Probability

$\therefore$ P (drawing red ball from B₁) = ${1 \over 3}$

$ \Rightarrow \left( {{{{n_1} - 1} \over {{n_1} + {n_2} - 1}}} \right)\left( {{{{n_1}} \over {{n_1} + {n_2}}}} \right) + \left( {{{{n_2}} \over {{n_1} + {n_2}}}} \right)\left( {{{{n_1}} \over {{n_1} + {n_2} - 1}}} \right) = {1 \over 3}$

$ \Rightarrow {{n_1^2 + {n_1}{n_2} - {n_1}} \over {({n_1} + {n_2})({n_1} + {n_2} - 1)}} = {1 \over 3}$

Clearly, options (c) and (d) satisfy.

Box I : red balls $\to$ n₁

black balls $\to$ n₂

Box II : red balls $\to$ n₃

black balls $\to$ n₄

$P(R) = {1 \over 2}.{{{n_1}} \over {{n_1} + {n_2}}} + {1 \over 2}.{{{n_3}} \over {{n_3} + {n_4}}}$

P(Box II / R) $ = {{{1 \over 2}.{{{n_3}} \over {{n_3} + {n_4}}}} \over {{1 \over 2}.{{{n_1}} \over {{n_1} + {n_2}}} + {1 \over 2}.{{{n_3}} \over {{n_3} + {n_4}}}}}$

$ = {1 \over {1 + \left( {{{{{{n_1}} \over {{n_1} + {n_2}}}} \over {{{{n_3}} \over {{n_3} + {n_4}}}}}} \right)}} = {1 \over 3}$

Thus, ${{{n_1}} \over {{n_1} + {n_2}}} = 2{{{n_3}} \over {{n_3} + {n_4}}}$

i.e. $2\left( {1 + {{{n_2}} \over {{n_1}}}} \right) = 1 + {{{n_4}} \over {{n_3}}}$ i.e. ${{{n_4}} \over {{n_3}}} - 2{{{n_2}} \over {{n_1}}} = 1$

$P(\overline {A \cup B} ) = {1 \over 6}$

or, $1 - P(A \cup B) = {1 \over 6}$

$\therefore$ $P(A \cup B) = 1 - {1 \over 6} = {5 \over 6};$ $P(A \cap B) = {1 \over 4}$; and $P(\overline A ) = {1 \over 4};$

$\therefore$ $P(A) = 1 - P(\overline A ) = 1 - {1 \over 4} = {3 \over 4}$

We know, $P(A \cup B) = P(A) + P(B) - P(A \cap B)$

or, ${5 \over 6} = {3 \over 4} + P(B) - {1 \over 4}$

or, $P(B) = {5 \over 6} - {1 \over 2} = {1 \over 3}$

Now, $P(A)\,.\,P(B) = {3 \over 4}.\,{1 \over 3} = {1 \over 4} = P(A \cap B)$

i.e., events A and B are mutually independent.

Since the probability of A and B are different, so they are not equally likely events.

Therefore, (A) is the correct option.

Given three Boxes:

Given, that a card is drow from each Box and $x_i$ denote the number on the card drawn from $i^{\text {tht }}$ box, $i=1,2,3$.

Sample space $={ }^3 \mathrm{C}_1 .{ }^5 \mathrm{C}_1 .{ }^7 \mathrm{C}_1=3 \cdot 5 \cdot 7=105$.

Let A be the event when $x_1+x_2+x_3$ is odd.

Case I: When all $x_1, x_2, x_3$ are odd

No. of ways $={ }^2 \mathrm{C}_1 .{ }^3 \mathrm{C}_1 .{ }^4 \mathrm{C}_1=2.3 \cdot 4=24$

Case II: When $x_1$ is odd and $x_2, x_3$ are even

No. of ways $={ }^2 \mathrm{C}_1 .{ }^2 \mathrm{C}_1 .{ }^3 \mathrm{C}_1=12$

Case III: When $x_2$ is odd and $x_1, x_3$ are even

No. of ways $={ }^1 \mathrm{C}_1 .{ }^3 \mathrm{C}_1 .{ }^3 \mathrm{C}_1=9$

Case IV: When $x_3$ is odd and $x_1, x_2$ are even

No. of ways $={ }^1 \mathrm{C}_1 .{ }^2 \mathrm{C}_1 . \mathrm{C}_1 \mathrm{C}_1=8$

No. of favorable cases for A is

$\begin{aligned} 24+12+9+8 & =53 \\ \text { Required Probability } & =\mathrm{P}(\mathrm{A})=\frac{53}{105} \end{aligned}$

Hint:

Recall that sum of three numbers is odd only when

(i) All three numbers are odd.

(ii) Two numbers are even and one is odd.

Given three Boxes :

Given, that a card is drow from each box and $x_i$ denote the number on the card drawn from $i^{\text {th }}$ box, $i=1,2,3$.

Sample Space $={ }^3 \mathrm{C}_1 .{ }^5 \mathrm{C}_1 .{ }^7 \mathrm{C}_1=3 \cdot 5 \cdot 7=105$.

Let B be the event when $x_1, x_2, x_3$ are in A.P.

