Given, A and B are two independent events such that
\(P(A \cap \bar{B})=\frac{1}{4} \text { and } P(\bar{A} \cap B)=\frac{1}{6}\)
Now, \(P(A \cap \bar{B})=\frac{1}{4}\)
\(\begin{aligned} \Rightarrow \quad P(A) \cdot P(\bar{B})=\frac{1}{4} \end{aligned}\)
\(\begin{aligned} \Rightarrow \quad P(A)[1-P(B)]=\frac{1}{4} \quad[\because P(B)+P(\bar{B})=1] \end{aligned}\)
\(\Rightarrow P(A)-P(A) P \cdot(B)=\frac{1}{4}\)    ...(i)
and \(P(\bar{A} \cap B)=\frac{1}{6} \Rightarrow P(\bar{A}) \cdot P(B)=\frac{1}{6}\)
\(\begin{aligned} \Rightarrow \quad[1-P(A)] P(B)=\frac{1}{6} \end{aligned}\)
\(\begin{aligned} \Rightarrow \quad P(B)-P(A) P(B)=\frac{1}{6} \end{aligned}\)  ...(ii)
On subtracting Eq. (ii) from Eq.(i), we get
\(P(A)-P(B)=\frac{1}{4}-\frac{1}{6} \Rightarrow P(A)=\frac{1}{12}+P(B)\)
Now, on substituting this value in Eq. (ii), we get
\(\begin{aligned} P(B)-\left(\frac{1}{12}+P(B)\right) P(B)=\frac{1}{6} \end{aligned}\)
Let \(\begin{aligned} \Rightarrow 12 x-x-12 x^2=2 \Rightarrow 12 x^2-11 x+2=0 \end{aligned}\)
\(\begin{aligned} \Rightarrow 12 x^2-8 x-3 x+2=0 \end{aligned}\)
\(\begin{aligned} \Rightarrow \quad(4 x-1)(3 x-2)=0 \end{aligned}\)
\(\begin{aligned} \Rightarrow x=\frac{1}{4} \text { or } x=\frac{2}{3} \Rightarrow P(B)=\frac{1}{4} \text { or } P(B)=\frac{2}{3} \end{aligned}\) [put x = P(B)]
Now, if P(B) = \(\frac{1}{4}\), then
\(P(A)=\frac{1}{12}+\frac{1}{4}=\frac{1+3}{12}=\frac{4}{12}=\frac{1}{3}\)
and if P(B) = \(\frac{2}{3}\), then \(P(A)=\frac{1}{12}+\frac{2}{3}=\frac{1+8}{12}=\frac{9}{12}=\frac{3}{4}\)

