Maths M.C.Q (1 Marks)

Four balls $1,2,3,4$ are randomly placed in boxes $B_1, B_2, B_3, B_4$.

Probability of exactly one box is empty is

$\frac{{ }^4 C_1 \times \frac{4 !}{1 ! \times 2 !} \times \frac{1}{2 !} \times 3 !}{4^4}=\frac{4 \times 6 \times 6}{4 \times 4 \times 4 \times 4}=\frac{9}{16}$

A fair coin is tossed $5$ times.

Total number of sample space $=2^5$

Probability of head does not occur two or more time in rows.

Possible cases

Case $I$ All tails $=\left(\frac{1}{2}\right)^5=\frac{1}{2^5}$

Case $II$ $4$ tails and one head

${ }^5 C_4\left(\frac{1}{2}\right)^5=\frac{5}{2^5}$

Case $III$ $3$ tails and $2$ head $\times T \times T \times T \times$

${ }^4 C_2 \times\left(\frac{1}{2}\right)^5=\frac{6}{2^5}$

Case $IV$ $2$ tails and $3$ heads $\times T \times T \times$

${ }^3 C_8 \times\left(\frac{1}{2}\right)^5=\frac{1}{2^5}$

$\therefore$ Required probability

$=\frac{1}{2^5}+\frac{5}{2^5}+\frac{6}{2^5}+\frac{1}{2^5}=\frac{13}{2^5}$

We have,

Total number of person $=32$

they seated in a around a circular table

Case $I$ No person between $X_1$ and $X_2$

Probability $=\frac{i 30 ! \times 2 !}{31 !}=\frac{2}{31}$

Case $II$ One person between $X_1$ and $X_2$

Probability $=\frac{30 C_1 \times 29 ! \times 2 !}{31 !}=\frac{2}{31}$

Case $III$ Two persons between $X_1$ and $X_2$

Probability $\frac{30 C_2 \times 28 ! \times 2 ! \times 2 !}{31 !}=\frac{2}{31}$

Case $IV$ Three persons between $X_1$ and

Total probability $=\frac{2}{31}+\frac{2}{31}+\frac{2}{31}+\frac{2}{31}=\frac{8}{31}$

$n=5, p=\frac{1}{4}, q=\frac{3}{4}$

$\therefore$ Required probability $={ }^5 C_2(p)^2 q^3 \times p$

$={ }^5 C_2\left(-\frac{1}{4}\right)^2\left(\frac{3}{4}\right)^3 \times \frac{1}{4}=\frac{270}{4096}=0.6591$

Consider the event

$A=$ Blue ball is drawn from first draw

$B=$ Red ball is drawn from first draw

$C=$ Blue ball is drawn from second draw

$P(A)=\frac{b}{b+r}, P\left(\frac{C}{A}\right)=\frac{b+1}{b+r+1}$

$P(B)=\frac{r}{b+r}, P\left(\frac{C}{B}\right)=\frac{b}{b+r+1}$

Required probability

$P(C)=P(A) \cdot P(C / A)+P(B) \cdot P(C / B)$

$=\frac{b}{b+r} \frac{b+1}{b+r+1}+\frac{r}{(b+r)} \frac{b}{b+r+1}$

$=\frac{b}{(b+r)(b+r+1)}(b+1+r)=\frac{b}{b+r}$

The $3-$digit number which is divisible by $4$ and $5$ both.

i.e. last digits are $00,20,40,60,80$

Now ending with $00$ are $(100,200,300 \ldots$ $900)=9$

If digit repeat other than $0^{\prime}$ then they are $(220,440,660,880)$ but $220$ number can be permuted according to condition as $(220,202).$

Similarly, for $440$ as $(440,404), 660$ and $880$ , so there are $8$ favourable cases.

If the number have no digit repeated like $120,120$ can be permuted in $4$ ways.

So, such number are $8 \times 4 \times 4=128$

We have,

$6$ boxes $B_1, B_2, B_3, B_4, B_5, B_6$ and

two dices $D_1$ and $D_2$.

Let $P\left(B_1\right)$ be the probability that $B_1$

contains at most one ball.