$\Rightarrow 2 x_2=x_1+x_3$

Here, $x_1+x_3$ is even number and for every even value of $x_1+x_3$ there is only one possible value of $x_2$

$x_1+x_2$ is even in two cases

Case I: When both $x_1$ and $x_2$ are even

No. of ways $={ }^1 \mathrm{C}_1$. ${ }^3 \mathrm{C}_{1=}=3$

Case II: When both $x_1$ and $x_2$ are odd

No. of ways $={ }^2 \mathrm{C}_1 .{ }^4 \mathrm{C}_1=2 \times 4=8$

Number of favorable cases for $\mathrm{B}=3+8=11$

Required probability $=\frac{11}{105}$

Hint:

$2 x_2=x_1+x_3$ indicate, $x_1+x_3$ is even and for every even value of $x_1+x_3$ there is only one possible value of $x_2$. Now, $x_1+x_3$ is even in two cases:

(i) When both $x_1$ and $x_2$ are even.

(ii) When both $x_1$ and $x_2$ are odd.

Given, 3 boys and 2 girls stand in a queue.

Sample space $n(s)=\left| \!{\underline {\, {5} \,}} \right.=120$

According to the given condition, following cases may arise.

BGGBB $\quad \left| \!{\underline {\, {3} \,}} \right. \left| \!{\underline {\, {2} \,}} \right.$

GGBBB $\quad \left| \!{\underline {\, {3} \,}} \right. \left| \!{\underline {\, {2} \,}} \right.$

GBGBB $\quad \left| \!{\underline {\, {3} \,}} \right. \left| \!{\underline {\, {2} \,}} \right.$

GBBGB $\quad \left| \!{\underline {\, {3} \,}} \right. \left| \!{\underline {\, {2} \,}} \right.$

BGBGB $\quad \left| \!{\underline {\, {3} \,}} \right. \left| \!{\underline {\, {2} \,}} \right.$

So, number of favorable ways $=5\left| \!{\underline {\, {3} \,}} \right. \left| \!{\underline {\, {2} \,}} \right.=60$

Required probability $=\frac{60}{120}=\frac{1}{2}$

Let $E$ be the event of one white and one red balls $A_1, A_2, A_3$ be the events of both balls are from box $\mathrm{B}_1, \mathrm{~B}_2, \mathrm{~B}_3$ respectively

Now, the probability of one white and one red both balls are drawn from box $B_2$ is $P\left(\frac{A_2}{E}\right)$.

$\Rightarrow P\left(\frac{A_2}{E}\right)=\frac{P\left(A_2 \cap E\right)}{P(E)}$

$\begin{aligned} & \Rightarrow P\left(\frac{A_2}{E}\right)=\frac{P\left(A_2\right) \cdot P\left(\frac{E}{A_2}\right)}{P\left(A_1\right) \cdot P\left(\frac{E}{A_1}\right)+P\left(A_2\right).\mathrm{P}\left(\frac{\mathrm{E}}{\mathrm{A}_2}\right)+\mathrm{P}\left(\mathrm{A}_3\right) \cdot \mathrm{P}\left(\frac{\mathrm{E}}{\mathrm{A}_3}\right)} \\ & \Rightarrow \mathrm{P}\left(\frac{\mathrm{A}_2}{\mathrm{E}}\right)=\frac{\frac{1}{3} \times \frac{{ }^2 \mathrm{C}_1 \cdot{ }^3 \mathrm{C}_1}{{ }^9 \mathrm{C}_2}}{\frac{1}{3} \times \frac{{ }^1 \mathrm{C}_1 \cdot{ }^3 \mathrm{C}_1}{{ }^6 \mathrm{C}_2}+\frac{1}{3} \times \frac{{ }^2 \mathrm{C}_1 \cdot{ }^3 \mathrm{C}_1}{{ }^9 \mathrm{C}_2}+\frac{1}{3} \times \frac{{ }^3 \mathrm{C}_1 \cdot{ }^4 \mathrm{C}_1}{{ }^{12} \mathrm{C}_2} } \\ & \Rightarrow \mathrm{P}\left(\frac{\mathrm{A}_2}{\mathrm{E}}\right)=\frac{\frac{1}{6}}{\frac{1}{5}+\frac{1}{6}+\frac{2}{11}} \\ & \Rightarrow \mathrm{P}\left(\frac{\mathrm{A}_2}{\mathrm{E}}\right)=\frac{55}{181} \\ \end{aligned}$

Hints:

$\Rightarrow$ Recall the Bay's theorem

$ \begin{gathered} P\left(\frac{A_2}{E}\right)=\frac{P\left(A_2\right) \cdot P\left(\frac{E}{A_2}\right)}{P\left(A_1\right) \cdot P\left(\frac{E}{A}\right)+P\left(A_2\right) \cdot P\left(\frac{E}{A_2}\right)+P\left(A_3\right) \cdot P\left(\frac{E}{A_3}\right)}\\ \end{gathered}$

$\begin{aligned} & \text { Let } \mathrm{P} \text { be the probability of same colour balls } \\ & \text { drawn from each boxes } \mathrm{B}_1, \mathrm{~B}_2, \mathrm{~B}_3 \\ & \therefore \quad \mathrm{P}=\mathrm{P} \text { (all are white) }+\mathrm{P} \text { (all are red) }+\mathrm{P} \text { (all } \\ & \quad \text { are Black) } \\ & \Rightarrow P=\frac{1}{6} \times \frac{2}{9} \times \frac{3}{12}+\frac{3}{6} \times \frac{3}{9} \times \frac{4}{12}+\frac{2}{6} \times \frac{4}{9} \times \frac{5}{12} \\ & \Rightarrow P=\frac{6}{648}+\frac{36}{648}+\frac{40}{648} \\ & \Rightarrow P=\frac{82}{648} \end{aligned}$

Hints:

The probability of one ball is drawn from each boxes $B_1, B_2, B_3$ is equal to the sum of probabilities of all are white balls, all are red balls and all are Black balls.