We have, the following equations of lines
\(\begin{aligned} \frac{x+2}{1}=\frac{y-3}{2}=\frac{z+1}{4} \end{aligned}\)
and \(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4} \)
Let \(\frac{x+2}{1}=\frac{y-3}{2}=\frac{z+1}{4}=\lambda\)
Thus, the general point on the first line is
\((\lambda-2,2 \lambda+3,4 \lambda-1)\)
Again, let \(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}=\mu\)
Thus, the general point on the second line is
\((2 \mu+1,3 \mu+2,4 \mu+3)\)
Now, the direction ratios of the required line are
\((\lambda-3,2 \lambda+2,4 \lambda-2)\)
Direction ratios of the same line may be
\(\begin{aligned} (2 \mu, 3 \mu+1,4 \mu+2) \end{aligned}\)
\(\begin{aligned} \therefore & \frac{\lambda-3}{2 \mu}=\frac{2 \lambda+2}{3 \mu+1}=\frac{4 \lambda-2}{4 \mu+2} \end{aligned}\)         ...(i)
Let   \(\frac{\lambda-3}{2 \mu}=\frac{2 \lambda+2}{3 \mu+1}=\frac{2 \lambda-1}{2 \mu+1}=k\)
\(\begin{aligned} \Rightarrow \quad \lambda-3=2 \mu k, 2 \lambda+2=(3 \mu+1) k \end{aligned}\),
\(\begin{aligned} \text { and } 2 \lambda-1=(2 \mu+1) k \end{aligned}\)
\(\Rightarrow \quad \frac{\lambda-3}{2}=\mu k, 2 \lambda+2=3 \times \frac{\lambda-3}{2}+k\)
and \(\begin{aligned} 2 \lambda-1 & =\lambda-3+k \end{aligned}\)
\(\begin{aligned} \Rightarrow k & =\frac{4 \lambda+4-3 \lambda+9}{2} \end{aligned}\)
and \(k=2 \lambda-1-\lambda+3\)
\(\begin{array}{ll} \Rightarrow & k=\frac{\lambda+13}{2} \text { and } k=\lambda+2 \\ \end{array}\)
\(\begin{array}{ll} \therefore & \frac{\lambda+13}{2}=\lambda+2 \\ \end{array}\)
\(\begin{array}{ll} \Rightarrow & 13+\lambda=2 \lambda+4 \Rightarrow \lambda=9 \end{array}\)
\(\begin{array}{ll} \Rightarrow & k=\lambda+2=9+2=11 \end{array}\)
\(\because \mu k=\frac{\lambda-3}{2} \Rightarrow \mu(11)=\frac{9-3}{2} \Rightarrow \mu=\frac{6}{22}=\frac{3}{11}\)
Now, the direction ratios of the required line are (6, 20, 34) or (3, 10, 17).
Hence, the equation of the required line is given by \(\frac{x-1}{3}=\frac{y-1}{10}=\frac{z-1}{17}\)

Let E_I = event of selecting box 
and E ₂ = event of selecting box II
Let A = event of drawing an orange ball.
Now, \(P\left(E_{1}\right)=\frac{1}{2} \text { and } P\left(E_{2}\right)=\frac{1}{2}\) 
Now, P(A / E₁) = Probability of selecting an orange ball from box I
\(=\frac{4}{8}=\frac{1}{2}\) 
\([\because \text { total balls }=8 \text { and orange balls }=4]\) 
and P(A / E₂) = Probability of selecting an orange ball from box II
= \(\frac{3}{8}\) [\(\therefore\)  total balls = 8 and orange balls = 3]
Now, P(E₁ / A) = Probability that drawn ball is from the first box
\(=\frac{P\left(E_{1}\right) \cdot P\left(A / E_{1}\right)}{P\left(E_{1}\right) \cdot P\left(A / E_{1}\right)+P\left(E_{2}\right) \cdot P\left(A / E_{2}\right)}\) 
\(=\frac{\frac{1}{2} \times \frac{1}{2}}{\left(\frac{1}{2} \times \frac{1}{2}\right)+\left(\frac{1}{2} \times \frac{3}{8}\right)}=\frac{\frac{1}{4}}{\frac{1}{4}+\frac{3}{16}}\) 
\(=\frac{\frac{1 / 4}{4+3}}{16}=\frac{1}{4} \times \frac{16}{7}=\frac{4}{7}\) 
Hence, the probability that the ball is drawn from the first box is \(\frac{4}{7}\) .

First of all, let us graph the feasible region of the system of inequalities (2) to (5). The feasible region (shaded). Observe that the feasible region is unbounded.
We now evaluate Z at the corner points.

From this table, we find that – 300 is the smallest value of Z at the corner point (6, 0). Can we say that minimum value of Z is – 300? Note that if the region would have been bounded, this smallest value of Z is the minimum value of Z (Theorem 2). But here we see that the feasible region is unbounded. Therefore, – 300 may or may not be the minimum value of Z. To decide this issue, we graph the inequality
– 50x + 20y < – 300 (see Step 3(ii) of corner Point Method.)
i.e., – 5x + 2y < – 30
and check whether the resulting open half plane has points in common with feasible region or not. If it has common points, then –300 will not be the minimum value of Z.
Otherwise, –300 will be the minimum value of Z. 
It has common points. Therefore, Z = –50 x + 20 y has no minimum value subject to the given constraints.
In the above example, can you say whether z = – 50 x + 20 y has the maximum value 100 at (0, 5)? For this, check whether the graph of – 50 x + 20 y > 100 has points in common with the feasible region.