$\therefore P\left(B_1\right)=P\left(B_1\right.$ contains 0 ball $)+P\left(B_1\right.$

contains 1 ball) $P\left(B_1\right)=\left(D_1\right.$ never show 1)

$+P\left(D_2\right.$ shows 1 once when $D_1$ show 1$)$

$P\left(B_1\right)=\left(\frac{5}{6}\right)^n+n\left(\frac{5^{n-1}}{6^{n-1}}\right)\left(\frac{1}{6^2}\right)$

When $n=10$, let $A_r$ be number of ways of selecting $r$ numbers.

Number of selection of $A$ is

$n\left(A_6\right)+n\left(A_1\right)+n\left(A_2\right)+n\left(A_3\right)+n\left(A_4\right)$

$=1+10+(7+6+5+\ldots+1)$

$+(4+3+2+1)$

$+(3+2+1)+(2+1)+1$

$=11+28+10+6+3+1=60$

$N(p)=n$ (Number of ways 1 is selected)

$=1+7+4+3+2+1+1=19$

$N(q)=n$ (Number of ways $2$ is selected)

$=1+6+3+2+1=13$

So, $\quad p=\frac{19}{60}, q=\frac{13}{60}$

$p > q, p-q=\frac{19-13}{60}=\frac{6}{60}=\frac{1}{10}$

A box contains coupons $\{1,2,3, \ldots, n\}$

Let a number $x=1$ is selected.

Then, favourable outcomes $(1,1)(1,2), . .$, .

$(1, n)=\left[\frac{n}{1}\right]$

Number of favourable outcomes, when $x=1, y=1$ is $2\left[\frac{n}{1}\right]-1$

Number of favourable outcomes, when $x=2, y=2, x \neq 1, y \neq 1=2\left[\frac{n}{2}\right]-1$

Similarly, for number of favourable outcomes, when

$x=k, y=k$ but $x, y \neq\{1,2, \ldots, k-1\}$

$=2\left[\frac{n}{k}\right]-1$

So, required probability

$-\frac{2 \sum\limits_{k=1}^n\left[\frac{n}{k}\right\rfloor-(1+1+\ldots n \text { times })}{n^2}$

$=\frac{2}{n^2} \sum\limits_{k=1}^n\left[\frac{n}{k}\right]-\frac{n}{n^2}$

$=\frac{-1}{n}+\frac{2}{n^2} \sum\limits_{k=1}^n\left[\frac{n}{k}\right]$

Consider the event

$E_1=$ Purse contains $4$ copper coins and $3$ silver coins.

$E_2=$ Purse contains $6$ copper coins and $4$ silver coins

$A=$ draw copper coins

$\therefore \quad P\left(E_1\right) =\frac{1}{2}, P\left(A / E_1\right)=4 / 7$

$P\left(E_2\right)=\frac{1}{2}, P\left(A / E_2\right)=6 / 10$

$\therefore$ Required probability

$P(A) =P\left(E_1\right) \times P\left(A / E_1\right)+P\left(E_2\right)$

$=\frac{1}{2} \times \frac{4}{7}+\frac{1}{2} \times \frac{6}{10}$

$=\frac{1}{2}\left(\frac{4}{7}+\frac{6}{10}\right)=\frac{1}{2}\left(\frac{82}{7}\right)=\frac{41}{70}$

A man tosses a coin $10$ times, scoring $1$ point for head and $2$ points for tails.

$P(k)$ be the probability of scoring at least $k$ points.

$\therefore$ Required probability

$=\frac{{ }^{10} C_0+{ }^{10} C_1+{ }^{10} C_2+\ldots+{ }^{10} C_{k-10}}{2^{10}}$

Given, $P(k) \geq \frac{1}{2}$

$\therefore^{10} C_0+{ }^{10} C_1+{ }^{10} C_2+\ldots+$

${ }^{10} C_{k-10} \geq 2^9=512$

$1+10+45+120+210+252+\ldots > 512$

$\therefore \quad k-10 \geq 5 \Rightarrow k \geq 15$

$\therefore \quad$ Largest value of $k$ is $15 .$

We have, $S_n=\sum \limits_{k=1}^n k$

$S_1=1$

$S_2=1+2=3$

$S_3=1+2+3=6$

$S_4=1+2+3+4=10$

$\quad S_5=1+2+3+4+5=15$

$\quad S_6=1+2+3+4+5+6=21$

Clearly, from two terms are even two term are odd.