Let A, B, C, D are the events of solving a problem.

Given: $\mathrm{P}(\mathrm{A})=\frac{1}{2}, \mathrm{P}(\mathrm{B})=\frac{3}{4}, \mathrm{P}(\mathrm{C})=\frac{1}{4}, \mathrm{P}(\mathrm{D})=\frac{1}{8}$

Also given A, B, C, D are independent events.

Let P is the probability of solved by at least one person.

$\Rightarrow \mathrm{P}=1-$ (Probability that none of the person can solve the problem)

$\begin{aligned} & \Rightarrow \mathrm{P}=1-\mathrm{P}(\overline{\mathrm{A}} \cap \overline{\mathrm{B}} \cap \overline{\mathrm{C}} \cap \overline{\mathrm{D}}) \\ & \Rightarrow \mathrm{P}=1-\mathrm{P}(\overline{\mathrm{A}}) \cdot \mathrm{P}(\overline{\mathrm{B}}) \cdot \mathrm{P}(\overline{\mathrm{C}}) \cdot \mathrm{P}(\overline{\mathrm{D}}) \\ & \Rightarrow \mathrm{P}=1-(1-\mathrm{P}(\mathrm{A}))(1-\mathrm{P}(\mathrm{B}))(1-\mathrm{P}(\mathrm{C}))(1-\mathrm{P}(\mathrm{D})) \\ & \Rightarrow \mathrm{P}=1-\left(1-\frac{1}{2}\right)\left(1-\frac{3}{4}\right)\left(1-\frac{1}{4}\right)\left(1-\frac{1}{8}\right) \\ & \Rightarrow \mathrm{P}=1-\frac{1}{2} \times \frac{1}{4} \times \frac{3}{4} \times \frac{7}{8} \\ & \Rightarrow \mathrm{P}=1-\frac{21}{256}\\ & \Rightarrow \mathrm{P}=\frac{235}{256} \end{aligned}$

Hints:

$\Rightarrow$ (i) If $A$ and $B$ are independent events, then

$\begin{aligned} & \rightarrow \mathrm{P}(\mathrm{A} \cap \mathrm{B})=\mathrm{P}(\mathrm{A}) \cdot \mathrm{P}(\mathrm{B}) \\ & \rightarrow \mathrm{P}(\overline{\mathrm{A}} \cap \overline{\mathrm{B}})=\mathrm{P}(\overline{\mathrm{A}}) \cdot \mathrm{P}(\overline{\mathrm{B}})=(1-\mathrm{P}(\mathrm{A}))(1-\mathrm{P}(\mathrm{B})) \\ & \rightarrow \mathrm{P}(\overline{\mathrm{A}} \cap \mathrm{B})=\mathrm{P}(\overline{\mathrm{A}}) \cdot \mathrm{P}(\mathrm{B})=(1-\mathrm{P}(\mathrm{A})) \mathrm{P}(\mathrm{B}) \\ & \rightarrow \mathrm{P}(\mathrm{A} \cap \overline{\mathrm{B}})=\mathrm{P}(\mathrm{A}) \cdot \mathrm{P}(\overline{\mathrm{B}})=\mathrm{P}(\mathrm{A}) \cdot(1-\mathrm{P}(\mathrm{B})) \end{aligned}$

(ii) Probability of at least one person can solve the problem $= 1-$ Probability that none of the persons can solve that problem.

For the given condition, the sample space $=6^4$

For favourable condition

Case I : For $D_4$ there are ${ }^6 C_1$ way. Now, it appears on any one of $D_1, D_2, D_3$ i.e. ${ }^3 C_1 \cdot 1$, for other two there are $5 \times 5$ ways.

$\Rightarrow{ }^6 \mathrm{C}_1 \cdot{ }^3 \mathrm{C}_1 \cdot 1 \cdot 5^2=450 \text { Ways }$

Case II : For $\mathrm{D}_4$ there are ${ }^6 \mathrm{C}_1$ ways. Now, it appears on any two of $D_1, D_2, D_3$ i.e. ${ }^3 \mathrm{C}_2 .1 .1$, for other one there are 5 ways.

$\Rightarrow{ }^6 \mathrm{C}_1 \cdot{ }^3 \mathrm{C}_2 \cdot 1^2 \cdot 5=90 \text { Ways }$

Case III : For $\mathrm{D}_4$ there are ${ }^6 \mathrm{C}_1$ ways. Now, it appears on all i.e. $1^3$.

