Here, sample space = {1, 2, 3,4,5, 6}, A ⋂ B = {5}, B ⋂ C = {2, 5}, A ⋂ C = {1, 5}, A ⋂ B ⋂ C = {5} and {A U B} ⋂ C = {1, 2, 5}
Also, \(P(A)=\frac{2}{6}, P(B)=\frac{3}{6}, P(C)=\frac{3}{6}\)
\(P(A \cap B)=\frac{1}{6}, P(B \cap C)=\frac{2}{6}, P(A \cap C)=\frac{2}{6} \)
\(P(A \cap B \cap C)=\frac{1}{6} \text { and } P((A \cup B) \cap C)=\frac{3}{6}\)
\((i) \ (\mathbf{b}): P(A \mid B)=\frac{P(A \cap B)}{P(B)}=\frac{1 / 6}{3 / 6}=\frac{1}{3}\)
\((ii) \ (a): P(B \mid C)=\frac{P(B \cap C)}{P(C)}=\frac{2 / 6}{3 / 6}=\frac{2}{3}\)
\((iii) \ (\mathrm{d}): P(A \cap B \mid C)=\frac{P(A \cap B \cap C)}{P(C)}=\frac{1 / 6}{3 / 6}=\frac{1}{3}\)
\((iv) \ (c): P(A \mid C)=\frac{P(A \cap C)}{P(C)}=\frac{2 / 6}{3 / 6}=\frac{2}{3}\)
\( (v) \ (\mathbf{d}): P(A \cup B \mid C)=\frac{P((A \cup B) \cap C)}{P(C)}=\frac{3 / 6}{3 / 6}=1\)

Let B, R, Y and G denote the events that ball drawn is blue, red, yellow and green respectively.
\(\therefore P(B)=\frac{12}{35}, P(R)=\frac{8}{35}, P(Y)=\frac{10}{35} \text { and } P(G)=\frac{5}{35}\)
\((i) \ (c): P(G \cap B)=P(B) \cdot P(G \mid B)=\frac{12}{35} \cdot \frac{5}{34}=\frac{6}{119}\)
\((ii) \ (\mathbf{b}): P(R \cap Y)=P(Y) \cdot P(R \mid Y)=\frac{10}{35} \cdot \frac{8}{34}=\frac{8}{119}\)
(iii) (a): Let E = event of drawing a first red ball and
F = event of drawing a second red ball
Here, \(P(E)=\frac{8}{35} \text { and } P(E)=\frac{7}{34}\)
\(\therefore \P(F \cap E)=P(E) \cdot P(F \mid E)=\frac{8}{35} \cdot \frac{7}{34}=\frac{4}{85}\)
\(\text {(iv) }(c): P\left(Y^{\prime} \cap G\right)=P(G) \cdot\left(Y^{\prime} \mid G\right)=\frac{5}{35} \cdot \frac{24}{34}=\frac{12}{119}\)
(v) (d): Let E = event of drawing a first non-blue ball and F = event of drawing a second non-blue ball
Here, \(P(E)=\frac{23}{35} \text { and } P(F)=\frac{22}{34}\)
\(\therefore \ P(F \cap E)=P(E) \cdot P(F \mid E)=\frac{23}{35} \cdot \frac{22}{34}=\frac{253}{595}\)

Here, P(A) = 0.6 and P(B) = 0.8
\(\text { (i) } \ (c): P(A \cap B)=P(A) \cdot P(B)=(0.6)(0.8)=0.48\)
\(\text { (ii) }(\text { a) }: P(A \cup B)=P(A)+P(B)-P(A \cap B)\)
= 0.6 + 0.8 - 0.48 = 0.92
(iii) (d): P(B I A) = P(B) (\(\because\)A and B are independent)
= 0.8
(iv) (a): p (A I B) = p(A) (\(\because\)A and B are independent)
= 0.6
(v) (c) : P(not A and not B) = \(P\left(A^{\prime} \cap B^{\prime}\right)=P(A \cup B)^{\prime}\)
\(=1-P(A \cup B)=1-0.92=0.08\)