$\therefore S_1, S_2, S_3, \ldots, S_{98}$

Among these $48$ terms are even and $48$ terms are odd

Woman has $10$ keys out of which only one opens a lock.

The first keys works with probability $\frac{1}{10}$

The conditional probability that the second keys works given that first failed $=\frac{1}{9}$

$\therefore$ Required probability (seventh key works)

$=P($ fails $) \cdot P(2$ nd $/$ I fails $) \cdot P$ (III $/ 2$ fails $)$

$P \text { (7th/6th fails) }$

$=\frac{9}{10} \times \frac{8}{9} \times \frac{7}{8} \times \frac{6}{7} \times \frac{5}{6} \times \frac{4}{5} \times \frac{1}{4}=\frac{1}{10}$

$Two$ point on a circumference of circle of radius $r$.

Distance between two points is atleast $r$.

$\therefore$ Angle subtends to chord of length

atleast $r$ is greater than or equal to $60^{\circ}$.

$\therefore$ Required probability $=1-\frac{60}{180}$

$=1-\frac{1}{3}=\frac{2}{3}$

$\mathrm{P}($ Manufactured at B / found standard quality $)=$ ?

$A$ : Found $S.Q$

$B$ : Manufacture $B$

$C$ : Manufacture $A$

$ P\left(E_1\right)=\frac{40}{100} $

$ P\left(E_2\right)=\frac{60}{100} $

$ P\left(A / E_1\right)=\frac{90}{100} $

$ P\left(A / E_2\right)=\frac{80}{100} $

$ \because P\left(E_1 / A\right)=\frac{P\left(A / E_1\right) P\left(E_1\right)}{P\left(A / E_1\right) P\left(E_1\right)+P\left(A / E_2\right) P\left(E_2\right)}=\frac{3}{7} $

$ \therefore 126 P=54$

5 one $\& 4$ five 4 one $\& 5$ five $3$ one $\& 6$ five

$P=\frac{4 / 9}{5 / 9+4 / 9+3 / 9}=\frac{4}{12}=\frac{1}{3}$

${E}_2: \mathrm{B}$ is selected

${E}:$ white ball is drawn

$P\left(E_1 / E\right)=$

$\frac{P(E) \cdot P\left(E / E_1\right)}{P\left(E_1\right) \cdot P\left(E / E_1\right)+P\left(E_2\right) \cdot P\left(E / E_2\right)}=\frac{\frac{1}{2} \times \frac{3}{10}}{\frac{1}{2} \times \frac{3}{10}+\frac{1}{2} \times \frac{3}{5}}$

$=\frac{3}{3+6}=\frac{1}{3}$

$ \frac{P(4 W 4 B) \times P(2 W 2 B / 4 W 4 B)}{P(2 W 6 B) \times P(2 W 2 B / 2 W 6 B)+P(3 W 5 B) \times P(2 W 2 B / 3 W 5 B)}  +\ldots \ldots \ldots+P(6 W 2 B) \times P(2 W 2 B / 6 W 2 B)$

$ =\frac{\frac{1}{5} \times \frac{{ }^4 \mathrm{C}_2 \times{ }^4 \mathrm{C}_2}{{ }^8 \mathrm{C}_4}}{\frac{1}{5} \times \frac{{ }^2 \mathrm{C}_2 \times{ }^6 \mathrm{C}_2}{{ }^8 \mathrm{C}_4}+\frac{1}{5} \times \frac{{ }^3 \mathrm{C}_2 \times{ }^5 \mathrm{C}_2}{{ }^8 \mathrm{C}_4}+\ldots+\frac{1}{5} \times \frac{{ }^6 \mathrm{C}_2 \times{ }^2 \mathrm{C}_2}{{ }^8 \mathrm{C}_4}}$

$ =\frac{2}{7}$

$7 \mathrm{R}, 5 \mathrm{~B}\ \ \ \   5 \mathrm{R}, 7 \mathrm{~B} \ \ \ \  6 \mathrm{R}, 6 \mathrm{~B}$

$ \mathrm{P}(\mathrm{B})=\frac{1}{3} \cdot \frac{5}{12}+\frac{1}{3} \cdot \frac{7}{12}+\frac{1}{3} \cdot \frac{6}{12} $