= \(2 \sqrt{3} \text { sq units }\)

Given, \(\overrightarrow{O A}=2 \vec{a}+\vec{b} \text { and } \overrightarrow{O B}=\vec{a}-3 \vec{b}\) 
Also, it is given that C is the point which divides the line joining A and B externally in the ratio 1: 2.
Then by using section formula of external division, we get

\(\overrightarrow{O C}=\frac{2 \overrightarrow{O A}-\overrightarrow{O B}}{2-1}\) 
\(\Rightarrow \overrightarrow{O C}=\frac{2(\overrightarrow{2 a}+\vec{b})-1(\vec{a}-3 \vec{b})}{1} \quad[\text { from Eq }\) 
\(=4 \vec{a}+2 \vec{b}-\vec{a}+3 \vec{b}=3 \vec{a}+5 \vec{b}\) 
i.e. to show \(\overrightarrow{O A}=\frac{\overrightarrow{O B}+\overrightarrow{O C}}{2}\) 
Consider, \(\frac{\overrightarrow{O B}+\overrightarrow{O C}}{2}=\frac{\vec{a}-3 \vec{b}+3 \vec{a}+5 \vec{b}}{2}\) 
[from Equation (i) and (ii)]
\(=\frac{4 \vec{a}+2 \vec{b}}{2}=2 \vec{a}+\vec{b}=\overrightarrow{O A} \quad[\text { from Eq. }(i)]\) 
Thus, \(\frac{\overrightarrow{O B}+\overrightarrow{O C}}{2}=\overrightarrow{O A}\) 
Hence, A is mid-point of line segment BC.

\(\text { Required vector }=\frac{5(\vec{a}+\vec{b})}{|\vec{a}+\vec{b}|}\) 
\(=\frac{15}{\sqrt{10}} \hat{i}+\frac{5}{\sqrt{10}} \hat{j}\)

Given line is \( \frac{x+2}{3}=\frac{y+1}{2}=\frac{z-3}{2}=\lambda\)
General point on the line is \( R(3 \lambda-2,2 \lambda-1,2 \lambda+3)\)
Distance of point R from P( 1, 3, 3) is 5 units.
\(\sqrt { ({ 3\lambda -2-1) }^{ 2 }+({ 2\lambda +3-3) }^{ 2 }+({ 2\lambda +3-3) }^{ 2 } } =5\)
\(\Rightarrow { (3\lambda -3) }^{ 2 }+{ (2\lambda -4) }^{ 2 }+{ (2\lambda ) }^{ 2 }=25\Rightarrow \lambda =0,2\)
Substituting in (i), we get point as R(-2, -1, 3) or R(4, 3, 7).

Let E be the event that it will rain on any particular day.
\(\therefore P(E)=50 \%=\frac{1}{2} \Rightarrow P(\bar{E})=1-P(E)=1-\frac{1}{2}=\frac{1}{2}\)
\(\therefore\) Required probability
\(\begin{aligned} =P(E) \cdot P(E) \cdot P(E) \cdot P(E) \cdot P(\bar{E}) \cdot P(\bar{E}) \cdot P(\bar{E}) \end{aligned}\)
\(\begin{aligned} =(P(E))^4(P(\bar{E}))^3 \end{aligned}\)
\(\begin{aligned} =\left(\frac{1}{2}\right)^4\left(\frac{1}{2}\right)^3=\left(\frac{1}{2}\right)^7 \end{aligned}\)