$\Rightarrow{ }^6 \mathrm{C}_1 \cdot 1^3=6$

Total number of favourable cases

$\begin{aligned} & =450+90+6 \\ & =546 \text { ways } \end{aligned}$

Thus, Probability $=\frac{546}{6^4}=\frac{91}{216}$

Let's analyze the given conditions and evaluate which options are correct.

Given:

1. $ P(X|Y) = \frac{1}{2} $


2. $ P(Y|X) = \frac{1}{3} $


3. $ P(X \cap Y) = \frac{1}{6} $

Now, let's proceed step by step through each option.

Option A: $ P(X \cup Y) = \frac{2}{3} $

We can use the formula for the union of two events:

$ P(X \cup Y) = P(X) + P(Y) - P(X \cap Y) $

To find $ P(X) $ and $ P(Y) $, we use the definitions of conditional probability:

$ P(X|Y) = \frac{P(X \cap Y)}{P(Y)} $

$ \frac{1}{2} = \frac{\frac{1}{6}}{P(Y)} $

$ P(Y) = \frac{1}{6} \times 2 = \frac{1}{3} $

Similarly,

$ P(Y|X) = \frac{P(X \cap Y)}{P(X)} $

$ \frac{1}{3} = \frac{\frac{1}{6}}{P(X)} $

$ P(X) = \frac{1}{6} \times 3 = \frac{1}{2} $

Now, substituting these values into the union formula:

$ P(X \cup Y) = \frac{1}{2} + \frac{1}{3} - \frac{1}{6} $

To add these fractions, we need a common denominator. The common denominator is 6:

$ P(X \cup Y) = \frac{3}{6} + \frac{2}{6} - \frac{1}{6} = \frac{4}{6} = \frac{2}{3} $

Hence, Option A is correct.

Option B: $ X $ and $ Y $ are independent

For two events $ X $ and $ Y $ to be independent, the following condition should hold:

$ P(X \cap Y) = P(X) \times P(Y) $

We already found that:

$ P(X) = \frac{1}{2} $

$ P(Y) = \frac{1}{3} $

$ P(X \cap Y) = \frac{1}{6} $

Checking the independence condition:

$ P(X) \times P(Y) = \frac{1}{2} \times \frac{1}{3} = \frac{1}{6} $

Since this is equal to $ P(X \cap Y) $, $ X $ and $ Y $ are indeed independent.

Hence, Option B is correct.

Option C: $ X $ and $ Y $ are not independent

From the previous analysis, we have shown that $ X $ and $ Y $ are independent.

Hence, Option C is not correct.

Option D: $ P((X^c) \cap Y) = \frac{1}{3} $

To find $ P((X^c) \cap Y) $, we use the fact that:

$ P(Y) = P((X \cap Y) \cup (X^c \cap Y)) $

Since $ (X \cap Y) $ and $ (X^c \cap Y) $ are disjoint events, we can write:

$ P(Y) = P(X \cap Y) + P(X^c \cap Y) $

Given:

$ P(Y) = \frac{1}{3} $

$ P(X \cap Y) = \frac{1}{6} $

Substituting these,

$ \frac{1}{3} = \frac{1}{6} + P(X^c \cap Y) $

Solving for $ P(X^c \cap Y) $:

$ P(X^c \cap Y) = \frac{1}{3} - \frac{1}{6} = \frac{2}{6} - \frac{1}{6} = \frac{1}{6} $

This contradicts Option D, as it says $ \frac{1}{3} $.

Hence, Option D is not correct.

In conclusion:

Option A and Option B are correct.

Option C and Option D are not correct.

$\begin{array}{r} \text { Given, } \mathrm{P}\left(\mathrm{X}_1\right)=\frac{1}{2} \Rightarrow \mathrm{P}\left(\overline{\mathrm{X}}_1\right)=1-\frac{1}{2}=\frac{1}{2} \\ \mathrm{P}\left(\mathrm{X}_2\right)=\frac{1}{4} \Rightarrow \mathrm{P}\left(\bar{X}_2\right)=1-\frac{1}{4}=\frac{3}{4} \\ \mathrm{P}\left(\mathrm{X}_3\right)=\frac{1}{4} \Rightarrow \mathrm{P}\left(\bar{X}_3\right)=1-\frac{1}{4}=\frac{3}{4} \end{array}$

$\begin{aligned} & \text { Now, } P(X)=P\left(E_1 E_2 E_3\right)+P\left(\bar{E}_1 E_2 E_3\right)+P\left(E_1\right. \\ & \left.\overline{\mathrm{E}}_2 \mathrm{E}_3\right)+\mathrm{P}\left(\mathrm{E}_1 \mathrm{E}_2 \overline{\mathrm{E}}_3\right) \\ & =\frac{1}{2} \cdot \frac{1}{4} \cdot \frac{1}{4}+\frac{1}{2} \cdot \frac{1}{4} \cdot \frac{1}{4}+\frac{1}{2} \cdot \frac{3}{4} \cdot \frac{1}{4}+\frac{1}{2} \cdot \frac{1}{4} \cdot \frac{3}{4} \\ \end{aligned}$

$\begin{aligned} & P(X)=\frac{1}{32}+\frac{1}{32}+\frac{3}{32}+\frac{3}{32} \\ & P(X)=\frac{1}{4} \end{aligned}$

(i) $\mathrm{P}\left(\frac{\mathrm{X}_1^c}{\mathrm{X}}\right)=\frac{\mathrm{P}\left(\mathrm{X}_1^c \cap \mathrm{X}\right)}{\mathrm{P}(\mathrm{X})}$

Now, $X_1^c \cap X$ indicates the event that the ship is operational but $\mathrm{E}_1$ is not functioning.

