Probability

Given, mean $ = np = \alpha $.

and variance $ = npq = {\alpha \over 3}$

$ \Rightarrow q = {1 \over 3}$ and $p = {2 \over 3}$

$P(X = 1) = n.{p^1}.{q^{n - 1}} = {4 \over {243}}$

$ \Rightarrow n.{2 \over 3}.{\left( {{1 \over 3}} \right)^{n - 1}} = {4 \over {243}}$

$ \Rightarrow n = 6$

$P(X = 4\,\mathrm{or}\,5) = {}^6{C_4}\,.\,{\left( {{2 \over 3}} \right)^4}\,.\,{\left( {{1 \over 3}} \right)^2} + {}^6{C_5}\,.\,{\left( {{2 \over 5}} \right)^5}\,.\,{1 \over 3}$

$ = {{16} \over {27}}$

$0 \le {{2 + 3P} \over 6} \le 1 \Rightarrow P \in \left[ { - {2 \over 3},{4 \over 3}} \right]$

$0 \le {{2 - P} \over 8} \le 1 \Rightarrow P \in [ - 6,2]$

$0 \le {{1 - P} \over 2} \le 1 \Rightarrow P \in [ - 1,1]$

$0 < P({E_1}) + P({E_2}) + P({E_3}) \le 1$

$0 < {{13} \over {12}} - {P \over 8} \le 1$

$P \in \left[ {{2 \over 3},{{26} \over 3}} \right]$

Taking intersection of all

$P \in \left[ {{2 \over 3},1} \right)$

${P_1} + {P_2} = {5 \over 3}$

$P(A) = {1 \over 3},\,P(B) = {1 \over 5}$ and $P\left( {A \cup B} \right) = {1 \over 2}$

$\therefore$ $P\left( {A \cap B} \right) = {1 \over 3} + {1 \over 5} - {1 \over 2} = {1 \over {30}}$

Now, $P\left( {A|B'} \right) + P\left( {B|A'} \right) = {{P\left( {A \cap B'} \right)} \over {P\left( {B'} \right)}} + {{P\left( {B \cap A'} \right)} \over {P\left( {A'} \right)}}$

$ = {{{9 \over {30}}} \over {{4 \over 5}}} + {{{5 \over {30}}} \over {{2 \over 3}}} = {5 \over 8}$

Given,

$n = 5$

and $P(X = 0) = P(X = 1)$

We know, $p(x = r) = {}^n{C_r}\,.\,{p^r}\,.\,{q^{n - r}}$

where $p + q = 1$

$\therefore$ $P(X = 0) = P(X = 1)$

$ \Rightarrow {}^5{C_0}\,.\,{p^0}\,.\,{q^5} = {}^5{C_1}\,.\,{p^1}\,.\,{q^4}$

$ \Rightarrow 1\,.\,1\,.\,{(1 - p)^5} = 5\,.\,p\,.\,{(1 - p)^4}$

$ \Rightarrow 1 - p = 5p$

$ \Rightarrow 6p = 1$

$ \Rightarrow p = {1 \over 6}$

$\therefore$ $q = 1 - p = 1 - {1 \over 6} = {5 \over 6}$

Now, ${{P(X = 2)} \over {P(X = 3)}} = {{{}^5{C_2}\,.\,{p^2}\,.\,{q^3}} \over {{}^5{C_3}\,.\,{p^3}\,.\,{q^2}}}$

$ = {{10\,.\,q} \over {10\,.\,p}}$

$ = {5 \over 6} \times {6 \over 1}$

$ = 5$

First 10 prime numbers are

={2, 3, 5, 7, 11, 13, 17, 19, 23, 29}

Let A is a 2 $\times$ 2 matrix,

$A = \left[ {\matrix{ a & b \cr c & d \cr } } \right]$

Given that matrix A is singular.

$\therefore$ | A | = 0

$ \Rightarrow \left| {\matrix{ a & b \cr c & d \cr } } \right| = 0$

$ \Rightarrow ad = bc$

Case I :

ad = bc condition satisfy when a = b = c = d.

