Maths Solve the Following Question.(3 Marks)

Let X : number of question answered correctly out of 5.
p = P (correct answer)
p = 1/4
q = 1 - p = 3/4 and n = 5
X ~ B( n= 5,p =1/4)
The p.m.f of X is given as
$p(X=x)=p(x)={ }^5 C_x\left(\frac{1}{4}\right)^x\left(\frac{3}{4}\right)^5-x$
where x = 0, 1, 2, 3, 4, 5
Probability a student will get at least one correct answer
= P(X ≥ 1) = 1 - P(X < 1) 
= 1 - P(X = 0)
= $1-{ }^5 C_0\left(\frac{1}{4}\right)^0\left(\frac{3}{4}\right)^5$
= $1-\frac{243}{1024}$
=781/1024=0.7627

Probability of recovery=P(R)= 0.5
Probability of non-recovery $=P(\bar{R})=1-0.5=0.5$
(a) If there are six patients, the probability that none recovers
$={ }^6 C_0 \times[P(R)]^0 \times[P(\bar{R})]^6=(0.5)^6=\frac{1}{64}$
(b) Of the six patients, the probability that half will recover 
$={ }^6 C_3 \times[P(R)]^3 \times[P(\bar{R})]^3=\frac{6 !}{3 ! 3 !} \times 0.5^3 \times 0.5^3=20 \times \frac{1}{64}=\frac{5}{16}$

Given, X ~ B (n, p) and E(X) = 6 and Var (X) = 4.2
$\begin{aligned} & \text { Now, } \frac{E(X)}{u(X)}=\frac{n p}{n p q} \\ & \Rightarrow \frac{6}{4.2}=\frac{1}{q} \\ & \Rightarrow q =\frac{4.2}{6}=0.7\end{aligned}$
Since,  p + q = 1
⇒ p + 0.7 = 1
⇒ p = 0.3
Now E(X) = np
⇒ 6 = n × 0.3
$\Rightarrow n =\frac{6}{0.3}=\frac{60}{3}=20$

Let X = no. of heads shows
$\begin{aligned} & n =9 \quad p =\frac{1}{2} \quad q =\frac{1}{2} \\ & P ( X = x )={ }^n C_x p^n \cdot(q)^{n-x} X =0,1 \ldots . . n \\ & P ( X =5)={ }^9 C_5\left(\frac{1}{2}\right)^5\left(\frac{1}{2}\right)^4\end{aligned}$
$\begin{aligned} & =\frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} \times \frac{1}{2^9} \\ & =\frac{3024}{24} \times \frac{1}{2^9} \\ & =\frac{126}{2^9} \\ & =\frac{126}{512} \\ & =0.2460\end{aligned}$

(a)
Let X = Number of heads
p = probability of getting head in one toss
$\begin{aligned} & p=\frac{1}{2} \\ & q=1-p=1-\frac{1}{2}=\frac{1}{2}\end{aligned}$
Given n = 8
$x \sim B\left(8, \frac{1}{2}\right)$
The p.m.f. of X is given as
$\begin{aligned} & P ( X = x )={ }^n C_x p^x q^{n-x} \\ & \text { i.e } P ( x )={ }^8 C_x\left(\frac{1}{2}\right)^x\left(\frac{1}{2}\right)^{8-x}, x=0,1,2,3, \ldots \ldots, 8\end{aligned}$
P(exactly 5 heads) = P[X = 5]
= P(5)
$\begin{aligned} & ={ }^8 C_5\left(\frac{1}{2}\right)^5\left(\frac{1}{2}\right)^{8-5} \\ & ={ }^8 C_3\left(\frac{1}{2}\right)^5\left(\frac{1}{2}\right)^3 \quad \ldots\left[\because{ }^n C_x={ }^n C_{n-x}\right]\end{aligned}$
$\begin{aligned} & =\frac{8 \cdot 7 \cdot 6}{6} \times\left(\frac{1}{2}\right)^8 \\ & =8 \times 7 \times \frac{1}{16 \times 16} \\ & =\frac{7}{32}\end{aligned}$
∴ P[X = 5] = 0.21875
Hence, the probability of getting exactly 5 heads is 0.21875
(b)
Let X = Number of heads
p = probability of getting head in one toss
$\begin{aligned} & p=\frac{1}{2} \\ & q=1-p=1-\frac{1}{2}=\frac{1}{2}\end{aligned}$
Given n = 8
$x \sim B\left(8, \frac{1}{2}\right)$
The p.m.f. of X is given as
$\begin{aligned} & P ( X = x )={ }^n C_x p^x q^{n-x} \\ & \text { i.e } P ( x )={ }^8 C_x\left(\frac{1}{2}\right)^x\left(\frac{1}{2}\right)^{8-x}, x=0,1,2,3, \ldots \ldots, 8\end{aligned}$
P (getting heads at least once)
P[X > = 1] = 1 – P[X = 0]
= 1 – P(0)
$\begin{aligned} & =1-{ }^8 C_0\left(\frac{1}{2}\right)^0\left(\frac{1}{2}\right)^{8-0} \\ & =1-\left(\frac{1}{2}\right)^8 \\ & =1-\frac{1}{256} \\ & =\frac{255}{256}\end{aligned}$
P[X > = 1] = 0.996