$\begin{aligned} & \therefore \quad \mathrm{P}\left(\mathrm{X}_1^c \cap \mathrm{X}\right)=\frac{1}{2} \times \frac{1}{4} \times \frac{1}{4}=\frac{1}{32} \\ & \therefore \mathrm{P}\left(\frac{\mathrm{X}_1^c}{\mathrm{X}}\right)=\frac{\frac{1}{32}}{\frac{1}{4}}=\frac{1}{8} \end{aligned}$

(ii) P (Exactly two engines of the ship are functioning $\mid X$)

$\begin{aligned} & =\frac{\mathrm{P}\left(\overline{\mathrm{E}}_1 \mathrm{E}_2 \mathrm{E}_3\right)+\mathrm{P}\left(\mathrm{E}_1 \overline{\mathrm{E}}_2 \mathrm{E}_3\right)+1\left(\mathrm{E}_1 \mathrm{E}_2 \overline{\mathrm{E}}_3\right)}{\mathrm{P}(\mathrm{X})} \\ & =\frac{\frac{1}{2} \times \frac{1}{4} \times \frac{1}{4}+\frac{1}{2} \times \frac{3}{4} \times \frac{1}{4}+\frac{1}{2} \times \frac{1}{4} \times \frac{3}{4}}{\frac{1}{4}} \\ & =\frac{7}{8} \end{aligned}$

$\begin{aligned} \text { (iii) } & P\left(X \mid X_2\right)=\frac{P\left(X \cap X_2\right)}{P\left(X_2\right)} \\ & =\frac{P\left(E_1 E_2 E_3\right)+P\left(\bar{E}_1 E_2 E_3\right)+P\left(E_1 E_2 \bar{E}_3\right)}{P\left(X_2\right)} \\ & =\frac{\frac{1}{2} \times \frac{1}{4} \times \frac{1}{4}+\frac{1}{2} \times \frac{1}{4} \times \frac{1}{4}+\frac{1}{2} \times \frac{1}{4} \times \frac{3}{4}}{\frac{1}{4}} \\ & =\frac{5}{8} \end{aligned}$

$\begin{aligned} \text { (iv) } & P\left(X \mid X_1\right)=\frac{P\left(X \cap X_1\right)}{P\left(X_1\right)} \\ = & \frac{P\left(E_1 E_2 E_3\right)+P\left(E_1 \bar{E}_2 E_3\right)+P\left(E_1 E_2 \bar{E}_3\right)}{P\left(X_1\right)} \\ = & \frac{\frac{1}{2} \times \frac{1}{4} \times \frac{1}{4}+\frac{1}{2} \times \frac{3}{4} \times \frac{1}{4}+\frac{1}{2} \times \frac{1}{4} \times \frac{3}{4}}{\frac{1}{2}} \\ = & \frac{7}{16} \end{aligned}$

$P\left( {{H \over W}} \right) = {{P(W/H) \times P(H)} \over {P(W/T)\,.\,P(T) + (W/H)\,.\,P(H)}}$

$ = {{{1 \over 2}\left( {{3 \over 5} \times 1 + {2 \over 5} \times {1 \over 2}} \right)} \over {23/30}} = {{12} \over {23}}$

H $\to$ One ball from U₁ to U₂.

T $\to$ Two balls from U₁ to U₂.

E : One ball drawn from U₂.

P/W from

${U_2} = {1 \over 2} \times \left( {{3 \over 5} \times 1} \right) + {1 \over 2} \times \left( {{2 \over 5} \times {1 \over 2}} \right) + {1 \over 2} \times \left( {{{{}^3{C_2}} \over {{}^5{C_2}}} \times 1} \right) + {1 \over 2} \times \left( {{{{}^2{C_2}} \over {{}^5{C_2}}} \times {1 \over 3}} \right) + {1 \over 2} \times \left( {{{{}^3{C_1}\,.\,{}^2{C_1}} \over {{}^5{C_2}}} \times {2 \over 3}} \right) = {{23} \over {30}}$

Let $P(E) = e$ and $P(F) = f$.

$P(E \cup F) - P(E \cap F) = {{11} \over {25}}$

$ \Rightarrow e + f - 2ef = {{11} \over {25}}$ ...... (1)

$P(\overline E \cap \overline F ) = {2 \over {25}}$

$ \Rightarrow (1 - e)(1 - f) = {2 \over {25}}$

$ \Rightarrow 1 - e - f + ef = {2 \over {25}}$ ...... (2)

From Eqs. (1) and (2), we get

$ef = {{12} \over {25}}$ and $e + f = {7 \over 5}$

Solving, we get

$e = {4 \over 5},\,f = {3 \over 5}$ or $e = {3 \over 5},\,f = {4 \over 5}$

Sample space A dice is thrown thrice, $n(s) = 6 \times 6 \times 6$.