Let E be the event that the doctor visit the patient late and let A₁, A₂, A₃, A₄ be the events that the doctor comes by cab, metro, bike and other means of transport respectively.
\(P\left(A_{1}\right)=0.3, P\left(A_{2}\right)=0.2, P\left(A_{3}\right)=0.1, P\left(A_{4}\right)=0.4\)
P(E I A₁) = Probability that the doctor arriving late when he comes by cab = 0.25
Similarly, P ( E I A₂) = 0.3, P (E I A₃) = 0.35 and P ( E I A₃) = 0.1
(i) (b): P(A₂ | E) = Probability that the doctor arriving late and he comes by metro
\(=\frac{P\left(A_{2}\right) P\left(E \mid A_{2}\right)}{\sum P\left(A_{i}\right) P\left(E \mid A_{i}\right)}\)
\(=\frac{(0.2)(0.3)}{(0.3)(0.25)+(0.2)(0.3)+(0.1)(0.35)+(0.4)(0.1)}\)
\(=\frac{0.06}{0.21}=\frac{2}{7}\)
(ii) (c): P(A₁ | E) = Probability that the doctor arriving late and he comes by cab
\(=\frac{P\left(A_{1}\right) P\left(E \mid A_{1}\right)}{\Sigma P\left(A_{i}\right) P\left(E \mid A_{i}\right)}\)
\(=\frac{(0.3)(0,25)}{(0.3)(0.25)+(0.2)(0.3)+(0.1)(0.35)+(0.4)(0.1)}\)
\(=\frac{0.075}{0.21}=\frac{5}{14}\)
(iii) (d): P(A₃| E) = Probability that the doctor arriving late and he comes by bike
\(=\frac{P\left(A_{3}\right) P\left(E \mid A_{3}\right)}{\sum P\left(A_{i}\right) P\left(E \mid A_{i}\right)}\)
\(=\frac{(0.1)(0.35)}{(0.3)(0.25)+(0.2)(0.3)+(0.1)(0.35)+(0.4)(0.1)}\)
\(=\frac{0.035}{0.21}=\frac{1}{6}\)
(iv) (c): P(A₄ | E) = Probability that the doctor arriving late and he comes by other means of transport
\(=\frac{P\left(A_{4}\right) P\left(E \mid A_{4}\right)}{\Sigma P\left(A_{i}\right) P\left(E \mid A_{i}\right)}\)
\(=\frac{(0.4)(0.1)}{(0.3)(0.25)+(0.2)(0.3)+(0.1)(0.35)+(0.4)(0.1)}\)
\(=\frac{0.04}{0.21}=\frac{4}{21}\)
(v) (a): Probability that the doctor is late by any means
\(=\frac{2}{7}+\frac{5}{14}+\frac{1}{6}+\frac{4}{21}=1\)

(i) (a): We know that \(\Sigma P_{i}=1\)
Then 0.2 + k + 2k + 3k + 2k + 0 = 1
\(\Rightarrow\) 8k = 1 - 0.2 = 0.8 \(\Rightarrow\) k = 0.1
(ii) (b): P(Average study time is not more than 1 hour)
= P(X ≤ 1) = P(X = 0) + P(X = 1) = 0.2 + 0.1 = 0.3
(iii) (a): P(Average study time is at least 3 hours)
P(X ≥ 3) = P(X = 3) + P(X = 4) = 0.3 + 0.2 = 0.5
(iv) (d): P(Average study time is exactly 2 hours)
= P(X = 2) = 0.2
(v) (c): P(Average study time is at least 1 hour)
= 1 - P(X = 0) = 1 - 0.2 = 0.8

Let E₁ be the event that Ravi solved the puzzle and
E₂ be the event that Priya solved the puzzle
Then, p(E₁) = 1/4 and P(E2) = 1/5
(i) (b): Since, E₁ and E₂ are independent events.
∴ P(both solved the puzzle) \(=P\left(E_{1} \cap E_{2}\right)\)
\(=P\left(E_{1}\right) \cdot P\left(E_{2}\right)=\frac{1}{4} \times \frac{1}{5}=\frac{1}{20}=\frac{1}{20} \times 100 \%=5 \%\)
(ii) (b): P(puzzle is solved by Ravi but not by Priya)
\(=P\left(\bar{E}_{2}\right) P\left(E_{1}\right)=\left(1-\frac{1}{5}\right) \cdot \frac{1}{4}=\frac{4}{5} \cdot \frac{1}{4}=\frac{1}{5}\)
(iii) (c) : P(puzzle is solved) = P(E₁ or E₂)
\(=P\left(E_{1} \cup E_{2}\right)=P\left(E_{1}\right)+P\left(E_{2}\right)-P\left(E_{1} \cap E_{2}\right)\)
\(=\frac{1}{4}+\frac{1}{5}-\frac{1}{20}=\frac{8}{20}=\frac{2}{5}\)
(iv) (c): P(Exactly one of them solved the puzzle)
\(=P\left[\left(E_{1} \text { and } \bar{E}_{2}\right) \operatorname{or}\left(E_{2} \text { and } \bar{E}_{1}\right)\right] \)
\(=P\left(E_{1} \cap \bar{E}_{2}\right)+P\left(E_{2} \cap \bar{E}_{1}\right) \)
\(=P\left(E_{1}\right) \times P\left(\bar{E}_{2}\right)+P\left(E_{2}\right) \times P\left(\bar{E}_{1}\right) \)
\(=\frac{1}{4} \times \frac{4}{5}+\frac{1}{5} \times \frac{3}{4} \quad\left[\because P\left(\bar{E}_{1}\right)=1-P\left(E_{1}\right)\right] \)
\(=\frac{4}{20}+\frac{3}{20}=\frac{7}{20}\)
(v) (b): P(none of them solved the puzzle)
\(=P\left(\bar{E}_{1} \cap \bar{E}_{2}\right)=P\left(\bar{E}_{1}\right) \cdot P\left(\bar{E}_{2}\right)=\frac{3}{4} \times \frac{4}{5}=\frac{3}{5}\)