$ \text { required probability }=\frac{\frac{1}{3} \cdot \frac{5}{12}}{\frac{1}{3} \cdot\left[\frac{5}{12}+\frac{7}{12}+\frac{6}{12}\right]}=\frac{5}{18}$

$ \frac{5}{6} \times \frac{1}{6}+\left(\frac{5}{6}\right)^3 \times \frac{1}{6}+\left(\frac{5}{6}\right)^5 \times \frac{1}{6}+\ldots \ldots $

$ =\frac{1}{6} \times \frac{\frac{5}{6}}{1-\frac{25}{36}}=\frac{5}{11}$

$ a+2 a+a+b+2 b+3 b=1 $

$ 4 a+6 b=1 $.................($i$)

$ E(x)=\text { mean }=\frac{46}{9}$

$ \sum \mathrm{P}_{\mathrm{i}} \mathrm{X}_{\mathrm{i}}=\frac{46}{9} \Rightarrow 4 \mathrm{a}+4 \mathrm{a}+4 \mathrm{~b}+12 \mathrm{~b}+24 \mathrm{~b}=\frac{46}{9} $

$ 8 \mathrm{a}+40 \mathrm{~b}=\frac{46}{9}$

$4 \mathrm{a}+20 \mathrm{~b}=\frac{23}{9}$

Subtract ($I$) from ($II$) we get

$ \mathrm{b}=\frac{1}{9} \& \mathrm{a}=\frac{1}{12} $

$ \text { Variance }=\mathrm{E}\left(\mathrm{x}_{\mathrm{i}}^2\right)-\mathrm{E}\left(\mathrm{x}_{\mathrm{i}}\right)^2 $

$ \mathrm{E}\left(\mathrm{x}_{\mathrm{i}}^2\right)=0^2 \times 9^2+2^2 \times 2 \mathrm{a}+4^2(\mathrm{a}+\mathrm{b})+6^2(2 \mathrm{~b})+8^2(3 \mathrm{~b}) $

$ =24 \mathrm{a}+280 \mathrm{~b}$

Put $\mathrm{a}=\frac{1}{12} \quad \mathrm{~b}=\frac{1}{9}$

$ \mathrm{E}\left(\mathrm{x}_{\mathrm{i}}^2\right)=2+\frac{280}{9}=\frac{298}{9} $

$ \therefore \sigma^2=\mathrm{E}\left(\mathrm{x}_{\mathrm{i}}^2\right)-\mathrm{E}\left(\mathrm{x}_{\mathrm{i}}\right)^2 $

$ =\frac{298}{9}-\left(\frac{46}{9}\right)^2 $

$ \sigma^2=\frac{298}{9}-\frac{2116}{81} $

$ =\frac{566}{81}$

$ \mu=\Sigma p_i x_i=\frac{18}{12} $

$ \Sigma p_i x_1^2=\frac{34}{12} $

$ \sigma^2=\Sigma p_i x_1^2-(\mu)^2 $

$ =\frac{34}{12}-\left(\frac{18}{12}\right)^2=\frac{17}{6}-\frac{9}{4} $

$ \frac{34-27}{12}=\frac{7}{12} $

$ 96 \sigma^2=96 \times \frac{7}{12}=56$

$ \mu=\frac{\alpha}{3}+\frac{1}{4}-\frac{3}{4} $

$ \mu=\frac{\alpha}{3}-\frac{1}{2} $

$ \sigma=\sqrt{\left(\alpha^2 \frac{1}{3}+\frac{1}{4}+9 \frac{1}{4}\right)-\left(\frac{\alpha}{3}-\frac{1}{2}\right)^2} $

$ \sigma=\sqrt{\frac{2 \alpha^2}{9}+\frac{\alpha}{3}+\frac{9}{4}} $

$ \sigma=\mu+2 $

$ \sigma^2=(\mu+2)^2 \Rightarrow \frac{2 \alpha^2}{9}+\frac{\alpha}{3}+\frac{9}{4}=\frac{\alpha^2}{9}+\frac{9}{4}+\alpha $

