ll} & \operatorname{Var}(X)=E\left(X^2\right)-[E(X)]^2 \\ \text { where, } & E(X)=\mu=\sum_\limits{i=1}^n x_i P_i(x i) \\ \text { and } & E\left(X^2\right)=\sum_\limits{i=1}^n x_i^2 P\left(x_i\right) \end{array}$$
$$\begin{array}{lrl} \therefore & E(X)=0+0.25+0.6+0.6+0.60=2.05 \\ & E\left(X^2\right)=0+0.25+1.2+1.8+2.40=5.65 \end{array}$$
$$\begin{aligned} \text{(i)}\quad V\left(\frac{X}{2}\right) & =\frac{1}{4} V(X)=\frac{1}{4}\left[5.65-(2.05)^2\right] \\ & =\frac{1}{4}[5.65-4.2025]=\frac{1}{4} \times 1.4475=0.361875 \end{aligned}$$
(ii) $V(X)=1.4475$
The probability distribution of a random variable $X$ is given below
| $X$ | 0 | 1 | 2 | 3 |
|---|---|---|---|---|
| $P(X)$ | $k$ | $\frac{k}{2}$ | $\frac{k}{4}$ | $\frac{k}{8}$ |
(i) Determine the value of $k$.
(ii) Determine $P(X \leq 2)$ and $P(X>2)$.
(iii) Find $P(X \leq 2)+P(X>2)$.
We have,
| $X$ | 0 | 1 | 2 | 3 |
|---|---|---|---|---|
| $P(X)$ | $k$ | $\frac{k}{2}$ | $\frac{k}{4}$ | $\frac{k}{8}$ |
$$\begin{aligned} &\text { (i) Since, }\\ &\sum_{i=1}^n P_i=1, i=1,2, \ldots, n \text { and } P_i \geq 0\\ &\begin{aligned} \therefore \quad& k+\frac{k}{2}+\frac{k}{4}+\frac{k}{8} =1 \\ \Rightarrow \quad & 8 k+4 k+2 k+k =8 \\ \therefore \quad & k =\frac{8}{15} \end{aligned} \end{aligned}$$
$$\begin{aligned} \text { (ii) } P(X \leq 2)=P(0)+P(1)+P(2) & =k+\frac{k}{2}+\frac{k}{4} \\ & =\frac{(4 k+2 k+k)}{4}=\frac{7 k}{4}=\frac{7}{4} \cdot \frac{8}{15}=\frac{14}{15} \end{aligned}$$
(iii) $P(X \leq 2)+P(X>2)=\frac{14}{15}+\frac{1}{15}=1$
For the following probability distribution determine standard deviation of the random variable $X$.
| $X$ | 2 | 3 | 4 |
|---|---|---|---|
| $P(X)$ | 0.2 | 0.5 | 0.3 |
We have,
| $X$ | 2 | 3 | 4 |
|---|---|---|---|
| $P(X)$ | 0.2 | 0.5 | 0.3 |
| $XP(X)$ | 0.4 | 1.5 | 1.2 |
| $X^2P(X)$ | 0.8 | 4.5 | 4.8 |
We know that, standard deviation of $X=\sqrt{\operatorname{Var} X}$
where,
$$\begin{aligned} \operatorname{Var} X & =E\left(X^2\right)-[E(X)]^2 \\ & =\sum_{i=1}^n x_i^2 P\left(x_i\right)-\left[\sum_{i=1}^n x_i P_i\right]^2 \end{aligned}$$
$$\begin{aligned} \therefore\quad\operatorname{Var} X & =[0.8+4.5+4.8]-[0.4+1.5+1.2]^2 \\ & =10.1-(3.1)^2=10.1-9.61=0.49 \end{aligned}$$
$\therefore$ Standard deviation of $X=\sqrt{\operatorname{Var} X}=\sqrt{0.49}=0.7$
A biased die is such that $P(4)=\frac{1}{10}$ and other scores being equally likely. The die is tossed twice. If $X$ is the 'number of fours seen', then find the variance of the random variable $X$.
Since, $X=$ Number of fours seen
On tossing two die, $X=0,1,2$.
Also, $$P_{(4)}=\frac{1}{10} \text { and } P_{(\text {not } 4)}=\frac{9}{10}$$
So,
$$\begin{aligned} & P(X=0)=P_{(\text {not } 4)} \cdot P_{(\text {not } 4)}=\frac{9}{10} \cdot \frac{9}{10}=\frac{81}{100} \\ & P(X=1)=P_{(\text {not } 4)} \cdot P_{(4)}+P_{(4)} \cdot P_{(\text {not } 4)}=\frac{9}{10} \cdot \frac{1}{10}+\frac{1}{10} \cdot \frac{9}{10}=\frac{18}{100} \\ & P(X=2)=P_{(4)} \cdot P_{(4)}=\frac{1}{10} \cdot \frac{1}{10}=\frac{1}{100} \end{aligned}$$
Thus, we get following table
| $X$ | 0 | 1 | 2 |
|---|---|---|---|
| $P(X)$ | $\frac{81}{100}$ | $\frac{18}{100}$ | $\frac{1}{100}$ |
| $XP(X)$ | 0 | 18/100 | 2/100 |
| $X^2P(X)$ | 0.8 | 18/100 | 4/100 |
$$\begin{aligned} \therefore \quad \operatorname{Var}(X) & =E\left(X^2\right)-[E(X)]^2=\Sigma X^2 P(X)-[\Sigma X P(X)]^2 \\ & =\left[0+\frac{18}{100}+\frac{4}{100}\right]-\left[0+\frac{18}{100}+\frac{2}{100}\right]^2 \\ & =\frac{22}{100}-\left(\frac{20}{100}\right)^2=\frac{11}{50}-\frac{1}{25} \\ & =\frac{11-2}{50}=\frac{9}{50}=\frac{18}{100}=0.18 \end{aligned}$$