Given, line is \(5 x-25=14-7 y=35 z\) 
\(\Rightarrow \frac{x-5}{1 / 5}=\frac{2-y}{1 / 7}=\frac{z}{1 / 35} \Rightarrow \frac{x-5}{1 / 5}=\frac{y-2}{-1 / 7}=\frac{z}{1 / 35}\)
 \(\Rightarrow \frac{x-5}{7}=\frac{y-2}{-5}=\frac{z-0}{1}\)      ...(i)
Hence, parallel vector of given line i.e.
\(\vec{b}=7 \hat{i}-5 \hat{j}+\hat{k}\)
Since, required line is parallel to given line (i).
\(\Rightarrow \vec{b}=7 \hat{i}-5 \hat{j}+\hat{k}\) will also be parallel vector of required line which passes through A(1, 2, -1).
\(\therefore\) The required vector equation of line is
\(\vec{r}=(\hat{i}+2 \hat{j}-\hat{k})+\lambda(7 \hat{i}-5 \hat{j}+\hat{k})\)

-10

= 5

Given vectors are \(\vec{a}=2 \hat{i}+\lambda \hat{j}+\hat{k} \text { and } \vec{b}=\hat{i}-2 \hat{j}+3 \hat{k} \).
Since, the vectors are perpendicular
\(\therefore \vec{a} \cdot \vec{b}=0 \)
\(\Rightarrow (2 \hat{i}+\lambda \hat{j}+\hat{k}) \cdot(\hat{i}-2 \hat{j}+3 \hat{k})=0 \)
\(\Rightarrow 2-2 \lambda+3=0 \)
\(\therefore \lambda=5 / 2 \)

P(E\(\cap\)F) = P(E) + P(F) - P(E\(\cup\)F)
\(=\frac { 6 }{ 11 } +\frac { 5 }{ 11 } -\frac { 7 }{ 11 } \)
\(=\frac { 4 }{ 11 } \)
(a)  \(P(E/F)=\frac { P(E\cap F) }{ P(F) } \)
\(=\frac { \frac { 4 }{ 11 } }{ \frac { 5 }{ 11 } } =\frac { 4 }{ 5 } \)
(b) P(E/E) = \(\frac { P(E\cap F) }{ P(E) } \)
\(=\frac { \frac { 4 }{ 11 } }{ \frac { 6 }{ 11 } } =\frac { 4 }{ 6 } \)
\(=\frac { 2 }{ 3 } \)

Let points be A(2, 3, 4), B(- 1, - 2, 1) and C(5, 8, 7)
Direction ratios or AB are 2 + 1, 3 + 2, 4 - 1, i.e. 3, 5, 3;
Direction ratios of Be are 5 + 1, 8 + 2, 7 - 1, i.e. 3,5,3
As \(\frac{3}{3}=\frac{5}{5}=\frac{3}{3}\)
⇒  AB is parallel to BC, B is common.
Hence, A, B, C are collinear.

Given, \(\vec{\alpha}=4 \hat{i}+5 \hat{j}-\hat{k}, \vec{\beta}=\hat{i}-4 \hat{j}+5 \hat{k}\)
and \(\vec{q}=3 \hat{i}+\hat{j}-\hat{k}\)
Also, vector \(\vec{p}\) is perpendicular to \(\alpha\) and \(\beta\).
Then, \(\vec{p}=\lambda(\vec{\alpha} \times \vec{\beta})\)       ...(i)
Now, \(\vec{\alpha} \times \vec{\beta}=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 4 & 5 & -1 \\ 1 & -4 & 5 \end{array}\right|\)
\(\begin{aligned} =\hat{i}(25-4)-\hat{j}(20+1)+\hat{k}(-16-5) \end{aligned}\)
\(\begin{aligned} =\hat{i}(21)-\hat{j}(21)+\hat{k}(-21) \end{aligned}\)
\(\Rightarrow \quad \vec{\alpha} \times \vec{\beta}=21 \hat{i}-21 \hat{j}-21 \hat{k}\)
So, \(\vec{p}=21 \lambda \hat{i}-21 \lambda \hat{j}-21 \lambda \hat{k}\) [from Eq. (i)]  ...(ii)
Also, given that \(\vec{p} \cdot \vec{q}=21\)
\(\begin{aligned} \therefore \quad(21 \lambda \hat{i}-21 \lambda \hat{j}-21 \lambda \hat{k}) \cdot(3 \hat{i}+\hat{j}-\hat{k})=21 \end{aligned}\)
\(\begin{aligned} \Rightarrow \quad 63 \lambda-21 \lambda+21 \lambda=21 \end{aligned}\)
\(\begin{aligned} \Rightarrow \quad 63 \lambda=21 \end{aligned}\)
\(\begin{aligned} \Rightarrow \quad \lambda=1 / 3 \end{aligned}\)
On putting \(\lambda=\frac{1}{3}\) in Eq. (ii), we get
\(\vec{p}=21 \times \frac{1}{3} \hat{i}-21 \times \frac{1}{3} \hat{j}-21 \times \frac{1}{3} \hat{k}\)
\(\therefore \quad \vec{p}=7 \hat{i}-7 \hat{j}-7 \hat{k}\)
which is the required vector.