$ \frac{\alpha^2}{9}-\frac{2 \alpha}{3}=0 $

$ \alpha=0,(\text { reject }) \text { or } \alpha=6 $

$ (\because \mathrm{x}=0 \text { is already given) } $

$ \Rightarrow \sigma+\mu=2 \mu+2 $

$ =5$

$ 7 \overline{\mathrm{x}}=\frac{{ }^5 \mathrm{C}_1{ }^4 \mathrm{C}_2+{ }^5 \mathrm{C}_2 \cdot{ }^4 \mathrm{C}_1 \times 2+{ }^5 \mathrm{C}_3 \cdot{ }^4 \mathrm{C}_0 \times 3}{{ }^9 \mathrm{C}_3} \times 7 $

$ \frac{30+80+30}{84} \times 7 $

$ =\frac{140}{12}=\frac{70}{6}=\frac{35}{3}$

$4 \bar{y}=\frac{40+60+12}{84} \times 4=\frac{112}{21}=\frac{16}{3}$

$ b=P(X \geq 3)=\frac{5}{6} \times \frac{5}{6} \times \frac{1}{6}+\left(\frac{5}{6}\right)^3 \cdot \frac{1}{6}+\left(\frac{5}{6}\right)^4 \cdot \frac{1}{6}+\ldots \ldots $

$=\frac{\frac{25}{216}}{1-\frac{5}{6}}=\frac{25}{216} \times \frac{6}{1}=\frac{25}{36} $

$ P(x \geq 6)=\left(\frac{5}{6}\right)^5 \cdot \frac{1}{6}+\left(\frac{5}{6}\right)^6 \cdot \frac{1}{6}+\ldots \ldots .$

$P(X \geq 6)=\left(\frac{5}{6}\right)^5 \cdot \frac{1}{6}+\left(\frac{5}{6}\right)^6 \cdot \frac{1}{6}+\ldots \ldots $

$ =\frac{\left(\frac{5}{6}\right)^5 \cdot \frac{1}{6}}{1-\frac{5}{6}}=\left(\frac{5}{6}\right)^5 $

$ c=\frac{\left(\frac{5}{6}\right)^5}{\left(\frac{5}{6}\right)^3}=\frac{25}{36} $

$ \frac{b+c}{a}=\frac{\left(\frac{5}{6}\right)^2+\left(\frac{5}{6}\right)^2}{\left(\frac{5}{6}\right)^2 \cdot \frac{1}{6}}=12$

Let $\mathrm{X}$ be no of bad apples

Then $P(X=0)=\frac{{ }^{15} C_2}{{ }^{18} C_2}=\frac{105}{153}$

${P}(\mathrm{X}=1)=\frac{{ }^3 \mathrm{C}_1 \times{ }^{15} \mathrm{C}_1}{{ }^{18} \mathrm{C}_2}=\frac{45}{153}$

${P}(\mathrm{X}=2)=\frac{{ }^3 \mathrm{C}_2}{{ }^{18} \mathrm{C}_2}=\frac{3}{153}$

$E(X)=0 \times \frac{105}{153}+1 \times \frac{45}{153}+2 \times \frac{3}{153}=\frac{51}{153}$

$=\frac{1}{3}$

$ \operatorname{Var}(X)=E\left(X^2\right)-(E(X))^2 $

$ =0 \times \frac{105}{153}+1 \times \frac{45}{153}+4 \times \frac{3}{153}-\left(\frac{1}{3}\right)^2$

$ =\frac{57}{153}-\frac{1}{9}=\frac{40}{153}$

$ x=\text { number of matches that team wins } $

$ y=\text { number of matches that team loses } $

$ |x-y| \leq 2 \text { and } x+y=10 $

$ |x-y|=0,1,2 \quad x, y \in N$

Case $-I$ : $|x-y|=0 \Rightarrow x=y $

$ \because x+y=10 \Rightarrow x=5=y $

Case $-II$ : $|x-y|=1 \Rightarrow x-y= \pm 1$

Case $-III$ : $|x-y|=2 \Rightarrow x-y= \pm 2$

$ x-y=2 \quad \text { OR } \quad x-y=-2 $

$ \because \mathrm{x}+\mathrm{y}=10 \quad \because \mathrm{x}+\mathrm{y}=10 $

$ x=6, y=4 \quad x=4, y=6 $

$ \mathrm{P}(|\mathrm{x}-\mathrm{y}|=2)={ }^{10} \mathrm{C}_6\left(\frac{1}{3}\right)^6\left(\frac{2}{3}\right)^4+{ }^{10} \mathrm{C}_4\left(\frac{1}{3}\right)^4\left(\frac{2}{3}\right)^6 $

