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, \(P(E)=\frac{23}{35} \text { and } P(F)=\frac{22}{34}\)
\(\therefore \ P(F \cap E)=P(E) \cdot P(F \mid E)=\frac{23}{35} \cdot \frac{22}{34}=\frac{253}{595}\)

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
\(=\frac{2}{7}+\frac{5}{14}+\frac{1}{6}+\frac{4}{21}=1\)

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) (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.

(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)
\(=\frac{25}{50} \times \frac{25}{49}=\frac{25}{98}\)