The vector equations of given lines are
\(\begin{aligned} \vec{r}=3 \hat{i}+2 \hat{j}-4 \hat{k}+\lambda(\hat{i}+2 \hat{j}+2 \hat{k}) \end{aligned}\)
and \(\begin{aligned} \vec{r}=5 \hat{i}-2 \hat{j}+\mu(3 \hat{i}+2 \hat{j}+6 \hat{k}) \end{aligned}\)
On comparing them with \(\vec{r}=\overrightarrow{a_1}+\lambda \overrightarrow{b_1}\) and \(\vec{r}=\overrightarrow{a_2}+\mu \overrightarrow{b_2}\), we get
\(\overrightarrow{a_1}=3 \hat{i}+2 \hat{j}-4 \hat{k}, \overrightarrow{a_2}=5 \hat{i}-2 \hat{j}, \vec{b}_1=\hat{i}+2 \hat{j}+2 \hat{k}\)
and \(\vec{b}_2=3 \hat{i}+2 \hat{j}+6 \hat{k}\)
\(\therefore \quad \overrightarrow{a_2}-\overrightarrow{a_1}=(5 \hat{i}-2 \hat{j})-(3 \hat{i}+2 \hat{j}-4 \hat{k})\)
\(=2 \hat{i}-4 \hat{j}+4 \hat{k}\)
\(\begin{aligned} \therefore \quad \vec{b}_1 \times \vec{b}_2 & =\left|\begin{array}{lll} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 2 \\ 3 & 2 & 6 \end{array}\right| \end{aligned}\)
\(\begin{aligned} =\hat{i}(12-4)-\hat{j}(6-6)+\hat{k}(2-6) \end{aligned}\)
\(\begin{aligned} =8 \hat{i}-4 \hat{k} \end{aligned}\)
\(\therefore\left(\vec{a}_2-\vec{a}_1\right) \cdot\left(\vec{b}_1 \times \vec{b}_2\right)=(2 \hat{i}-4 \hat{j}+4 \hat{k}) \cdot(8 \hat{i}-4 \hat{k})\)
= 16 + 0 - 16 = 0
\(\therefore\) The lines are intersecting and the shortest distance between the lines is 0.
Now, the position vectors of arbitrary points on the given lines are \((3+\lambda) \hat{i}+(2+2 \lambda) \hat{j}+(-4+2 \lambda) \hat{k}\) and \((5+3 \mu) \hat{i}+(-2+2 \mu) \hat{j}+6 \mu \hat{k}\), respectively.
Since, lines intersect then they have a common point.
\(\therefore \begin{aligned} 3+\lambda & =5+3 \mu \end{aligned}\)         ...(i)
\(\begin{aligned} 2+2 \lambda & =-2+2 \mu \end{aligned}\)        ...(ii)
\(\begin{aligned} -4+2 \lambda & =6 \mu \end{aligned}\)               ...(iii)
On solving Eqs. (i) and (ii), we get
\(\lambda=-4 \text { and } \mu=-2\)
\(\therefore\) Point of intersection is (3 - 4, 2 - 8, -4 - 8)
i.e. (-1, -6, -12). 