For ex when a = 2, b = 2, c = 2, d = 2, then ad = bc satisfy.

Now there are 10 prime numbers.

We can choose any one of the 10 prime number in ${}^{10}C_{1}$ = 10 ways and put them in the four positions of the matrix and matrix will be singular.

$\therefore$ In this case, total favorable case = 10

Case 2 :

ad = bc condition satisfies when

(1)

a = 2, d - 3 then

(a) b = 2, c = 3

(b) b = 3, c = 2

or

a = 3, d = 2 then

(a) b = 2, c = 3

(b) b = 3, c = 2

So you can see for two different prime number for a and d there are 4 possible value of b and c which satisfy ad = bc condition.

Two different values of a and d can be chosen from 10 prime numbers = ${}^{10}C_{2}$ ways

And for each combination of a and d there are 4 possible values of b and c.

$\therefore$ Total possible values = ${}^{10}C_{2}$ $\times$ 4

From case I and case II total possible values of 10 prime numbers which satisfy ad = bc condition

= 10 + ${}^{10}C_{2}$ $\times$ 4

For sample space,

Number of ways to fill element a of matrix A = chose any prime number among 10 available prime number = ${}^{10}C_{1}$ ways

Similarly,

For element b of matrix A = ${}^{10}C_{1}$ ways

For element c of matrix A = ${}^{10}C_{1}$ ways

For element d of matrix A = ${}^{10}C_{1}$ ways

$\therefore$ Sample space = ${}^{10}C_{1}$ $\times$ ${}^{10}C_{1}$ $\times$ ${}^{10}C_{1}$ $\times$ ${}^{10}C_{1}$ = 10⁴

$\therefore$ Probability $ = {{10 + {{}^{10}C_{2}} \times 4} \over {{{10}^4}}}$

$ = {{10 + 180} \over {{{10}^4}}}$

$ = {{190} \over {{{10}^4}}}$

$ = {{19} \over {{{10}^3}}}$

The required probability

$ = {\text{Area of Region PQCAP} \over \text{Area of Region ABCA}}$

$ = {{{1 \over 2} \times 8 \times 6 - {1 \over 2} \times 2 \times 4} \over {{1 \over 2} \times 8 \times 6}}$

$ = {5 \over 6}$

Given P(X = 3) = 5P(X = 4) and n = 7

$ \Rightarrow {}^7{C_3}{p^3}{q^4} = 5\,.\, \Rightarrow {}^7{C_4}{p^4}{q^3}$

$ \Rightarrow q = 5p$ and also $p + q = 1$

$ \Rightarrow p = {1 \over 6}$ and $q = {5 \over 6}$

Mean $ = {7 \over 6}$ and variance $ = {{35} \over {36}}$

Mean + Variance $ = {7 \over 6} + {{35} \over {36}} = {{77} \over {36}}$

Coin is tossed 5 times, so n = 5

Let, p = probability of getting heads

q = probability of getting tails.

$\therefore$ p + q = 1 ...... (1)

$\therefore$ Probability of getting 4 heads

= ⁵C₄ . p⁴ . q

And probability of getting 5 heads

= ⁵C₅ . p⁵

Given, ⁵C₄ . p⁴ . q = ⁵C₅ . p⁵

$\Rightarrow$ 5q = p ....... (2)

From equation (1) and (2), we get,

5q + q = 1

$\Rightarrow$ 6q = 1

$\Rightarrow$ q = ${1 \over 6}$

$\therefore$ p = 1 $-$ ${1 \over 6}$ = ${5 \over 6}$

Now, probability of getting atmost two heads

= p (x = 0) + p (x = 1) + p (x = 2)

p (x = 0) = Getting zero head in 5 trials

= ⁵C₀ . p⁰ . q⁵

p (x = 1) = Getting one head in 5 trials

= ⁵C₁ . p¹ . q⁴

p (x = 2) = Getting two heads in 5 trials

= ⁵C₂ . p² . q³

= ⁵C₀ . q⁵ + ⁵C₁ . pq⁴ + ⁵C₂ . p²q³

$ = {\left( {{1 \over 6}} \right)^5} + 5\,.\,{5 \over 6}\,.\,{\left( {{1 \over 6}} \right)^4} + 10\,.\,{\left( {{5 \over 6}} \right)^2}\,.\,{\left( {{1 \over 6}} \right)^3}$