Probability of recovery=P(R)= 0.5
Probability of non-recovery $=P(\bar{R})=1-0.5=0.5$
(a) If there are six patients, the probability that none recovers
$={ }^6 C_0 \times[P(R)]^0 \times[P(\bar{R})]^6=(0.5)^6=\frac{1}{64}$
(b) Of the six patients, the probability that half will recover 
$={ }^6 C_3 \times[P(R)]^3 \times[P(\bar{R})]^3=\frac{6 !}{3 ! 3 !} \times 0.5^3 \times 0.5^3=20 \times \frac{1}{64}=\frac{5}{16}$

Let X = Number of heads
p = probability of getting head in one toss
$\begin{aligned} & p=\frac{1}{2} \\ & q=1-p=1-\frac{1}{2}=\frac{1}{2}\end{aligned}$
Given n = 8
$x \sim B\left(8, \frac{1}{2}\right)$
The p.m.f. of X is given as
$\begin{aligned} & P ( X = x )={ }^n C_x p^x q^{n-x} \\ & \text { i.e } P ( x )={ }^8 C_x\left(\frac{1}{2}\right)^x\left(\frac{1}{2}\right)^{8-x}, x=0,1,2,3, \ldots \ldots, 8\end{aligned}$
P (getting heads at least once)
P[X > = 1] = 1 – P[X = 0]
= 1 – P(0)
$\begin{aligned} & =1-{ }^8 C_0\left(\frac{1}{2}\right)^0\left(\frac{1}{2}\right)^{8-0} \\ & =1-\left(\frac{1}{2}\right)^8 \\ & =1-\frac{1}{256} \\ & =\frac{255}{256}\end{aligned}$
P[X > = 1] = 0.996

$\begin{aligned} & p=0.6, q=1-p=1-0.6=0.4 \\ & n=4, r=2 \\ & p(X=r)={ }^n C_r p^r q^{n-r} \\ & p(X=2)={ }^4 C_2(0.6)^2(0.4)^2 \\ & =\frac{4 \times 3}{1 \times 2}(0.36)(0.16) \\ & =0.3456\end{aligned}$

$\begin{aligned} & p=\frac{2}{3}, q=1-\frac{2}{3}=\frac{1}{3}, n=5 \\ & P(x \leq 3)=1-P(x>3) \\ & =1-[P(4) \mp P(5)]\end{aligned}$
$\begin{aligned} & =1-\left[{ }^5 C_4\left(\frac{2}{3}\right)^4\left(\frac{1}{3}\right)^1+{ }^5 C_5\left(\frac{2}{3}\right)^5\right] \\ & =1-\left[5 \times \frac{16}{81} \times \frac{1}{3}+1 \times \frac{32}{343}\right] \\ & =1-\frac{16}{243} \times 7 \\ & =1-\frac{112}{243}\end{aligned}$
$\begin{aligned} & =\frac{243-112}{243} \\ & =\frac{131}{243} \\ & =0.539\end{aligned}$

X = Number of families who own a television set.
P = Probability of families who own a television set.
$\begin{aligned} & P=80 \%=\frac{80}{100}=\frac{4}{5} \\ & q=1-p=1-\frac{4}{5}=\frac{1}{5} \\ & \text { Given } n=5, X \sim B\left(5, \frac{4}{5}\right)\end{aligned}$
The p.m.f. or X is given as
$\begin{aligned} & P(X=x)={ }^n C_x p^x q^{n-x} \\ & ={ }^n C_x p^x q^{5-x}\end{aligned}$
a. P(families own television set)
$\begin{aligned} & =P(X=3) \\ & ={ }^5 C_3\left(\frac{4}{5}\right)^3\left(\frac{1}{5}\right)^{5-3} \\ & =\frac{128}{625} \\ & =0.2048\end{aligned}$
b. P(At least two families own television set)
$\begin{aligned} & P(X \geq 2)=1-P(X<2) \\ & =1-[P(X=0)+P(X=1)] \\ & =1-\left[{ }^5 C_0\left(\frac{4}{5}\right)^0\left(\frac{1}{5}\right)^5+{ }^5 C_1\left(\frac{4}{5}\right)^1 \times\left(\frac{1}{5}\right)^4\right] \\ & =1-\left(\frac{1}{55}+\frac{20}{55}\right) \\ & =1-\left(\frac{21}{55}\right)=\frac{34}{55} .\end{aligned}$