Let T be the image of the point P(2, -1, 5). Q is the foot of perpendicular drawn from point P on the line AB.
Given, equation of line AB is
\(\frac{x-11}{10}=\frac{y+2}{-4}=\frac{z+8}{-11}\)                         ...(i)

Let \(\frac{x-11}{10}=\frac{y+2}{-4}=\frac{z+8}{-11}=\lambda\) (say)
\(\Rightarrow \quad x=10 \lambda+11, y=-4 \lambda-2\)
and \(z=-11 \lambda-8\)
Then, coordinates of Q are
\(=(10 \lambda+11,-4 \lambda-2,-11 \lambda-8)\)     ....(ii)
Now, DR's of line PQ
\(\begin{aligned} =10 \lambda+11-2-4 \lambda-2+1,-11 \lambda-8-5 \end{aligned}\)
\(\begin{aligned} =10 \lambda+9,-4 \lambda-1,-11 \lambda-13 \end{aligned}\)
Since, line \(P Q \perp A B\)
\(\therefore\) a₁a₂ + b₁b₂ + c₁c₂ = 0
where \(a_1=10 \lambda+9, b_1=-4 \lambda-1, c_1=-11 \lambda-13\)
and a₂ = 10, b₂ = -4, c₂ = -11
\(\begin{aligned} \therefore(10 \lambda+9)(10)+(-4 \lambda-1)( & -4) +(-11 \lambda-13)(-11)=0 \end{aligned}\)
\(\begin{aligned} \Rightarrow & & 100 \lambda+90+16 \lambda+4+121 \lambda+143 & =0 \end{aligned}\)
\(\begin{aligned} \Rightarrow & & 237 \lambda+237 & =0 \end{aligned}\)
\(\begin{aligned} \Rightarrow & & \lambda & =-1 \end{aligned}\)
On putting \(\lambda\) = - 1 in Eq. (ii), we get
Q = (10(-1)+11), -4(-1)-2, -11(-1)-8)
=(-10+11, 4-2, 11-8)
= (1, 2, 3)
Let image of a point P be T(x, y, z).
Then, Q will be the mid-point of PT.
By using mid-point formula
Q = Mid-point of P(2, -1,5) and T(x, y, z).
\(=\left(\frac{x+2}{2}, \frac{y-1}{2}, \frac{z+5}{2}\right)\)
But Q = (1, 2, 3)
\(\therefore\left(\frac{x+2}{2}, \frac{y-1}{2}, \frac{z+5}{2}\right)=(1,2,3)\)
On equating corresponding coordinates, we get
\(\begin{aligned} \frac{x+2}{2} & =1, \frac{y-1}{2}=2, \frac{z+5}{2}=3 \end{aligned}\)
\(\begin{aligned} \Rightarrow & \quad x =2-2, y=4+1, z=6-5 \end{aligned}\)
\(\therefore\) x = 0, y = 5, z = 1
\(\therefore\) Coordinates of T = (x, y, z) = (0, 5, 1)
Hence, the coordinate image of point P(2, -1, 5) is T(0, 5, 1).

Let, E₁: Event that lost card is a spade
E₂: Event that lost card is a not spade
A: Event that three spades are drawn without replacement from 51 cards
\(P({ E }_{ 1 })=\frac { 13 }{ 52 } =\frac { 1 }{ 4 } ,\quad P({ E }_{ 2 })=1-\frac { 1 }{ 4 } =\frac { 3 }{ 4 } \)
\(P(A/{ E }_{ 1 })=\frac { { 12 }_{ C_{ 3 } } }{ { 51 }_{ { C }_{ 3 } } } ,\quad P(A/{ E }_{ 2 })=\frac { 13_{ { C }_{ 3 } } }{ { 51 }_{ { C }_{ 3 } } } \)
\(P({ E }_{ 1 }/A)=\frac { \frac { 1 }{ 4 } .\frac { { 12 }_{ C_{ 3 } } }{ { 51 }_{ { C }_{ 3 } } } }{ \frac { 1 }{ 4 } .\frac { { 12 }_{ C_{ 3 } } }{ { 51 }_{ { C }_{ 3 } } } +\frac { 3 }{ 4 } .\frac { { 12 }_{ C_{ 3 } } }{ { 51 }_{ { C }_{ 3 } } } } \)
\(=\frac { 10 }{ 49 } \)

