Here, $$n=5, p=\left(\frac{1}{6}+\frac{1}{6}+\frac{1}{6}\right)=\frac{1}{2} \text { and } q=1-p=1-\frac{1}{2}=\frac{1}{2}$$

Also, $r=3$

$$\begin{aligned} \therefore \quad P(X & =r)={ }^n C_r(p)^r(q)^{n-r}={ }^5 C_3\left(\frac{1}{2}\right)^3\left(\frac{1}{2}\right)^{5-3} \\ & =\frac{5!}{3!2!} \cdot \frac{1}{8} \cdot \frac{1}{4}=\frac{10}{32}=\frac{5}{16} \end{aligned}$$

In this case, we have to find out the probability of getting atleast 8 heads. Let $X$ is the random variable for getting a head.

<-p>$$\begin{aligned} \text{Here, }\quad n & =10, r \geq 8 \\ \text{i.e.,}\quad r & =8,9,10, p=\frac{1}{2}, q=\frac{1}{2} \\ \text{We know that,}\quad P(X & =r)={ }^n C_r p^r q^{n-r} \end{aligned}$$

$$\begin{aligned} \therefore \quad P(X & =r)=P(r=8)+P(r=9)+P(r=10) \\ & ={ }^{10} C_8\left(\frac{1}{2}\right)^8\left(\frac{1}{2}\right)^{10-8}+{ }^{10} C_9\left(\frac{1}{2}\right)^9\left(\frac{1}{2}\right)^{10-9}+{ }^{10} C_{10}\left(\frac{1}{2}\right)^{10}\left(\frac{1}{2}\right)^{10-10} \\ & =\frac{10!}{8!2!}\left(\frac{1}{2}\right)^{10}+\frac{10!}{9!1!}\left(\frac{1}{2}\right)^{10}+\frac{10!}{0!10!}\left(\frac{1}{2}\right)^{10} \\ & =\left(\frac{1}{2}\right)^{10}\left[\frac{10 \times 9}{2}+10+1\right] \\ & =\left(\frac{1}{2}\right)^{10} \cdot 56=\frac{1}{2^7 \cdot 2^3} \cdot 56=\frac{7}{128} \end{aligned}$$

$$\begin{array}{l} \text { Here, } & n=7 p=0.25=\frac{1}{4}, q=1-\frac{1}{4}=\frac{3}{4} r \geq 2, \\ \text { where, } & P(X)={ }^n C_r(p)^r(q)^{n-r} \end{array}$$

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.

$$\therefore \quad P(X=r)=1-[P(r=0)+P(r=1)]$$

$$\begin{aligned} & =1-\left[{ }^7 C_0\left(\frac{1}{4}\right)^0\left(\frac{3}{4}\right)^{7-0}+{ }^7 C_1\left(\frac{1}{4}\right)^1\left(\frac{3}{4}\right)^{7-1}\right] \\ & =1-\left[\frac{7!}{0!7!}\left(\frac{3}{4}\right)^7+\frac{7!}{1!6!}\left(\frac{1}{4}\right)\left(\frac{3}{4}\right)^6\right] \\ & =1-\left[\left(\frac{3}{4}\right)^6\left(\frac{3}{4} \cdot 1+\frac{1}{4} \cdot 7\right)\right] \\ & =1-\left[\frac{3^6}{4^6}\left(\frac{10}{4}\right)\right]=1-\left[\frac{3^6 \times 10}{4^7}\right]=1-\left[\frac{27 \cdot 27 \cdot 10}{64 \cdot 256}\right] \\ & =1-\left[\frac{7290}{16384}\right]=1-\frac{3645}{8192}=\frac{4547}{8192} \end{aligned}$$

Probability of defective watch from a lot of 100 watches $=\frac{10}{100}=\frac{1}{10}$

$$\begin{aligned} \therefore \quad p=1 / 10, q & =\frac{9}{10}, n=8 \text { and } r \geq 1 \\ \therefore \quad P(r \geq 1) & =1-P(r=0)=1-{ }^8 C_0\left(\frac{1}{10}\right)^0\left(\frac{9}{10}\right)^{8-0} \\ & =1-\frac{8!}{0!8!} \cdot\left(\frac{9}{10}\right)^8=1-\left(\frac{9}{10}\right)^8 \end{aligned}$$

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) $V(X)=1.4475$