$ = {{1 + 25 + 250} \over {{6^5}}} = {{276} \over {{6^5}}}$ = ${{46} \over {{6^4}}}$

There are only two ways to get sum 48, which are (32, 8, 8) and (16, 16, 16)

So, required probability

$ = 3\left( {{2 \over {32}}\,.\,{1 \over 8}\,.\,{1 \over 8}} \right) + \left( {{1 \over {16}}\,.\,{1 \over {16}}\,.\,{1 \over {16}}} \right)$

$ = {3 \over {{2^{10}}}} + {1 \over {{2^{12}}}}$

$ = {{13} \over {{2^{12}}}}$

$P\left( {{{{E_1}} \over {{E_2}}}} \right) = {1 \over 2} \Rightarrow {{P({E_1} \cap {E_2})} \over {P({E_2})}} = {1 \over 2}$

$P\left( {{{{E_2}} \over {{E_1}}}} \right) = {3 \over 4} \Rightarrow {{P({E_2} \cap {E_1})} \over {P({E_1})}} = {3 \over 4}$

$P({E_1} \cap {E_2}) = {1 \over 8}$

$P({E_2}) = {1 \over 4},\,P({E_1}) = {1 \over 6}$

(A) $P({E_1} \cap {E_2}) = {1 \over 8}$ and $P({E_1})\,.\,P({E_2}) = {1 \over {24}}$

$ \Rightarrow P({E_1} \cap {E_2}) \ne P({E_1})\,.\,P({E_2})$

(B) $P(E{'_1} \cap E{'_2}) = 1 - P({E_1} \cup {E_2})$

$ = 1 - \left[ {{1 \over 4} + {1 \over 6} - {1 \over 8}} \right] = {{17} \over {24}}$

$P(E{'_1}) = {3 \over 4} \Rightarrow P(E{'_1})P({E_2}) = {3 \over {24}}$

$ \Rightarrow P(E{'_1} \cap E{'_2}) \ne P(E{'_1})\,.\,P({E_2})$

(C) $P({E_1} \cap E{'_2}) = P({E_1}) - P({E_1} \cap {E_2})$

$ = {1 \over 6} - {1 \over 8} = {1 \over {24}}$

$P({E_1})\,.\,P({E_2}) = {1 \over {24}}$

$ \Rightarrow P({E_1} \cap E{'_2}) = P({E_1})\,.\,P({E_2})$

(D) $P(E{'_1} \cap {E_2}) = P({E_2}) - P({E_1} \cap {E_2})$

$ = {1 \over 4} - {1 \over 8} = {1 \over 8}$

$P({E_1})P({E_2}) = {1 \over {24}}$

$ \Rightarrow P(E{'_1} \cap {E_2}) \ne P({E_1})\,.\,P({E_2})$

$\because$ x is a random variable

$\therefore$ $k + 2k + 4k + 6k + 8k = 1$

$\therefore$ $k = {1 \over {21}}$

Now, $P(1 < x < 4\,|\,x \le 2) = {{4k} \over {7k}} = {4 \over 7}$

Two cases

(i) A spade king + a spade card Total ways $=12$

(ii) A non spade king + a spade card Total ways $=3 \times 13$

$ \begin{aligned} \therefore \text { Required probability } & =\frac{3 \times 13+12}{{ }^{52} C_2} \\ & =\frac{51}{52 \times 51} \times 2=\frac{2}{52}=\frac{1}{26} \end{aligned} $

$P_1=P$ (first card is queen) $\cdot P$ (second card is diamond)

$ =\frac{4}{52} \times \frac{13}{52}=\frac{1}{52} $

$P_2$ : Case $I$ queen is not diamond

$ \rightarrow \frac{3}{52} \times \frac{13}{51} $

Case II queen is diamond $\rightarrow \frac{1}{52} \times \frac{12}{51}$

$ \begin{aligned} \therefore \quad P_2 & =\frac{3}{52} \times \frac{13}{51}+\frac{1}{52} \times \frac{12}{51} \\ & =\frac{39+12}{52 \times 51}=\frac{51}{52 \times 51}=\frac{1}{52} \end{aligned} $

Therefore, $\frac{P_1}{P_2}=\frac{1 / 52}{1 / 52}=1$

Let $E_1=$ Event that the ball is drawn from bag $A$.