Let, E₁: Two white balls are transferred
E₂: Two red balls are transferred
E₃: One red and one white ball are transferred.
 A: The ball drawn from the bag II is red.
\(P({ E }_{ 1 })=\frac { { 4 }_{ C_{ 2 } } }{ { 9 }_{ { c }_{ 2 } } } =\frac { 4\times 3 }{ 9\times 8 } =\frac { 1 }{ 6 } \)
\({ P({ E } }_{ 2 })=\frac { { 5 }_{ C_{ 2 } } }{ { 9 }_{ C_{ 2 } } } =\frac { 4\times 3 }{ 9\times 8 } =\frac { 5 }{ 18 } \)
\(P(E_{ 3 })=\frac { { 5 }_{ { C }_{ 1 } }\times { 4 }_{ C_{ 1 } } }{ { 9 }_{ C_{ 2 } } } =\frac { 4\times 5\times 2 }{ 9\times 8 } =\frac { 5 }{ 9 } \)
\(P(A/{ E }_{ 1 })=\frac { 3 }{ 8 } ,P(A/{ E }_{ 2 })=\frac { 5 }{ 8 } ,\)
\(P(A/{ E }_{ 3 })=\frac { 4 }{ 8 } \)
The required probability, P(E₃/A), by Bayes' Theorem
\(=\frac { P({ E }_{ 3 }).P(A/{ E }_{ 3 }) }{ P({ E }_{ 1 }).P(A/{ E }_{ 1 })+P({ E }_{ 2 }).P(A/{ E }_{ 2 })+P({ E }_{ 3 }).P(A/{ E }_{ 3 }) } \)
\(=\frac { \frac { 5 }{ 9 } \times \frac { 4 }{ 8 } }{ \frac { 1 }{ 6 } \times \frac { 3 }{ 8 } +\frac { 5 }{ 18 } \times \frac { 5 }{ 8 } +\frac { 5 }{ 9 } \times \frac { 4 }{ 8 } } \)
\(=\frac { 20 }{ 37 } \) 

As P(A/B) = \(\frac { P(A\cap B) }{ P(B) } \)
= \(\frac{70}{100}\times \frac{100}{85}=\frac{14}{17}\)

Given, 70% customers pay their first month bill in time, therefore, the customers who do not pay bill in time is 30%.
(i) Let E₁ and E₂, denote the event of customer paying or not paying the fìrst month bill in time, respectively.
Then, P(E₁) = 70% = 0.7
and P(E₂) = 30% = 0.3
(ii) Given that the probability of customers who pays the bill in time next month is 0.8 and who does not pay in time is 0.4.
Let A denotes the event of customer paying second month's bill in time.
\(\therefore P\left(\frac{A}{E_1}\right)=P\) (customer paying second month bill in time when they pay first month bill)
= 0.8
and \(P\left(\frac{A}{E_2}\right)=P\) (customer paying second month bill in time when they do not pay first month bill in time)
= 0.4
(ii) The probability of customer paying second month's bill in time = 0.7 \(\times\) 0.8 = 0.56
Or
The probability of customer paying first month's bill in time, if it is found that customer has paid the second month's bill in time is \(P\left(\frac{E_1}{A}\right)\)
\(\begin{aligned} P\left(\frac{E_1}{A}\right) & =\frac{P\left(E_1\right) \cdot P\left(\frac{A}{E_1}\right)}{P\left(E_1\right) \cdot P\left(\frac{A}{E_1}\right)+P\left(E_2\right) \cdot P\left(\frac{A}{E_2}\right)} \end{aligned}\)
\(\begin{aligned} =\frac{0.7 \times 0.8}{0.7 \times 0.8+0.3 \times 0.4} \end{aligned}\)
\(\begin{aligned} =\frac{0.56}{0.56+0.12}=\frac{0.56}{0.68} \end{aligned}\)
= 0.824

