Let H₁, H₂, H₃ and H₄ be the events of hitting the plane by the anti-aircraft gun in the first second, third and fourth shot respectively.
Let H be the event that anti-aircraft gun hits the plane. Therefore \(\bar{H}\) is the event that the plane is not shot down. Given that
P(H₁) = 0.2 ⇒ P(\({ \bar { H } }_{ 1 }\)) = 1 - P(H₁) = 0.8
P(H₂) = 0.4 ⇒  P(\({ \bar { H } }_{ }\)) = 1-P(H₂) = 0.6
P(H₃) = 0.2  ⇒  P(\({ \bar { H } }_{ 3 }\)) = 1-P(H₃) = 0.8
P(H₄) = 0.1  ⇒  P(\({ \bar { H } }_{ 4 }\)) = 1-P(H₄) = 0.9
The probability that the gun hits the plane is
P(H) = 1 - P\((\bar { H } )\) = 1 - \((\overline { { H }_{ 1 }\cup { H }_{ 2 }\cup { H }_{ 3 }\cup { H }_{ 4 } } )\)
= 1 - P\(\left( { \bar { H } }_{ 1 }\cap { \bar { H } }_{ 2 }{ \bar { H } }_{ 3 }\cap { \bar { H } }_{ 4 } \right) \)
= 1 - P\(P(\bar { { H }_{ 1 } } )P(\bar { { H }_{ 2 } } )P(\bar { { H }_{ 3 } } )P(\bar { { H }_{ 4 } } )\)
= 1 - (0.8)(0.6)(0.8)(0.9) = 1 - 0.3456
P(H) = 0.6544

Let A₁ be the event of selecting urn-I and A₂ be the event of selecting urn-II.
Let B be the event of selecting 2 red balls.
We have to find the total probability of event B. That is, P(B).
Clearly A₁ and A₂A₁ are mutually exclusive and exhaustive events.

	Red balls	Blue balls	Total
Urn-I	8	4	12
Urn-II	5	10	15
Total	13	14	27

We have, \(P(A_1)={1\over2},P(B/A_1)={8c_2\over 12c_2}={14\over 33}\)
\(P(A_2)={1\over2},P(B/A_2)={5c_2\over 15c_2}={2\over21}\)

We know P(B) = P(A₁).P(B/A₁) + P(A₂).P(B/A₂)
P(B) =\({1\over2}.{14\over33}+{1\over2}.{2\over21}={20\over77}.\)

Let A₁, A₂ and A₃ be the event of A, B, C becoming managers of the company respectively. Let X be the event that the office canteen will be improved.
Then, \(P\left(A_1\right)=\frac{5}{10}=0.5 \)
\(P\left(A_2\right)=\frac{3}{10}=0.3 \)
\(P\left(A_3\right)=\frac{2}{10}=0.2 \)
\(P\left(X / A_1\right)=0.4 \)
\(P\left(X / A_2\right)=0.5\)
\(P\left(X / A_3\right)=0.3\)
\(P\left(A_2 / X\right)=\frac{P\left(A_2\right) P\left(X / A_2\right)}{P\left(A_1\right) P\left(X / A_1\right)+P\left(A_2\right) P\left(X / A_2\right)}+P\left(A_3\right) P\left(X / A_3\right)\)
\(=\frac{0.3(0.5)}{0.5(0.4)+0.3(0.5)+0.2(0.3)} \)
\(=\frac{0.15}{0.2+0.15+0.06} \)
\(=\frac{0.15}{0.41}=\frac{15}{41}\)

Let A₁, A₂ and A₃ be the event that the units of PVC pipes produced by plants X, Y, Z respectively.
Let B be the event of selected item is defective.
Then \(P\left(A_1\right)=\frac{2000}{10,000}=0.2, P\left(B / A_1\right)=0.03\)
\(P\left(A_2\right)=\frac{3000}{10,000}=0.3, P\left(B / A_2\right)=0.04\)
\(P\left(A_3\right)=\frac{5000}{10,000}=0.5, P\left(B / A_3\right)=0.02\)
(i) P (the selected pipe is defective) = P(B)
\(=P\left(A_1\right) \cdot P\left(B / A_1\right)+P\left(A_2\right) \cdot P\left(B / A_2\right)+P\left(A_3\right) \cdot P\left(B / A_3\right)\)
\(=0.2(0.03)+0.3(0.04)+0.5(0.02)\)
\(=0.006+0.012+0.010\)
\(=0.028=\frac{28}{1000}=\frac{7}{250}\)
(i) By Bayes theorem
\(P\left(A_2 / B\right)=\frac{P\left(A_2\right) \cdot P\left(B / A_2\right)}{P\left(A_1\right) P\left(B / A_1\right)+P\left(A_2\right) P\left(B / A_2\right)} +P\left(A_3\right) P\left(B / A_3\right)\)
\(=\frac{0.3(0.04)}{0.2(0.03)+0.3(0.04)+0.5(0.02)}\)
\(=\frac{0.012}{0.028}=\frac{12}{28}=\frac{3}{7}\)