$E_2=$ Event that the ball is drawn from bag B.

$E_3=$ Event that the ball is drawn from bag $C$. .

Let $E$ be the event that the ball drawn is black.

$ \begin{aligned} P\left(E_1\right)= & P\left(E_2\right)=P\left(E_3\right)=\frac{1}{3} \\ P\left(\frac{E}{E_1}\right)= & \frac{2}{6}, P\left(\frac{E}{E_2}\right)=\frac{3}{6}, P\left(\frac{E}{E_3}\right)=\frac{4}{6} \\ P(E)= & P\left(E_1\right) \cdot P\left(\frac{E}{E_1}\right)+P\left(E_2\right) \cdot P\left(\frac{E}{E_2}\right) \\ & +P\left(E_3\right) \cdot P\left(\frac{E}{E_3}\right) \\ = & \frac{1}{3} \times \frac{2}{6}+\frac{1}{3} \times \frac{3}{6}+\frac{1}{3} \times \frac{4}{6} \\ = & \frac{2+3+4}{18}=\frac{9}{18}=\frac{1}{2} \end{aligned} $

$ \begin{array}{lccccccccc} \hline X=x_i & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\ \hline P\left(x=x_i\right) & 10 k & 9 k & 8 k & 8 k & 6 k & 5 k & 4 k & 3 k & k \\ \hline \end{array} $

$ \begin{aligned} &\text { We know, } \Sigma P=1\\ &\begin{aligned} \Rightarrow 10 k+9 k+8 k+8 k & +6 k+5 k +4 k+3 k+k=1 \end{aligned} \end{aligned} $

$ \begin{aligned} & \Rightarrow \quad 54 k=1 \Rightarrow k=\frac{1}{54} \\ & A=\left\{x_i \mid x_i\right. \text { is a prime number } \\ & B=\left\{x_i \mid x_i>5\right\} \\ & P(A)=9 k+8 k+6 k+4 k \\ &(\because 2,3,5,7 \text { are prime numbers }) \\ &=27 \times \frac{1}{54}=\frac{27}{54} \\ & P(B)=5 k+4 k+3 k+k \\ &(\because 6,7,8,9 \text { are more than } 5) \\ &=13 \times \frac{1}{54}=\frac{13}{54} \end{aligned} $

$ P(A \cap B)=4 k $

( $\because 7$ is only prime which is more than 5 in the, given problem)

$ =4 \times \frac{1}{54}=\frac{4}{54} $

$ \begin{aligned} P(A \cup B) & =P(A)+P(B)-P(A \cap B) \\ & =\frac{27+13-4}{54} \\ & =\frac{36}{54}=\frac{2}{3} \end{aligned} $

$ \begin{aligned} &\\ &\begin{aligned} & \text { } \frac{5}{3} k=P(X=2)=P(X=3) \\ & P(X=x)=\frac{\lambda^x e^{-\lambda}}{x!} \\ & P(X=2)=P(X=3) \\ & \Rightarrow \quad \frac{\lambda^2 e^{-\lambda}}{2!}=\frac{\lambda^3 e^{-\lambda}}{3!} \Rightarrow \frac{1}{2}=\frac{\lambda}{6} \\ &\Rightarrow \lambda=3 \\ & \text { Now, } P\left(X=9=\frac{\lambda^5 e^{-\lambda}}{5!}\right. \\ & =\lambda^3\left(\frac{\lambda^2 e^{-\lambda}}{2!}\right) \cdot \frac{2!}{5!} \\ & =(3)^3\left(\frac{5}{3} k\right) \cdot \frac{2 \times 1}{5 \times 4 \times 3 \times 2 \times 1} \\ & =\frac{3}{4} k \end{aligned} \end{aligned} $