$ \mathrm{p}={ }^{10} \mathrm{C}_5 \frac{2^5}{3^{10}}+{ }^{10} \mathrm{C}_6 \frac{2^4}{3^{10}}+{ }^{10} \mathrm{C}_4 \frac{2^6}{3^{10}} $

$ 3^9 \mathrm{p}=\frac{1}{3}\left({ }^{10} \mathrm{C}_5 2^5+{ }^{10} \mathrm{C}_6 2^4+{ }^{10} \mathrm{C}_4 2^6\right) $

$ =8288 $

$ a, b, c \in\{1,2,3,4,5,6,7,8\} $

 Repeated roots $ D=0 $

$ \Rightarrow b^2-4 a c=0 \Rightarrow b^2=4 a c $

Prob $=\frac{8}{8 \times 8 \times 8}=\frac{1}{64} $

$ \Rightarrow(a, b, c) $

$ (1,2,1) ;(2,4,2) ;(1,4,4) ;(4,4,1) ;(3,6,3) ; $

$ (2,8,8) ;(8,8,2) ;(4,8,4) $

$ 8 $ case

$ \mathrm{~b}^2>4 \mathrm{ac} $

$ \mathrm{b}=3:(\mathrm{a}, \mathrm{c})=(1,1)(1,2)(2,1) $

$ \mathrm{b}=4:(\mathrm{a}, \mathrm{c})=(1,1)(1,2)(2,1)(1,3)(3,1) $

$ \mathrm{b}=5:(\mathrm{a}, \mathrm{c})=(1,1)(1,2)(2,1)(1,3)(3,1)(1,4)(4,1) $

$ (1,5)(5,1)(1,6)(6,1)(2,3)(3,2)(2,2) $

$ \mathrm{b}=6:(\mathrm{a}, \mathrm{c})=(1,1)(1,2)(2,1)(1,3)(3,1)(1,4)(4,1) $

$ (1,5)(5,1)(1,6)(6,1)(2,3)(3,2)(2,4)(4,2)(2,2) $

 fav. cases $=38 $

 Prob $. :  \frac{38}{6 \times 6 \times 6} $

$ A M>G M \Rightarrow x y \leq 144 $

$ x y \geq 108$

Favorable pairs of $(\mathrm{x}, \mathrm{y})$ are

$ (13,11),(12,12),(14,10),(15,9),(16,8), $

$ (17,7),(18,6),(6,18),(7,17),(8,16),(9,15), $

$ (10,14),(11,13)$

i.e. $13$ cases

Total choices for $x+y=24$ is 23

Probability $=\frac{13}{23}=\frac{\mathrm{m}}{\mathrm{n}}$

$\mathrm{n}-\mathrm{m}=10$

$Image$

Tetrahedral dice

$a x^2+b x+c=0$

has all real roots

$ \Rightarrow \mathrm{D} \geq 0 $

$ \Rightarrow \mathrm{b}^2-4 \mathrm{ac} \geq 0$

Let $b=1 \Rightarrow 1-4 a c \geq 0$ (Not feasible)

$ \mathrm{b}=2 \Rightarrow 4-4 \mathrm{ac} \geq 0 $

$ 1 \geq \mathrm{ac} \Rightarrow \mathrm{a}=1, \mathrm{c}=1, $

$ \mathrm{~b}=3 \Rightarrow 9-4 \mathrm{ac} \geq 0 $

$ \frac{9}{4} \geq \mathrm{ac} $

$ \Rightarrow \mathrm{a}=1, \mathrm{c}=1 $

$ \Rightarrow \mathrm{a}=1, \mathrm{c}=2 $

$ \Rightarrow \mathrm{a}=2, \mathrm{c}=1 $

$ \mathrm{~b}=4 \Rightarrow 16-4 \mathrm{ac} \geq 0 $

$ 4 \geq \mathrm{ac} $

$ \Rightarrow \mathrm{a}=1, \mathrm{c}=1 $

$ \Rightarrow \mathrm{a}=1, \mathrm{c}=2 \quad \Rightarrow \mathrm{a}=2, \mathrm{c}=1 $

$ \Rightarrow \mathrm{a}=1, \mathrm{c}=3 \quad \Rightarrow \mathrm{a}=3, \mathrm{c}=1 $