Favorable events ${\omega ^{{r_1}}} + {\omega ^{{r_2}}} + {\omega ^{{r_3}}} = 0$

i.e. $({r_1},{r_2},{r_3})$ are ordered 3-triples which can take values,

$\left. {\matrix{ {(1,2,3),} & {(1,5,3),} & {(4,2,3),} & {(4,5,3)} \cr {(1,2,6),} & {(1,5,6),} & {(4,2,6),} & {(4,5,6)} \cr } } \right\}$

i.e. 8 ordered pairs and each can be arranged in 3! ways = 6

$\therefore$ $n(E) = 8 \times 6$

$ \Rightarrow P(E) = {{8 \times 6} \over {6 \times 6 \times 6}} = {2 \over 9}$

From the tree-diagram it follows that

$P({B_G}) = {{46} \over {80}}$

$P({B_G}|G) = {{10} \over {16}} = {5 \over 8}$

$\therefore$ $P({B_G} \cap G) = {5 \over 8} \times {4 \over 5} = {1 \over 2}$

$P(G|{B_G}) = {{{1 \over 2}} \over {P({B_G})}} = {1 \over 2} \times {{80} \over {46}} = {{20} \over {23}}$

The required probability is $P(X = 3)$

$\left( {{5 \over 6}} \right)\left( {{5 \over 6}} \right){1 \over 6} = {{25} \over {216}}$

The probability $P(X \ge 3)$ is nothing but the probability of $P(X \le 2)$:

${1 \over 6} + {5 \over 6} \times {1 \over 6} = {{11} \over {36}}$

The required probability is

$1 - {{11} \over {36}} = {{25} \over {36}}$

$X \ge 6$: The probability for

${{{5^5}} \over {{6^6}}} + {{{5^6}} \over {{6^7}}}\, + \,...\, + \,\infty = {{{5^5}} \over {{6^6}}}\left( {{1 \over {1 - 5/6}}} \right) = {\left( {{5 \over 6}} \right)^5}$

For $X>3$, we have

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

Hence, the conditional probability is

${{{{(5/6)}^6}} \over {{{(5/6)}^3}}} = {{25} \over {36}}$

Given,

An experiment has 10 equally likely outcomes,

Let B have n no of outcomes,

So, $P(B) = {n \over {10}}$

$P(A) = {4 \over {10}}$

$P(A \cap B) = {4 \over {10}} \times {n \over {10}} = {{{{2n} \over 5}} \over {10}}$

Since, A and B are independent

n is an integer

n = 5 or 10

We have,

$ax + by = 0$

$cx + dy = 0$

since, the system of homogeneous equation is always consistent and has a solution.

Therefore, statement 2 is true.

Now, $\Delta = \left| {\matrix{ a & b \cr c & d \cr } } \right|$ and $a,b,c,d \in \{ 0,1\} $

$ = ad - bc$

No. of ways of selecting $a,b,c,d$ from the set {0, 1} is 2 $\times$ 2 $\times$ 2 $\times$ 2 = 16

If the system has unique solution,

Then $\Delta\ne0$

$\Rightarrow$ either $ad = 1,bc = 0$ or $ad = 0,bc = 1$ favourable case = 6

Therefore, probability that system of equation has unique solution is ${6 \over {16}} = {3 \over 8}$. Statement 1 is True.

Hence, the Statement 2 is True but is not a correct explanation of statement 1.

Probability of getting score 9 in a single throw

$ = {4 \over {36}} = {1 \over 9}.$

Probability of getting score 9 exactly twice

$ = {}^3{C_2} \times {\left( {{1 \over 9}} \right)^2} \times {8 \over 9} = {8 \over {243}}.$

The desired probability

= (0.7) (0.2) + (0.7) (0.8) (0.7) (0.2) + (0.7) (0.8) (0.7) (0.8) (0.7) (0.2) + .......

= 0.14 [1 + (0.56) + (0.56)² + .......]

= 0.14 $\left( {{1 \over {1 - 0.56}}} \right) = {{0.14} \over {0.44}} = {7 \over {22}}$ = 0.32

$\begin{aligned} & \mathrm{E}_{1} f_{1} G \text { are pairwise independent events. } \\\\ & \therefore \quad \mathrm{P}(\mathrm{E} \cap \mathrm{F})=\mathrm{P}(\mathrm{E}) \cdot \mathrm{P}(\mathrm{F}) \\\\ & \mathrm{P}(\mathrm{F} \cap \mathrm{G})=\mathrm{P}(\mathrm{F}) \cdot \mathrm{P}(\mathrm{G}) \\\\ & \mathrm{P}(\mathrm{G} \cap \mathrm{E})=\mathrm{P}(\mathrm{G}) \cdot \mathrm{P}(\mathrm{E}) \\\\ & \mathrm{P}\left(\frac{\mathrm{E}^{\mathrm{c}} \cap \mathrm{F}^{\mathrm{c}}}{\mathrm{G}}\right)=\mathrm{P}\left(\frac{\left(E^{\mathrm{c}} \cap \mathrm{F}^{\mathrm{c}}\right) \cap G}{\mathrm{P}(\mathrm{G})}\right) \\\\ &=\frac{\mathrm{P}(\mathrm{G})-\mathrm{P}(\mathrm{G} \cap \mathrm{E})-\mathrm{P}(\mathrm{G} \cap \mathrm{F})}{\mathrm{P}(\mathrm{G})} \\\\ &=1-\mathrm{P}(\mathrm{E})-\mathrm{P}(\mathrm{F}) \\\\ &=\mathrm{P}\left(\mathrm{E}^{\mathrm{c}}\right)-\mathrm{P}(\mathrm{F}) . \end{aligned}$