9 identical black balls numbered from 1 to 9

4 identical white balls numbered from 1 to 4

3 balls are drawn

$P$ (at least 1 white ball)

$ \begin{aligned} & =P(1 W, 2 R)+P(2 W, 1 R)+P(3 W, 0 R) \\ & =\frac{\left(4{ }^4 C_1 \times{ }^9 C_2\right)+\left({ }^4 C_2 \times{ }^9 C_1\right)+\left({ }^4 C_3 \times{ }^9 C_0\right)}{{ }^{13} C_3} \\ & =\frac{(4 \times 36)+(6 \times 9)+(4 \times 1)}{\frac{13 \times 12 \times 11}{3 \times 2}}=\frac{101}{143} \end{aligned} $

Let $P(A)=\frac{1}{4}, P(B)=\frac{1}{5}$ probability of target hit

$ \begin{aligned} & =P(A) \cdot P(\bar{B})+P(\bar{A}) \cdot P(B)+P(A) \cdot P(B) \\ & =\frac{1}{4} \times \frac{4}{5}+\frac{3}{4} \times \frac{1}{5}+\frac{1}{4} \times \frac{1}{5}=\frac{8}{20}=\frac{2}{5} \end{aligned} $

3 dice are thrown

$A=$ Sum of the number appeared to be 15

Triplets $(6,6,3),(6,3,6),(3,6,6),(6,5,4) (6,4,5),(4,6,5),(5,6,4),(4,5,6),(5,4,6)$, $(5,5,5)$

$B=$ The number 5 does not appear on any of the dice

$ \begin{aligned} & =\{(6,6,3),(6,3,6),(3,6,6)\} \\ \therefore \quad P\left(\frac{B}{A}\right) & =\frac{n(A \cap B)}{n(A)}=\frac{3}{10} \end{aligned} $

We know that sum of total probability is 1 .

$ \begin{array}{ll} \Rightarrow & 0+k+2 k+5 k^2+2 k^2+3 k=1 \\ \Rightarrow & 7 k^2+6 k-1=0 \\ \Rightarrow & 7 k^2+7 k-k-1=0 \\ \Rightarrow & (7 k-1)(k+1)=0 \Rightarrow k=\frac{1}{7}, k \neq-1 \end{array} $

Mean of

$ \begin{aligned} X & =0+2 k+8 k+30 k^2+16 k^2+30 k \\ & =40 k+46 k^2=k(40+46 k) \\ & =\frac{1}{7}\left(40+\frac{46}{7}\right)=\frac{1}{7} \times \frac{326}{7}=\frac{326}{49} \end{aligned} $

Let probability of success is $p$ and that of failure is $q$.

Given, $p=5 q$

We know that, $p+q=1 \Rightarrow 5 q+q=1$

$ \Rightarrow \quad q=\frac{1}{6}, p=\frac{5}{6} $

p(atmost one success)

$ \begin{aligned} & =p(0 \text { success })+p(1 \text { success }) \\ & ={ }^5 C_0 p^0 q^5+{ }^5 C_1 p^1 q^4 \\ & =1 \times\left(\frac{5}{6}\right)^0 \times\left(\frac{1}{6}\right)^5+5 \times \frac{5}{6} \times\left(\frac{1}{6}\right)^4=\frac{26}{6^5} \end{aligned} $

There are total 9 balls in which 3 balls are white and 6 balls are red.

