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

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