(i) (b): Required probability = \(\frac{{ }^{12} C_{2}}{{ }^{51} C_{2}}\)
\(=\frac{12 \times 11}{51 \times 50}=\frac{22}{425}\)
(ii) (a): Required probability = \(\frac{{ }^{13} C_{2}}{{ }^{51} C_{2}}=\frac{13 \times 12}{51 \times 50}=\frac{26}{425}\)
(iii) (b): Clearly, P(A l E₁) = \(\frac{{ }^{12} C_{2}}{{ }^{51} C_{2}}=\frac{22}{425}\)
\(P\left(A \mid E_{2}\right)=\frac{{ }^{13} C_{2}}{{ }^{51} C_{2}}=\frac{26}{425} \)
\(P\left(A \mid E_{3}\right)=P\left(A \mid E_{4}\right)=\frac{26}{425}\)
\(\therefore \sum_{i=1}^{4} P\left(A \mid E_{i}\right)=\frac{22}{425}+\frac{26}{425}+\frac{26}{425}+\frac{26}{425}=\frac{100}{425}=0.24\)
(iv) (b): P(getting both king) = \(\frac{{ }^{4} C_{2}}{{ }^{52} C_{2}}=\frac{4 \times 3}{52 \times 51}=\frac{1}{221}\)
(v) (a): P(drawing a king) = \(\frac{4}{52}=\frac{1}{13}\)
\( \therefore \quad P(\text { not drawing a king })=1-\frac{1}{13}=\frac{12}{13} \)
\(\therefore \quad \text { Required probability }=\frac{12}{13} \times \frac{12}{13}=\frac{144}{169}\)

(i) (d): Since, it rained only 6 days each year, therefore, probability that it rains on chosen day is \(\frac{6}{366}=\frac{1}{61}\)
(ii) (c): The probability that it does not rain on chosen day = \(1-\frac{1}{61}=\frac{60}{61}=\frac{360}{366}\)
(iii) (c): It is given that, when it actually rains, the weatherman correctly forecasts rain 80% of the time.
Required probability \( =\frac{80}{100}=\frac{8}{10}=\frac{4}{5}\)
(iv) (a): Let A₁ be the event that it rains on chosen day, A₂ be the event that it does not rain on chosen day and E be the event the weatherman predict rain.
Then we have, P(A₁) \(=\frac{6}{366}, P\left(A_{2}\right)=\frac{360}{366},\)
\(P\left(E \mid A_{1}\right)=\frac{8}{10} \text { and } P\left(E \mid A_{2}\right)=\frac{2}{10}\)
Required probability = P(A₁ I E)
\(=\frac{P\left(A_{1}\right) \cdot P\left(E \mid A_{1}\right)}{P\left(A_{1}\right) \cdot P\left(E \mid A_{1}\right)+P\left(A_{2}\right) \cdot P\left(E \mid A_{2}\right)}\)
\(=\frac{\frac{6}{366} \times \frac{8}{10}}{\frac{6}{366} \times \frac{8}{10}+\frac{360}{366} \times \frac{2}{10}}=\frac{48}{768}=0.0625\)
(v) (a): Required probability = 1 - P(A₁ I E)
= 1 - 0.0625 = 0.9375 ≈ 0.94.

Let B and G denote the boy and girl respectively. If a family has 4 children then each of four children can either boy or girl.
Sample space is given by
S = {BBBB, BBBG, BBGB, BGBB, BBGG, BGBG, BGGB, BGGG,GBBB,GBBG,GBGB,GBGG,GGBB,GGBG, GGGB, GGGG}
(i) (d): Let E = All children are girls.
ஃ E = {GGGG} i.e., n(E) = 1
F = Elder child is a boy
ஃ F = {BBBB, BBBG, BBGB, BGBB, BBGG, BGBG, BGGB, BGGG} i.e., n(F) = 8
Now, n(EറF) = 0
\(\therefore P(E \mid F)=\frac{(E \cap F)}{n(F)}=0\)
(ii) (a): Let E = All are boys
ஃ E = {BBBB} i.e., n(E) = 1
F = Two elder children are boys
ஃ F = {BBBB, BBBG, BBGB, BBGG} i.e., n(F) = 4
Now, n(EറF) = 1
\(\therefore P(E \mid F)=\frac{n(E \cap F)}{n(F)}=\frac{1}{4}\)
(iii) (c) : Let E = Two middle children are boys.
ஃ E = {BBBB, BBBG, GBBB, GBBG} i.e., n(E) = 4
F = Eldest child is a girl
ஃ F = {GBBB, GBBG, GBGB, GBGG, GGBB, GGBG, GGGB, GGGG} i.e., n(F) = 8
Now, n(EറF) = 2
\(\therefore P(E \mid F)=\frac{2}{8}=\frac{1}{4}\)
(iv) (b): Let E = All are boys.
ஃ E = {BBBB} i.e., n(E) = 1
F = At most one child is girl.
ஃ F = {BBBB, BBBG, BBGB, BGBB, GBBB}
i.e., n(F) = 5
now, \(n(E \cap F)=1\)
\(\therefore P(E \mid F)=\frac{1}{5}\)
(v) (a): Let E = All are boys.
ஃ  E = {BBBB} i.e., n (E) = 1
F = At least three of the children are boys.
ஃ F = {BBBB, BBBG, BBGB, BGBB, GBBB} i.e., n(F) = 5
Now, n(EറF) = 1
\(\therefore P(E \mid F)=\frac{1}{5}\)