Total number of observation $={ }^9 C_4$

Number of favourable outcomes $=(2$

Red 2 Black

$ \begin{gathered} +3 \text { Red } \times 1 \text { Black }+4 \text { Red) balls } \\ ={ }^6 C_2 \times{ }^3 C_2+{ }^6 C_3 \times{ }^3 C_1+{ }^6 C_4 \end{gathered} $

$P$ (atleast two red balls)

$ \begin{aligned} & =\frac{{ }^6 C_2 \times{ }^3 C_2+{ }^6 C_3 \times{ }^3 C_1+{ }^6 C_4}{{ }^9 C_4} \\ & =\frac{15 \times 3+20 \times 3+15}{126} \\ & =\frac{45+60+15}{126}=\frac{120}{126}=\frac{60}{63} \end{aligned} $

$P(A)=\frac{2}{5}$ and $P(B)=\frac{1}{3}$

$A$ and $B$ are independent events,

$ \begin{aligned} P(A \cap B) & =P(A) \times P(B) \\ & =\frac{2}{5} \times \frac{1}{3}=\frac{2}{15} \\ P(A-B) & =P(A)-P(A \cap B) \\ & =\frac{2}{5}-\frac{2}{15}=\frac{4}{15} \end{aligned} $

$ \begin{aligned} P(\bar{A} \cup B) & =1-P(A-B)=1-\frac{4}{15}=\frac{11}{15} \\ P(A \cup B) & =P(A)+P(B)-P(A \cap B) \\ & =\frac{2}{5}+\frac{1}{3}-\frac{2}{15} \\ & =\frac{6+5-2}{15}=\frac{3}{5} \end{aligned} $

$ \begin{aligned} P\left(\frac{A}{\bar{B}}\right) & =\frac{P(A \cap \bar{B})}{P(\bar{B})}=\frac{P(A-B)}{1-P(B)} \\ & =\frac{4}{15}+\left(1-\frac{1}{3}\right)=\frac{4}{15} \times \frac{3}{2}=\frac{2}{5} \\ P\left(\frac{\bar{B}}{A}\right) & =\frac{P(\bar{B} \cap A)}{P(A)}=\frac{P(A-B)}{P(A)} \\ & =\frac{4}{15}+\frac{2}{5}=\frac{2}{3} \end{aligned} $

The total possible outcomes are (H1, H2, H3, H4, H5, H6, T1, T2, T3, T4, T5, T6).

∴ Total number of outcomes $=12$

$ \begin{aligned} & \text { Probability of winning }=\frac{1}{12} \\ & \text { and Probability of losing }=\frac{11}{12} \end{aligned} $

$P$ (B wins)

$ \begin{aligned} = & P(\bar{A} B+\overline{A B A} B+\overline{A B A B A} B+\ldots) \\ = & \frac{11}{12} \times \frac{1}{12}+\frac{11}{12} \times \frac{11}{12} \times \frac{11}{12} \times \frac{1}{12} \\ & +\frac{11}{12} \times \frac{11}{12} \times \frac{11}{12} \times \frac{11}{12} \times \frac{11}{12} \times \frac{1}{12}+\ldots \infty \\ = & \frac{11}{144}\left[1+\left(\frac{11}{12}\right)^2+\left(\frac{11}{12}\right)^4+\ldots \infty\right] \end{aligned} $

which form a GP.

$ =\frac{11}{144}\left[\frac{1}{1-\frac{121}{144}}\right]=\frac{11}{144} \times \frac{144}{23}=\frac{11}{23} $

When two dices are thrown there are total 36 outcomes.

Here, $X$ denotes the sum of the number that shows the faces of dice.

The sum of two dice can be

$ 2,3,4,5,6,7,8,9,10,11,12 $

$ \begin{aligned} & P(X=2)=\frac{1}{36} \\ & P(X=3)=\frac{2}{36} \\ & P(X=4)=\frac{3}{36} \\ & P(X=5)=\frac{4}{36} \\ & P(X=6)=\frac{5}{36} \\ & P(X=7)=\frac{6}{36} \\ & P(X=8)=\frac{5}{36} \\ & P(X=9)=\frac{4}{36} \\ & P(X=10)=\frac{3}{36} \\ & P(X=11)=\frac{2}{36} \\ & P(X=12)=\frac{1}{36} \end{aligned} $