$ \Rightarrow \mathrm{a}=1, \mathrm{c}=4 \quad \Rightarrow \mathrm{a}=4, \mathrm{c}=1 $

$ \Rightarrow \mathrm{a}=2, \mathrm{c}=2$

$ \text { Probability }=\frac{12}{(4)(4)(4)}=\frac{3}{16}=\frac{m}{m} $

$ m+n=19$

$P ( T )=\frac{1}{4}$

$\text { Mean } \overline{ X }=\frac{3}{4}+\frac{3}{8}+3\left(\frac{1}{64}+\frac{3}{64}\right)$

$=\frac{3}{4}+\frac{3}{8}+\frac{3}{16}$

$=3\left(\frac{7}{16}\right)$

$=\frac{21}{16}$

$=\frac{25}{35}=\frac{5}{7}$

$P \left( E _1\right)=\frac{1}{4}$

$E _2: \text { non-smokers }$

$P \left( E _2\right)=\frac{3}{4}$

$E :$ diagnosed with lung cancer

$P \left( E / E _1\right)=\frac{27}{28}$

$P \left( E / E _2\right)=\frac{1}{28}$

$P \left( E _1 / E _2\right)=\frac{ P \left( E _1\right) P \left( E / E _1\right)}{ P ( E )}$

$=\frac{\frac{1}{4} \times \frac{27}{28}}{\frac{1}{4} \times \frac{27}{28}+\frac{3}{4} \times \frac{1}{28}}=\frac{27^9}{30_{10}}=\frac{9}{10}$

$K =9$

$P (\text { Defective/A })=\frac{3}{100}, P (\text { Defective } / B )=\frac{4}{100}, P (\text { Defective } / C )=\frac{2}{100}$

$P(E)=\frac{5 / 10^2 / 100}{\frac{2}{10} \times \frac{3}{100}+\frac{3}{10} \times \frac{4}{100}+\frac{5}{10} \times \frac{2}{100}}=\frac{10}{6+12+10}$

$=\frac{10}{28}$

$=\frac{5}{14}$

$\sigma^2=\sum x^2 P(x)-\mu^2$

$\sigma^2+\mu^2=0+\frac{1}{2}+\frac{12}{10}+\frac{9}{30}=2$

$10\left(\sigma^2+\mu^2\right)=20 \text { Ans. }$

$\frac{ n }{9}=\left(\frac{ n -1}{ n }\right)\left(1+2\left(\frac{1}{ n }\right)+3\left(\frac{1}{ n }\right)^2 \ldots \ldots\right)$

$\frac{ n }{9}=\left(\frac{ n -1}{ n }\right)\left(1-\frac{1}{ n }\right)^{-2}=\left(\frac{ n -1}{ n }\right) \cdot \frac{ n ^2}{( n -1)^2}$

$\frac{ n }{9}=\frac{ n }{ n -1} \Rightarrow n =10$

$n p(1+q)=5, n^2 p^2 q=6$

$n^2 p^2(1+q)^2=25, n^2 p^2 q=6$

$\frac{6}{q}(1+q)^2=25$

$6 q^2+12 q+6=25 q$

$6 q^2-13 q+6=0$

$6 q^2-9 q-4 q+6=0$

$(3 q-2)(2 q-3)=0$

$q=\frac{2}{3}, \frac{3}{2}, q=\frac{2}{3} \text { is accepted }$

$p=\frac{1}{3} \Rightarrow n \cdot \frac{1}{3}+n \cdot \frac{1}{3} \cdot \frac{2}{3}=5$

$\frac{3 n+2 n}{9}=5$

$n=9$

So $6(n+p-q)=6\left(9+\frac{1}{3}-\frac{2}{3}\right)=52$

$\Rightarrow np ^2=1$

$2^{ n } C _2 p ^2 q ^{ n-2 }=3^{ n } C _1 p q^{n-1}$

$\Rightarrow n p-p=3 q \quad(\therefore q=1-p)$

$\Rightarrow p =\frac{1}{2}$

Hence $n=4$

$P(x > 1)=1-(p(x=0)+p(x=1)$

$=1-\left({ }^4 C _0\left(\frac{1}{2}\right)^4+{ }^4 C_1\left(\frac{1}{2}\right)^1\left(\frac{1}{2}\right)^3\right)=\frac{11}{16}$