(i) (b): Total number of tickets = 50
Let event A = First ticket shows even number and B = Second ticket shows even number
Now, P(Both tickets show even number) = P(A)・P(B | A)
\(=\frac{25}{50} \cdot \frac{24}{49}=\frac{12}{49}\)
(ii) (c): Let the event A = First ticket shows odd number and B = Second ticket shows odd number
P (Both tickets show odd number)
\(=\frac{25}{50} \times \frac{24}{49}=\frac{12}{49}\)
(iii) (c): Required probability = P(one number is a multiple of 4 and other is a multiple of 5)
= P(multiple of 5 on first ticket and multiple of 4 on second ticket) + P(multiple of 4 on first ticket and multiple of 5 on second ticket)
\(=\frac{10}{50} \cdot \frac{12}{49}+\frac{12}{50} \times \frac{10}{49}=\frac{12}{245}+\frac{12}{245}=\frac{24}{245}\)
(iv) (d): Required probability = P(one ticket with prime number and other ticket with a multiple of 4)
\(=2\left(\frac{15}{50} \times \frac{12}{49}\right)=\frac{36}{245}\)
(v) (b): Let the event A = First ticket shows even number and B = Second ticket shows odd number.
Now, P(First ticket shows an even number and second ticket shows an odd number) = P(A)・P(B | A)
\(=\frac{25}{50} \times \frac{25}{49}=\frac{25}{98}\)

(i) (b): Since, it is given that during prime time husband is watching T.V. 70%of the time
∴ Required probability = 1 - P(husband is watching television during prime time)
= 1 - 0.7 = 0.3
(ii) (b): Let H be the event that husband is watching TV, W be the event that wife is watching TV.
Then, P(H) = 0.7, P(H) = 0.3
P(W I H) = 0.3 arltl P(W I H) = 0.4
∴ Required probability = P(H I W)
\(=\frac{P(\mathrm{H}) \cdot P(\mathrm{~W} \mid \mathrm{H})}{P(\mathrm{H}) \cdot P(\mathrm{~W} \mid \mathrm{H})+P(\overline{\mathrm{H}}) P(\mathrm{~W} \mid \overline{\mathrm{H}})} \)
\(=\frac{0.7 \times 0.3}{0.7 \times 0.3+0.4 \times 0.3}=\frac{0.21}{0.33}=\frac{7}{11}\)
(iii) (a) : Required probability = P(H \(\cap\) W)
= P(H)P(W I H) = 0.7 x 0.3 = 0.21
(iv) (b): Required probability = P(W) = \(\frac{P(\mathrm{H} \cap \mathrm{W})}{P(\mathrm{H} \mid \mathrm{W})}\)
\(=\frac{0.21}{7 / 11}=\frac{21}{100} \times \frac{11}{7}=\frac{33}{100}=0.33\)
(v) (b): Required probability = P(H I W)
\(=1-P(\mathrm{H} \mid \mathrm{W})=1-\frac{7}{11}=\frac{4}{11}\)

(i) (c): It is given that if India loose any match, then the probability that it wins the next match is 0.3.
∴ Required probability = 0.3
(ii) (d): It is given that, if India loose any match, then the probability that it wins the next match is 0.3.
∴ Required probability = 1 - 0.3 = 0.7
(iii) (b): Required probability = P(lndia losing first match) . P(India losing second match when India has already lost first match)
= 0.4 x 0.7 = 0.28
(iv) (d): Required probability = P(lndia winning first match) . P(India winning second match if India has already won first match) P(lndia winning third match if India has already won first two matches)
= 0.6 x 0.4 x 0.4 = 0.096
(v) (c): Required probability = P(Win 1^st match) P(Lose 2^nd match) P(Lose 3^rd match) + P (Lose 1^st match) P(Win 2^nd match) P(Lose 3^rd match) + P(Lose 1^st match) P(Lose 2^nd match) P(Win 3^rd match)
= 0.6 x (1 - 0.4)・(1- 0.3) + (1 - 0.6)・(0.3)(1 - 0.4) + (1 - 0.6) (1 - 0.3) (0.3)
= 0.6 x 0.6 x 0.7 + 0.4 x 0.3 x 0.6 + 0.4 x 0.7 x 0.3
= 0.252 + 0.072 + 0.084 = 0.408