$ \begin{aligned} & \text { Mean }=\Sigma x_i p_i \\ & \begin{aligned} =\frac{2 \times 1}{36}+ & \frac{3 \times 2}{36}+\frac{4 \times 3}{36}+\frac{5 \times 4}{36}+\frac{6 \times 5}{36} \\ & +\frac{7 \times 6}{36}+\frac{8 \times 5}{36}+\frac{9 \times 4}{36}+\frac{10 \times 3}{36} \\ & +\frac{11 \times 2}{36}+\frac{12 \times 1}{36} \\ = & \frac{2+6+12+20+30+42+40+36+30+22+12}{36} \\ = & \frac{252}{36}=7 \end{aligned} \end{aligned} $

Let $p=P$ (chosen an engineering student) $=\frac{1}{5}$

$q=p$ (not chosen an engineering student)

$ \begin{aligned} & =1-\frac{1}{5}=\frac{4}{5} \\ P(X \leq 2) & =P(X=0)+P(X=1)+P(X=2) \\ & ={ }^8 C_0\left(\frac{1}{5}\right)^0\left(\frac{4}{5}\right)^8+{ }^8 C_1\left(\frac{1}{5}\right)^1\left(\frac{4}{5}\right)^7 +{ }^8 C_2\left(\frac{1}{5}\right)^2\left(\frac{4}{5}\right)^6 \\ & =\frac{4^6}{5^8}[16+32+28] \\ & =\frac{4^6}{5^8} \times 76=19 \times \frac{4^7}{5^8} \end{aligned} $

The total number of outcomes is 36.

Favourable outcomes $=(1,1)(1,2)$, $(1,3)(1,4),(2,1),(2,2),(2,3),(2,4)$, $(3,1),(3,2),(3,3),(3,4),(4,1)(4,2) (4,3)$.

Number of favourable outcomes $=15 P\left(\right.$ ordered pair such that $\left.x^2+y^2 \leq 25\right) =\frac{15}{36}=\frac{5}{12}$

There are total 8 cards, which are multiples of 5 i.e. 4 cards of 5 and 4 cards of 10 .

There are total 16 cards of prime number i.e. 4 cards each of $2,3,5$ and 7.

Case I When 5 number cards are selected from multiple of 5 .

Number of ways of selecting the card

$ ={ }^4 C_1 \times{ }^{15} C_1=4 \times 15=60 $

Case II When 10 number cards are selected from multiple of 5 , number of ways of selecting the card $={ }^4 C_1 \times{ }^{16} C_1$

$ =4 \times 16=64 $

$\therefore P$ (card having prime number and a card having a number which is multiple of 5)

$ \begin{aligned} & =\frac{60+64}{{ }^{52} C_2}=\frac{124 \times 2}{52 \times 51} \\ & =\frac{62}{663} \end{aligned} $

$ \begin{aligned} & \text {Given, } P(\bar{A})=\frac{2}{3} \\ & \Rightarrow \quad P(A)=\frac{1}{3} \Rightarrow P(B)=\frac{4}{15} \\ & \Rightarrow \quad P(\bar{B})=\frac{11}{15} \end{aligned} $

$ \begin{aligned} & \text { Given, } P(A \cap \bar{B})=\frac{1}{5} \\ & \therefore P(A)-P(A \cap B)=\frac{1}{5} \\ & P(A \cap B)=\frac{1}{3}-\frac{1}{5}=\frac{2}{15} \end{aligned} $

$ \begin{aligned} &\begin{aligned} P(A \cup B) & =P(A)+P(B)-P(A \cap B) \\ & =\frac{1}{3}+\frac{4}{15}-\frac{2}{15}=\frac{7}{15} \end{aligned}\\ &\begin{aligned} P(A \cup \bar{B}) & =1-P(B-A) \\ & =1-P(B)+P(A \cap B) \\ & =1-\frac{4}{15}+\frac{2}{15}=\frac{13}{15} \end{aligned}\\ &\begin{aligned} P(B \mid A \cup \bar{B}) & =\frac{P(B \cap(A \cup \bar{B}))}{P(A \cup \bar{B})} \\ & =\frac{P[(B \cap A) \cup(B \cap \bar{B})}{P(A \cup \bar{B})} \\ & =\frac{P(B \cap A)}{P(A \cup \bar{B})}=\frac{2}{13} \end{aligned}\\ &\text { So, } \sqrt{195[P(B \mid A \cup \bar{B})+P(A \cup B)]}\\ &\begin{aligned} & =\sqrt{195\left[\frac{2}{13}+\frac{7}{15}\right]}=\sqrt{195\left[\frac{30+91}{195}\right]} \\ & =\sqrt{121}=11 \end{aligned} \end{aligned} $