Taking $D _{(15)}$ as $\frac{15 \text { ! }}{e}$

$D _{(14)}$ as $\frac{14 !}{e}$

$D _{(13)}$ as $\frac{13 !}{e}$

We get, $1-\left(\frac{\frac{15 !}{e}+15 \cdot \frac{14 !}{e}+\frac{15 \times 14}{2} \times \frac{13 !}{e}}{15 !}\right)$

$=1-\left(\frac{1}{e}+\frac{1}{e}+\frac{1}{2 e}\right)=1-\frac{5}{2 e} \approx .08$

As $\sum \limits_{ k =1}^{\infty} P \left( w _{ k }\right)=1 \Rightarrow \frac{\lambda}{1-\frac{1}{2}}=1 \Rightarrow \lambda=\frac{1}{2}$

So, $P \left( w _{ n }\right)=\frac{1}{2^{ n }}$

$A =\{2 k +3 \ell ; k , \ell \in N \}=\{5,7,8,9,10 \ldots .\}$

$B =\left\{ w _{ n }: n \in A \right\}$

$B =\left\{ w _5, w _7, w _8, w _9, w _{10}, w _{11}, \ldots .\right\}$

$A = N -\{1,2,3,4,6\}$

$\therefore P ( B )=1-\left[ P \left( w _1\right)+ P \left( w _2\right)+ P \left( w _3\right)+ P \left( w _4\right)+ P \left( w _6\right)\right]$

$=1-\left[\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{64}\right]$

$=1-\frac{32+16+8+4+1}{64}=\frac{3}{64}$

$k \left(1+2 \cdot 3^{-1}+3 \cdot 3^{-2}+4 \cdot 3^{-3}+\ldots \infty\right)=1$

$\text { Let } \quad s =1+\frac{2}{3}+\frac{3}{3^2}+\frac{4}{3^3}+\ldots \infty$

$\frac{s}{3}=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^2}+\ldots \infty$

$\frac{2 s}{3}=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^2}+\ldots \infty$

$\frac{2 s}{3}-\frac{1}{1-\frac{1}{3}}=\frac{3}{2}$

$s=\frac{9}{4}$

$k=\frac{4}{9}$

$P(x \geq 2)=1-P(x=0)-P(x=1)$

$=1-\frac{4}{9}\left(1+\frac{2}{3}\right)$

$=\frac{7}{27}$

$D =25 N ^2-256 < 0$

$\Rightarrow N ^2 < \frac{256}{25} \Rightarrow N < \frac{16}{5}$

$\therefore N =1,2,3$

$\therefore \text { Probability }=\frac{1}{4}+\frac{3}{4} \times \frac{1}{4}+\frac{3}{4} \times \frac{3}{4} \times \frac{1}{4}=\frac{37}{64}$

$\therefore q - p =27$

$=\frac{105}{18}=\frac{35}{6}$

$\mu=\sum(x)=0$ as data is symmetric

$\sigma^2=\sum\left( x ^2\right)=\sum x ^2 P ( x )=\frac{35}{6} P =35=5 \times 7$

$\text { Sum of divisors }=\left(5^0+5^1\right)\left(7^0+7^1\right)=6 \times 8=48$

$\Rightarrow 4 \leq N \leq 12$

Probability that $2^N \geq N$ !

$\text { Now } P(N=2)+P(N=3)=\frac{1}{36}+\frac{2}{36}=\frac{3}{36}=\frac{1}{12}$

$\text { Required probability }=1-\frac{1}{12}=\frac{11}{12}=\frac{m}{n}$

$4 m -3 n =8$

$B :$ Red ball transferred

$P\left(\frac{B}{A}\right)=\frac{P(A \cap B)}{P(A)}$

$= \frac{3}{\frac{3}{9} \times \frac{5}{10}+\frac{4}{9} \times \frac{5}{10}+\frac{6}{9} \times \frac{5}{10}}$

$= \frac{15}{15+24+15}=\frac{15}{54}=\frac{5}{18}$

$=\frac{ P (1< x \leq 2)}{ P ( x \leq 2)}=\frac{ P ( x =2)}{ P ( x \leq 2)}$

$=\frac{4 k }{ k +2 k +4 k }$

$=\frac{4}{7}$