The probability of a man hitting a target is 0.25 . If he shoots 7 times, then what is the probability of his hitting atleast twice?
$$\begin{array}{l} \text { Here, } & n=7 p=0.25=\frac{1}{4}, q=1-\frac{1}{4}=\frac{3}{4} r \geq 2, \\ \text { where, } & P(X)={ }^n C_r(p)^r(q)^{n-r} \end{array}$$
In this case for easy approach we shall first find out the probability of his hitting atmost once (i.e., $r=0,1$ ) and then subtract this probability from 1 to get the desired probability.
$$\therefore \quad P(X=r)=1-[P(r=0)+P(r=1)]$$
$$\begin{aligned} & =1-\left[{ }^7 C_0\left(\frac{1}{4}\right)^0\left(\frac{3}{4}\right)^{7-0}+{ }^7 C_1\left(\frac{1}{4}\right)^1\left(\frac{3}{4}\right)^{7-1}\right] \\ & =1-\left[\frac{7!}{0!7!}\left(\frac{3}{4}\right)^7+\frac{7!}{1!6!}\left(\frac{1}{4}\right)\left(\frac{3}{4}\right)^6\right] \\ & =1-\left[\left(\frac{3}{4}\right)^6\left(\frac{3}{4} \cdot 1+\frac{1}{4} \cdot 7\right)\right] \\ & =1-\left[\frac{3^6}{4^6}\left(\frac{10}{4}\right)\right]=1-\left[\frac{3^6 \times 10}{4^7}\right]=1-\left[\frac{27 \cdot 27 \cdot 10}{64 \cdot 256}\right] \\ & =1-\left[\frac{7290}{16384}\right]=1-\frac{3645}{8192}=\frac{4547}{8192} \end{aligned}$$
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 $X=\sqrt{\operatorname{Var} X}=\sqrt{0.49}=0.7$