Let A be the event that the student is selected for LIC AAO, B be the event that the student is selected for SSC CGL and C be the event that the student is selected for Bank P.O.
Then, P(A) = a; P(B) = band P(C) = c
(i) (b): We have, \(P(A \cup B \cup C)=0.5\)
\(\Rightarrow \ 1-P(\overline{A \cup B \cup C})=0.5 \)
\(\Rightarrow 1-P(\bar{A} \cap \bar{B} \cap \bar{C})=0.5 \)
\(\Rightarrow 1-P(\bar{A}) \cdot P(\bar{B}) \cdot P(\bar{C})=0.5 \)
\(\Rightarrow\) 1 - (I - a) (1 - b) (1 - c) = 0.5
\(\Rightarrow\) 1 - [ (I - a - b + ab )(1 - c)] = 0.5
\(\Rightarrow\) 1 - [1 - c - a + ac - b + bc + ab - abc] = 0.5
\(\Rightarrow\) a + b + c - ab - bc - ca + abc = 0.5 .......(i)
(ii) (c): We have, P(selection in at least two competitive exams) = 0.4
\( \Rightarrow \ P(A \cap B \cap \bar{C})+P(\bar{A} \cap B \cap C)+P(A \cap \bar{B} \cap C)+ P(A \cap B \cap C)=0.4 \)
\(\Rightarrow\) ab(1 - c) + (1 - a)bc + a(1 - b)c + abc = 0.4
\(\Rightarrow\) ab - abc + bc - abc + ac - abc + abc = 0.4
\(\Rightarrow\) ab + bc + ac - 2abc= 0.4 ......................(ii)
(iii) (a) : We have, P(selection in exactly two examinations) = 0.3
\(\Rightarrow P(A \cap B \cap \bar{C})+P(\bar{A} \cap B \cap C)+P(A \cap \bar{B} \cap C)=0.3\)
\(\Rightarrow\) ab(1 - c) + (1 - a)bc + a(1 - b)c = 0.3
\(\Rightarrow\) ab + bc + ac - 3abc = 0.3 ...................(iii)
On subtracting (iii) from (ii), we get
abc = 0.1
(iv) (b): On substituting the value of abe in (ii), we get
ab + bc + ac = 0.6
(v) (a): On substituting the value of ab + bc + ac and
abc in (i), we get
a + b + c - 0.6 + 0.1 = 0.5
\(\Rightarrow\) a + b + c = 1

Let E₁, E₂, E₃ be the events of drawing a bolt produced by machine A, B and C, respectively.
Then \(P\left(E_{1}\right)=\frac{30}{100}=\frac{3}{10}, P\left(E_{2}\right)=\frac{20}{100}=\frac{1}{5}\) and \(P\left(E_{3}\right)=\frac{50}{100}=\frac{1}{2}\)
Also, let E be the event of drawing a defective bolt.
Then \(P\left(E \mid E_{1}\right)=\frac{5}{100}=\frac{1}{20}, P\left(E \mid E_{2}\right)=\frac{2}{100}=\frac{1}{50}\)
\(P\left(E \mid E_{3}\right)=\frac{4}{100}=\frac{1}{25}\)
(i) (b): Probability that the defective bolt is manufactured by machine A = P(E₁ | E)
\(=\frac{P\left(E \mid E_{1}\right) \times P\left(E_{1}\right)}{P\left(E \mid E_{1}\right) \cdot P\left(E_{1}\right)+P\left(E \mid E_{2}\right) \cdot P\left(E_{2}\right)+P\left(E \mid E_{3}\right) \cdot P\left(E_{3}\right)}\) [Using Bayes' Theorem]
\(=\frac{\frac{1}{20} \times \frac{3}{10}}{\frac{1}{20} \times \frac{3}{10}+\frac{1}{50} \times \frac{1}{5}+\frac{1}{25} \times \frac{1}{2}}=\frac{3}{200} \times \frac{1000}{39}=\frac{5}{13}\)
(ii) (b): Probability that the defective bolt is manufactured by machine B = P(E₂ I E)
\( =\frac{P\left(E \mid E_{2}\right) \cdot P\left(E_{2}\right)}{P\left(E \mid E_{1}\right) \cdot P\left(E_{1}\right)+P\left(E \mid E_{2}\right) \cdot P\left(E_{2}\right)+P\left(E \mid E_{3}\right) \cdot P\left(E_{3}\right)} \)
\(=\frac{\frac{1}{50} \times \frac{1}{5}}{\left(\frac{1}{20} \times \frac{3}{10}\right)+\left(\frac{1}{50} \times \frac{1}{5}\right)+\left(\frac{1}{25} \times \frac{1}{2}\right)} \)
\(=\frac{1}{250} \times \frac{1000}{39}=\frac{4}{39} \approx 0.1\)
(iii) (c) : Probability that the defective bolt is manufactured by machine C = P(E₃ I E)
\(=\frac{P\left(E \mid E_{3}\right) \cdot P\left(E_{3}\right)}{P\left(E \mid E_{1}\right) \cdot P\left(E_{1}\right)+P\left(E \mid E_{2}\right) \cdot P\left(E_{2}\right)+P\left(E \mid E_{3}\right) \cdot P\left(E_{3}\right)}\)
\(=\frac{\frac{1}{25} \times \frac{1}{2}}{\left(\frac{1}{20} \times \frac{3}{10}\right)+\left(\frac{1}{50} \times \frac{1}{5}\right)+\left(\frac{1}{25} \times \frac{1}{2}\right)}\)
\(=\frac{1}{50} \times \frac{1000}{39}=\frac{20}{39}\)
(iv) (a): Probability that the defective bolt is not manufactured by machine B i.e., it is manufactured by machine A or C \(=\frac{15}{39}+\frac{20}{39}=\frac{35}{39}\)
(v) (c): Required probability \(=\frac{15}{39}+\frac{4}{39}=\frac{19}{39} \approx 0.5\)