(i) (a): Eliminating 't' from the given equations, we get equation of path as, \(\frac{x}{2}=\frac{y}{-4}=\frac{z}{4} \text { or } \frac{x}{1}=\frac{y}{-2}=\frac{z}{2}\).
Thus, the path of the rocket represents a straight line.
(ii).(b) : Since, only (1, -2, 2) satisfy the equation of path of rocket therefore (1, -2, 2) lie on the path of rocket.
(iii) (b): For t = 10 sec, we have x = 20,y = -40, z = 40 
Now, required distance = \(\sqrt{x^{2}+y^{2}+z^{2}}\)
\(=\sqrt{20^{2}+(-40)^{2}+(40)^{2}}=\sqrt{400+1600+1600}\)
\(=\sqrt{3600}=60 \mathrm{~km}\)
(iv) (d) : Clearly, height of rocket from the ground = z-coordinate of given position = 6 km
(v) (b) : Let Q be the image of point P(1, -2, 2) in the plane 3x - y + 4z = 2. Then, equation of PQ is
\(\frac{x-1}{3}=\frac{y+2}{-1}=\frac{z-2}{4}\)
Let the coordinates of Q be (3r +1, -r - 2, 4r + 2).
Let R be the mid -point of PQ. Then, coordinates of R are
\(\left(\frac{3 r+2}{2}, \frac{-r-4}{2}, \frac{4 r+4}{2}\right) \text { or }\left(\frac{3}{2} r+1, \frac{-r}{2}-2,2 r+2\right)\)
Since, R lies on 3x - y + 4z = 2.
\( \therefore \quad 3\left(\frac{3}{2} r+1\right)-\left(\frac{-r}{2}-2\right)+4(2 r+2)=2 \)
\(\Rightarrow \frac{9 r}{2}+3+\frac{r}{2}+2+8 r+8=2 \\ \)
\(\Rightarrow 13 r+13=2 \Rightarrow r=\frac{-11}{13}\)
Hence, the coordinates of Q are
\(\left(\frac{-33}{13}+1, \frac{11}{13}-2, \frac{-44}{13}+2\right) \text { i.e., }\left(\frac{-20}{13}, \frac{-15}{13}, \frac{-18}{13}\right)\)

(i) (b): Let B(x, y) be the point of intersection of the given lines
2x + 5y = 100 ....(i)
and \(\frac{x}{25}+\frac{y}{40}=1 \Rightarrow 8 x+5 y=20\)...(ii)
Solving (i) and (ii), we get
\(x=\frac{50}{3}, y=\frac{40}{3}\)
ஃ The point of intersection \(B(x, y)=\left(\frac{50}{3}, \frac{40}{3}\right)\)
(ii) (d): The corner points of the feasible region shown in the given graph are
\((0,0), A(25,0), B\left(\frac{50}{3}, \frac{40}{3}\right), C(0,20)\)
(iii) (a): Here Z = x + y

Thus, max Z = 30 occurs at point \(\left(\frac{50}{3}, \frac{40}{3}\right)\)
(iv) (b): 

(v) (c):

(a) If two coins are tossed simultaneously, then sample space is given by
S = {(H, H), (H, T), (T, H), (T, T)}
Now, let A be the event of getting two heads.
\(\therefore \quad P(A)=\frac{1}{4}\)
and B be the event of getting atleast one head.
\(\therefore \quad P(B)=\frac{3}{4}\)
Thus, \(P\left(\frac{A}{B}\right)=\frac{P(A \cap B)}{P(B)}=\frac{1 / 4}{3 / 4}=\frac{1}{4} \times \frac{4}{3}=\frac{1}{3}\)
Hence, Assertion is true.
Reason is also true.

(d) Assertion is incorrect, Reason is correct.