Let E = event when each American man is seated adjacent to his wife and

A = event when Indian man is seated adjacent to his wife.

Now,

$n(A\cap E)=(4!)\times(2!)^5$

Event when each American man is seated adjacent to his wife.

Again $n(E)=(5!)\times(2!)^4$

$p\left( {{A \over E}} \right) = {{n(A \cap E)} \over {n(E)}}$

$ = {{(4!) \times {{(2!)}^5}} \over {(5!) \times {{(2!)}^4}}} = {2 \over 5}$

Statement 1 : If P(H$_i$ $\cap$ E) = O for some, then

$P\left( {{{{H_i}} \over E}} \right) = P\left( {{E \over {{H_i}}}} \right) = 0$

If P(H$_i$ $\cap$ E) $\ne o~\forall i=1,2,3.......,n$ then

$P\left( {{{{H_i}} \over E}} \right) = {{P({H_i} \cap E)} \over {P({H_i})}} \times {{P({H_i})} \over {P(E)}}$

$ = {{P\left( {{E \over {{H_i}}}} \right) \times P({H_i})} \over {P(E)}}$

$ > P\left( {{E \over {{H_i}}}} \right) \times P({H_i})$

Hence, statement 1 may not always be true.

Statement 2 : Clearly, we can write as

H$_1$ $\cup$ H$_2$ $\cup$ H$_1$ $\cup$ ------------------ $\cup$ H$_n$ = S

$\Rightarrow$ P(H$_1$) + P(H$_2$) + ---------------- + P(H$_n$) = 1

Hence, statement 2 is true.

Required probability

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

$ = {{{e^{ - 5}}} \over {0!}}{5^0} + {{{e^{ - 5}}} \over {1!}}{5^1}$

$ = {e^{ - 5}} + 5{e^{ - 5}}$

$ = {6 \over {{e^5}}}$

$\because \mathrm{P}\left(u_{i}\right) \propto i$

$\mathrm{P}\left(u_{i}\right)=\mathrm{K} i$

We know that $\sum \mathrm{P}\left(\mathrm{u}_{i}\right)=1$

$\begin{aligned} \mathrm{K} \frac{n(n+1)}{2} & =1 \\ \mathrm{~K} & =\frac{2}{n(n+1)} \\ \mathrm{P}\left(u_{i}\right) & =\frac{2 i}{n(n+1)} \\ \mathrm{P}\left(\frac{w}{u_{2 i}}\right) & =\frac{2 i}{n+1} \end{aligned}$

Now, $\quad \mathrm{P}(w)=\sum_\limits{1^{i}=1}^{n} \mathrm{P}\left(u_{i}\right) \cdot \mathrm{P}\left(\frac{w}{u_{i}}\right)$

$ = \sum\limits_{{1^i} = 1}^n {{{2i} \over {n(n + 1)}}{i \over {(n + 1)}}} $

$ = \sum\limits_{{1^i} = 1}^n {{{2{i^2}} \over {n{{(n + 1)}^2}}}} $

$ = {2 \over {n{{(n + 1)}^2}}}\sum\limits_{{1^i} = 1}^n {{i^2}} $

$ = {2 \over {n{{(n + 1)}^2}}}{{n(n + 1)(2n + 1)} \over 6}$

$ = {{2n + 1} \over {3(n + 1)}}$

Now, $\mathop {\lim }\limits_{n \to \infty } {{2n + 1} \over {3(n + 1)}} = \mathop {\lim }\limits_{n \to \infty } {{\left( {2 + {1 \over n}} \right)} \over {3\left( {{1 \over n} + 1} \right)}} = {2 \over 3}$

Since, $\mathrm{P}\left(u_{i}\right)=\mathrm{C}$ means

All events are equally likely

$\mathrm{P}\left(u_{i}\right)=\mathrm{C}=\frac{1}{n} \quad$ for $i=1,2,3, \ldots . n$

$\mathrm{P}(u_n)=\frac{1}{n}$

$\mathrm{P}\left(\frac{w}{u_{n}}\right)=\frac{n}{n+1}$

$\begin{aligned} \mathrm{P}(w) & =\sum \mathrm{P}\left(u_{i}\right) \cdot \mathrm{P}\left(\frac{w}{u_{n}}\right) \\ & =\sum\left(\frac{1}{n} \frac{2^{i}}{(n+1)}\right)=\frac{1}{n(n+1)} \cdot \frac{n(n+1)}{2}=\frac{1}{2} \\ \mathrm{P}\left(\frac{u_{\mathrm{n}}}{w}\right) & =\frac{\mathrm{P}\left(u_{\mathrm{n}}\right) \times \mathrm{P}\left(\frac{w}{u_{\mathrm{n}}}\right)}{\mathrm{P}(w)} \\ & =\frac{\frac{1}{n} \times \frac{n}{(n+1)}}{\frac{1}{2}}=\frac{2}{n+1} \end{aligned}$

Total number of ball

Even numbered urns $=\frac{n}{2}(n+1)$

Total no. of white balls = $2+4+6+....+n$.