We have, \(P\left(E_{1}\right)=P\left(E_{2}\right)=P\left(E_{3}\right)=\frac{1}{3}\)
(i) (a) : P(E l E₁) = Probability of drawing red and white ball, if box I is selected.
= P(red) x P(white) + P(white) x P(red)
\(=\frac{2}{6} \times \frac{1}{6}+\frac{1}{6} \times \frac{2}{6}=\frac{4}{36}=\frac{1}{9}\)
(ii) (a): P(E l E₁) = Probability of drawing red and white balls, if box II is selected
= P(red) x P(white) + P(white) x P(red)
\(=\frac{3}{6} \times \frac{2}{6}+\frac{2}{6} \times \frac{3}{6}=\frac{12}{36}=\frac{1}{3}\)
(iii) (c): p(E IE3) = Probability of drawing red and white balls, if box III is selected.
= P(red) x P(white) + P(white) x P(red)
\(=\frac{1}{6} \times \frac{3}{6}+\frac{3}{6} \times \frac{1}{6}=\frac{6}{36}=\frac{1}{6}\)
(iv) (d): \(\sum_{i=1}^{3} P\left(E \mid E_{i}\right)=P\left(E \mid E_{1}\right)+P\left(E \mid E_{2}\right)+P\left(E \mid E_{3}\right)\)
\(=\frac{4}{36}+\frac{12}{36}+\frac{6}{36}=\frac{4+12+6}{36}=\frac{22}{36}=\frac{11}{18}\)
(v) (b): Using Bayes' theorem
\( P\left(E_{2} \mid E\right)=\frac{P\left(E \mid E_{2}\right) \cdot P\left(E_{2}\right)}{P\left(E \mid E_{1}\right) \cdot P\left(E_{1}\right)+P\left(E \mid E_{2}\right) \cdot P\left(E_{2}\right)+P\left(E \mid E_{3}\right) \cdot P\left(E_{3}\right)} \)
\(=\frac{\frac{1}{3} \times \frac{1}{3}}{\left(\frac{1}{9} \times \frac{1}{3}\right)+\left(\frac{1}{3} \times \frac{1}{3}\right)+\left(\frac{1}{6} \times \frac{1}{3}\right)} \)
\(=\frac{\frac{1}{3}}{\frac{1}{9}+\frac{1}{3}+\frac{1}{6}}=\frac{\frac{1}{3}}{\frac{11}{18}}=\frac{1}{3} \times \frac{18}{11}=\frac{6}{11}\)

Let A be the event that Nisha is selected and B be the event that Ayushi is selected. Then, we have
\(P(A)=\frac{1}{3}\)
\(\Rightarrow P(\bar{A})=1-\frac{1}{3}=\frac{2}{3}=P(\text { Nisha is not selected })\)
\(P(B)=\frac{1}{2}\)
\(\Rightarrow P(\bar{B})=1-\frac{1}{2}=\frac{1}{2}=P \text { (Ayushi is not selected) }\)
(i) (c): P (both are selected) = \(P(A \cap B)=P(\bar{A}) \cdot P(\bar{B})\)
\(=\frac{1}{3} \times \frac{1}{2}=\frac{1}{6}\)
(ii) (d): P (both are rejected) = \(=P(\bar{A} \cap \bar{B})=P(\bar{A}) \cdot P(\bar{B})\)
\(=\frac{2}{3} \times \frac{1}{2}=\frac{1}{3}\)
(iii) (d): P (only one of them is selected)
\( =P(A \cap \bar{B})+P(\bar{A} \cap B)=P(A) \cdot P(\bar{B})+P(\bar{A}) \cdot P(B)\)
\(=\frac{1}{3} \times \frac{1}{2}+\frac{2}{3} \times \frac{1}{2}=\frac{1}{6}+\frac{2}{6}=\frac{3}{6}=\frac{1}{2}\)
(iv) (a): P (at least one of them is selected)
= 1 - P (Both are rejected) = \(1-\frac{1}{3}=\frac{2}{3}\)
(v) (a): Let E₁ be the event that Nisha is selected for post I and E₂ be the event that Nisha is selected for post II.
\(\therefore\) P (Nisha is selected for atleast one post)
\(=P\left(E_{1} \cup E_{2}\right)=P\left(E_{1}\right)+P\left(E_{2}\right)-P\left(E_{1} \cap E_{2}\right)\)
\(=\frac{1}{6}+\frac{1}{5}-\frac{1}{6} \times \frac{1}{5}=\frac{10}{30}=\frac{1}{3}\)