Given, $P(X=r)=k(1+r) \times 3^{-r}$

$ \begin{gathered} P(X=0)+P(X=1)+P(X=2)+\ldots=1 \\ \frac{k(1+0)}{3^0}+\frac{k(1+1)}{3}+\frac{k(1+2)}{3^2}+\ldots=1 \\ k+\frac{2 k}{3}+\frac{3 k}{3^2}+\ldots=1 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{gathered} $

Multiply by 3, we get

$ 3 k+2 k+\frac{3 k}{3}+\frac{4 k}{3^2}+\frac{5 k}{3^3}+\ldots=3 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

$ \begin{aligned} &\text { Subtracting Eq. (i) from Eq. (ii), }\\ &\begin{gathered} 3 k+k+\frac{k}{3}+\frac{k}{3^2}+\ldots=2 \\ 3 k+\frac{k}{1-\frac{1}{3}}=2 \\ {\left[\because k+\frac{k}{3}+\frac{k}{3^2}+\ldots \text { form a GP }\right]} \\ 3 k+\frac{3 k}{2}=2 \\ \Rightarrow\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{9 k}{2}=2 \Rightarrow k=\frac{4}{9} \\ \quad P(X=0)+P(X=1)+P(X=2) \\ k+\frac{2 k}{3}+\frac{3 k}{3^2} \\ k\left[1+\frac{2}{3}+\frac{1}{3}\right] \\ \frac{4}{9}[1+1]=\frac{8}{9} \end{gathered} \end{aligned} $

Let $p=\frac{\alpha}{\beta}, q=1-\frac{\alpha}{\beta}=\frac{\beta-\alpha}{\beta}$

$P$ (he fails at least $n-1$ times)

$ \begin{aligned} = & P(X=0)+P(X=1) \\ = & { }^n C_0\left(\frac{\alpha}{\beta}\right)^0\left(\frac{\beta-\alpha}{\beta}\right)^n +{ }^n C_1\left(\frac{\alpha}{\beta}\right)\left(\frac{\beta-\alpha}{\beta}\right)^{n-1} \\ = & \frac{(\beta-\alpha)^{n-1}}{\beta^n} \quad[\beta-\alpha+n \alpha] \end{aligned} $

Getting prime on one die $=1 / 2$ Getting composite on one die =1/3 ∴ Probability of getting prime on one die and composite on the other

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

As, $A, B$ and $C$ are pairwise independent, $\bar{B} \cap \bar{C}$ and $A$ are independent.

$ \begin{aligned} P\left(\frac{\bar{B} \cap \bar{C}}{A}\right) & =\frac{P(\bar{B} \cap \bar{C}) \cap A}{P(A)} \\ & =\frac{P(\bar{B} \cap \bar{C}) P(A)}{P(A)} \\ & =P(\bar{B} \cap \bar{C}) \end{aligned} $

$ \begin{aligned} & \text { Now, } P(\overline{B \cup \bar{C}})=1 / 2 \\ & \Rightarrow \quad P(\overline{B \cap C})=1 / 2 \\ & \Rightarrow \quad P(B \cap C)=1-\frac{1}{2}=\frac{1}{2} \\ & \therefore \quad P(B \cup C)=P(B)+P(C)-P(B \cap C) \\ & \begin{array}{l} \Rightarrow \quad c-\frac{1}{2} \\ \therefore \quad P(\bar{B} \cup \bar{C})=1-P(B \cup C) \\ =1-b-c+\frac{1}{2}=\frac{3}{2}-b-c \end{array} \end{aligned} $