$=2(1+2+3+...+m)$

where $m=\frac{n}{2}$

$=2 \times \frac{m(m+1)}{2}$

$=\frac{n}{2} \cdot\left(\frac{n}{2}+1\right)=\frac{n(n+2)}{4}$

$\therefore \quad \mathrm{P}\left(\frac{w}{\mathrm{E}}\right)=\frac{\frac{n}{4}(n+2)}{\frac{n}{2}(n+1)}$

$=\frac{(n+2)}{2(n+1)}$

Person 1st has three options to apply.

Similarly, person 2nd has three options t apply

and person 3rd has three options to apply.

Total cases = 3³

Now, favourable cases = 3 (An either all has applied for house 1 or 2 or 3)

So, probability $ = {3 \over {{3^3}}} = {1 \over 9}$.

Given that,

$P(\overline {A \cup B} ) = {1 \over 6}$, $P(A \cap B) = {1 \over 4}$, $P(\overline A ) = {1 \over 4}$

$\because$ $P(\overline {A \cup B} ) = {1 \over 6}$

$ \Rightarrow 1 - P(A \cup B) = {1 \over 6}$

$ \Rightarrow 1 - P(A) - P(B) + P(A \cap B) = {1 \over 6}$

$ \Rightarrow P(\overline A ) - P(B) + {1 \over 4} = {1 \over 6}$

$ \Rightarrow P(B) = {1 \over 4} + {1 \over 4} - {1 \over 6}$

$ \Rightarrow P(B) = {1 \over 3}$ and $P(A) = {3 \over 4}$

Clearly, $P(A \cap B) = P(A)P(B)$,

so, the events A and B are independent events but not equally likely.

In a position distribution,

$P(X = r) = {{{e^{ - \lambda }}{\lambda ^r}} \over {r!}}$ ($\lambda$ = mean).

Now, $P(X = r > 1.5) = P(2) + P(3) + $ ..... $\infty$

$ = 1 - \{ P(0) + P(1)\} $

$ = 1 - \left( {{e^{ - 2}} + {{{e^{ - 2}} \times 2} \over 1}} \right) = 1 - {3 \over {{e^2}}}$

$\begin{aligned} P\left(E_{\text {car }}\right)=\frac{1}{7} \quad & P\left(E_{S_{\text {cooter }}}\right)=\frac{3}{2} \\ P\left(E_{\text {bus}}\right)=\frac{2}{7} \quad & P\left(E_{\text {train }}\right)=\frac{1}{7} \\ P\left(\frac{E_{\text {late }}}{E_{\text {car }}}\right) & =\frac{2}{9} \\ P\left(\frac{E_{\text {late }}}{E_{\text {scooter }}}\right) & =\frac{1}{9} \\ E\left(\frac{E_{\text {late }}}{E_{\text {bus }}}\right) & =\frac{4}{9} \\ E\left(\frac{E_{\text {late }}}{E_{\text {train }}}\right) & =\frac{1}{9}\end{aligned}$

Using Total probability theorem.

$\begin{aligned} & P\left(E_{\text {late }}\right)=P\left(E_{\text {car }}\right) \times P\left(\frac{E_{\text {late }}}{E_{\text {car }}}\right)+P\left(E_{\text {scooter }}\right) \\ & \times P\left(\frac{E_{\text {late }}}{E_{\text {scooter }}}\right)+P\left(E_{\text {Bus }}\right) \times P\left(\frac{E_{\text {late }}}{E_{\text {Bus }}}\right) \\ & +P\left(E_{\text {train }}\right) \times P\left(\frac{E_{\text {late }}}{E_{\text {train }}}\right) \\ & =\frac{1}{7} \times \frac{2}{9}+\frac{3}{2} \times \frac{1}{9}+\frac{2}{7} \times \frac{4}{9}+\frac{1}{7} \times \frac{1}{9} \\ & =\frac{2}{63}+\frac{3}{18}+\frac{8}{63}+\frac{1}{63}=\frac{11}{63}+\frac{3}{18}=\frac{43}{126} \\ & \mathrm{P}\left(\mathrm{E}_{\text {ontime }}\right)=1-\frac{43}{126}=\frac{83}{126} \\ & P\left(\frac{E_{\text {car }}}{E_{\text {ontime }}}\right)=\frac{P\left(E_{\text {car }} \cap E_{\text {ontime }}\right)}{P\left(E_{\text {ontime }}\right)} \\ & =\frac{P\left(E_{\text {car }}\right) \times P\left(E_{\text {ontime }}\right)}{P\left(E_{\text {ontime }}\right)}=P\left(E_{\text {car }}\right)=\frac{1}{7} \end{aligned}$