(i) (b): P(X = 2) = (Probability of not getting six at first throw) \(\times\) (Probability of getting six at second throw)
\(=\frac{5}{6} \times \frac{1}{6}=\frac{5}{6^{2}}\)
(ii) (c): P(X =4) = \(\frac{5}{6} \times \frac{5}{6} \times \frac{5}{6} \times \frac{1}{6}=\frac{5^{3}}{6^{4}}\)
(iii) (c) : P(X ≥  2) = 1 - P(X < 2)
\(=1-P(X=1)=1-\frac{1}{6}=\frac{5}{6}\)
(iv) (a): \(P(X \geq 6)=\left(\frac{5}{6}\right)^{5} \times \frac{1}{6}+\left(\frac{5}{6}\right)^{6} \times \frac{1}{6}+\ldots \infty\)
\(=\frac{5^{5}}{6^{6}}\left[1+\frac{5}{6}+\left(\frac{5}{6}\right)^{2}+\ldots \infty\right]=\frac{5^{5}}{6^{6}}\left[\frac{1}{1-\frac{5}{6}}\right]=\left(\frac{5}{6}\right)^{5}\)
(v) (d): \(P(X \geq 4)=\left(\frac{5}{6}\right) \cdot \frac{1}{6}+\left(\frac{5}{6}\right)^{4} \cdot \frac{1}{6}+\ldots \ldots\)
\( =\frac{5^{3}}{6^{4}}\left[1+\left(\frac{5}{6}\right)+\left(\frac{5}{6}\right)^{2}+\ldots . .\right] \)
\(=\frac{5^{3}}{6^{4}}\left[\frac{1}{1-\left(\frac{5}{6}\right)}\right]=\left(\frac{5^{3}}{6^{4}}\right) \cdot 6=\frac{5^{3}}{6^{3}}\)

Let E denote the event that student has failed in Economics and M denote the event that student has failed in Mathematics.
\(\therefore \quad P(E)=\frac{50}{100}=\frac{1}{2}, P(M)=\frac{35}{100}=\frac{1}{20} \text { and } P(E \cap M)=\frac{25}{100}=\frac{1}{4} \)
(i) (d): Required probability = P(E | M)
\(=\frac{P(E \cap M)}{P(M)}=\frac{1 / 4}{7 / 20}=\frac{1}{4} \times \frac{20}{7}=\frac{5}{7}\)
(ii) (c): Required probability = P(M | E)
\(=\frac{P(M \cap E)}{P(E)}=\frac{1 / 4}{1 / 2}=\frac{1}{2}\)
(iii) (c) : Required probability\(=P\left(E^{\prime} \cup M^{\prime}\right)\)
\(=P(E \cap M)^{\prime}\)
\(=1-P(E \cap M)=1-\frac{1}{4}=\frac{3}{4} \)
(iv) (a): Required probability = P(E \(\cup\) M)
= P(E) + P(M) - P(E \(\cap\) M)
\(=\frac{5}{10}+\frac{7}{20}-\frac{1}{4}=\frac{12}{20}=\frac{3}{5}\)
(v) (d): Required probability = P(M' I E)
\(=\frac{P\left(M^{\prime} \cap E\right)}{P(E)}=\frac{P(E)-P(E \cap M)}{P(E)}=\frac{\frac{1}{2}-\frac{1}{4}}{\frac{1}{2}}=\frac{1}{2}\)

Sample space is given by {MFSD, MFDS, MSFD, MSDF, MDFS, MDSF, FMSD, FMDS, FSMD, FSDM, FDMS, FDSM, SFMD, SFDM, SMFD, SMDF, SDMF, SDFM DFMS, DFSM, DMSF, DMFS, DSMF, DSFM}, where F, M, D and S represent father, mother, daughter and son respectively.
∴ n(S) = 24
(i) (a): Let A denotes the event that daughter is at one end.
∴ n(A) = 12
and B denotes the event that father and mother are in the middle.
∴ n(B) = 4
Also, n(A \(\cap\) B) = 4
\(\therefore \ P(A \mid B)=\frac{P(A \cap B)}{P(B)}=\frac{4 / 24}{4 / 24}=1\)
(ii) (b): Let A denotes the event that mother is at right end.
∴ n(A) = 6
and B denotes the event that son and daughter are together.
∴ n(B) = 12
Also, n(A \(\cap\) B) = 4
\(\therefore \ P(A \mid B)=\frac{P(A \cap B)}{P(B)}=\frac{4 / 24}{12 / 24}=\frac{1}{3}\)
(iii) (c) : Let A denotes the event that father and mother are in the middle.
∴ n(A) = 4
and B denotes the event that son is at right end.
∴ n(B) = 6
Also, n(A \(\cap\) B) = 2
\(\therefore \ P(A \mid B)=\frac{P(A \cap B)}{P(B)}=\frac{2 / 24}{6 / 24}=\frac{1}{3}\)
(iv) (d): Let A denotes the event that father and son are standing tbgether.
∴ n(A) = 12
and B denotes the event that mother and daughter are standing together.
∴ n(B) = 12
Also, n(A \(\cap\) B) = 8
\(\therefore \ P(A \mid B)=\frac{P(A \cap B)}{P(B)}=\frac{8 / 24}{12 / 24}=\frac{2}{3}\)
(v) (a): Let A denotes the event that father and mother are on either of the ends.
∴ n(A) = 4
and B denotes the event that son is at second position from the right end.
∴ n(B) = 6
Also, n(A \(\cap\) B) = 2
\(\therefore \ P(A \mid B)=\frac{P(A \cap B)}{P(B)}=\frac{2 / 24}{6 / 24}=\frac{1}{3}\)

(i) (a): Clearly, P(T₂ winning a match against T₁)
= P(T₁ losing) = \(\frac{1}{5}\)
(ii) (d): Clearly, P(T₂ drawing a match against T₁)
= P(T₁ drawing) = \(\frac{3}{10}\)
(iii) (d): According to given information, we have the following possibilities for the values of X and Y.

Now, P(X > Y) = P(X = 4, Y = 0) + P(X = 3, Y = 1)
= P(T₁ win) P(T₁ win) + P(T₁ win) P(match draw) + P(match draw) P(T₁ win)
\(=\frac{1}{2} \cdot \frac{1}{2}+\frac{1}{2} \cdot \frac{3}{10}+\frac{3}{10} \cdot \frac{1}{2}=\frac{5+3+3}{20}=\frac{11}{20}\)
(iv) (c): P(X = Y) = P(X = 2, Y = 2)
= P(T₁ win) P(T₂ win) + P(T₂ win) P(T₁ win) + P(match draw) P(match draw)
\(=\frac{1}{2} \cdot \frac{1}{5}+\frac{1}{5} \cdot \frac{1}{2}+\frac{3}{10}, \frac{3}{10}=\frac{1}{10}+\frac{1}{10}+\frac{9}{100}=\frac{29}{100}\)
(v) (a): From the given information, it is clear that maximum sum of X and Y can be 4, therefore P(X + Y= 8) = 0

Let A be the event of commiting an error and E₁, E₂ and E₃ be the events that Vinay, Sonia and Iqbal, processed the form.
(i) (b): Required probability = P(A I E₂)
\(=\frac{P\left(A \cap E_{2}\right)}{P\left(E_{2}\right)}=\frac{\left(0.04 \times \frac{20}{100}\right)}{\left(\frac{20}{100}\right)}=0.04\)
(ii) (c): Required probability = P(A  \(\cap\) E₂)
\(=0.04 \times \frac{20}{100}=0.008\)
(iii) (b): Total probability is given by
\( P(A) =P\left(E_{1}\right) \cdot P\left(A \mid E_{1}\right)+P\left(E_{2}\right) \cdot P\left(A \mid E_{2}\right)+P\left(E_{3}\right) \cdot P\left(A \mid E_{3}\right) \)
\(=\frac{50}{100} \times 0.06+\frac{20}{100} \times 0.04+\frac{30}{100} \times 0.03=0.047\)
(iv) (d): Using Bayes' theorem, we have
\( P\left(E_{1} \mid A\right)=\frac{P\left(E_{1}\right) \cdot P\left(A \mid E_{1}\right)}{P\left(E_{1}\right) \cdot P\left(A \mid E_{1}\right)+P\left(E_{2}\right) \cdot P\left(A \mid E_{2}\right)} +P\left(E_{3}\right) \cdot P\left(A \mid E_{3}\right) \)
\(=\frac{0.5 \times 0.06}{0.5 \times 0.06+0.2 \times 0.04+0.3 \times 0.03}=\frac{30}{47}\)
(v) (d): \(\sum_{i=1}^{3} P\left(E_{i} \mid A\right)=P\left(E_{1} \mid A\right)+P\left(E_{2} \mid A\right)+P\left(E_{3} \mid A\right)\)
= 1 [ ∵ Sum of posterior probabilities is 1 ] .