Let A₁ be the event of selecting balls from urn I and A₂ be the event of selecting urn II.
Let B be the event of a black ball is drawn from it.
Then we have \(P\left(A_1\right)=P\left(A_2\right)=\frac{1}{2}\)
\(P\left(B / A_1\right)=\frac{6 C_1}{10 C_1} ; P\left(B / A_2\right)=\frac{2 C_1}{1 C_1}\)
\(\text {(i) } P(B)=P\left(A_1\right) \cdot P\left(B / A_1\right)+P\left(A_2\right): P\left(B / A_2\right)\)
\(=\frac{1}{2} \cdot \frac{6 C_1}{10 C_1}+\frac{1}{2} \cdot \frac{2 C_1}{4 C_1} \)
\(=\frac{1}{2}\left[\frac{6}{10}+\frac{2}{4}\right]=\frac{1}{2}\left[\frac{3}{5}+\frac{1}{2}\right]=\frac{1}{2}\left[\frac{6+5}{10}\right]\)
\(=\frac{11}{20} \)
(ii) P(I/B)
\(=\frac{P(B/I)P(I)}{P(B)}\)
\(=\frac{\frac{6}{10}\times \frac{1}{2}}{\frac{11}{20}}=\frac{6}{20}/\frac{11}{20}=\frac{6}{11}\)

Let A₁ be the event that the items are produced by Machine-1, A₂ be the event that items are produced by Machine-II.
Let B be the event of drawn a defective item.
Then, \(P(B)=P\left(A_1\right) \cdot P\left(B / A_1\right)+P\left(A_2\right) \cdot P\left(B / A_2\right)\)
We have, \(P\left(A_1\right)=0.60, P\left(B / A_1\right)=0.02\)
\(P\left(A_2\right)=0.40, P\left(B / A_2\right)=0.04\)
\(P(B)=P\left(A_1\right) \cdot P\left(B / A_1\right)+P\left(A_2\right) \cdot P\left(B / A_2\right)\)
\(=0.60 \times 0.02+0.40 \times 0.04 \)
\(=0.012+0.016 \)
\(=0.028 \)
\(=\frac{.28}{1000}=\frac{7}{250}\)

The sample space is S = {H, T} \(\times\) {H, T}
\(S=\{(H, H),(H, T),(T, H),(T, T)\}\)
and E = {{H, H), (H, T)}
F = {(H, H), (T, H)}
E\(\cup \)F = {(H, H), (H, T), (T, H)}
E\(\cap \)F = {(H, H)}
(i) P(E\(\cup \)F) = P(E) + P(F) - P(E\(\cap \)F) or \(\left( \frac { n(E\cup F) }{ n(S) } \right) \)
=\(\frac { 2 }{ 4 } +\frac { 2 }{ 4 } -\frac { 1 }{ 4 } =\frac { 3 }{ 4 } \)
(ii) P(E/F) =\(\frac { P(E\cap F) }{ P(F) } =\frac { (1/4) }{ (2/4) } =\frac { 1 }{ 2 } \)
(iii) \(P(\bar { E } /F)=\frac { P(\bar { E } \cap F) }{ P(F) } \)
=\(\frac { P(F)-P(E\cap F) }{ P(F) } \)
=\(\frac { (2/4)-(1/4) }{ (2/4) } \)
=\(\frac { 1 }{ 2 } \)
We have \(P(E\cap F)=\frac { 1 }{ 4 } \)
P(E) = \(\frac{2}{4}\), P(F) = \(\frac{2}{4}\)
P(E)P(F) =  \(\frac { 2 }{ 4 } .\frac { 2 }{ 4 } =\frac { 2 }{ 5 } \)
⇒ P(E\(\cap \)F) =P(E).P(F)
Therefore E and F are independent events.

Let L be the' event of select a leap year
Then, \(P(L)=\frac{1}{4}, P(\bar{L})=\frac{3}{4}\)
If A is the year containing 53 Sundays
Then, \(P(A / L)=\frac{2}{7} \text { and } P(A / \bar{L})=\frac{1}{7}\)
\(\text {(i) } P(A)=P(L \cap A)+P(\bar{L} \cap A)\)
\(=P(L) \cdot P(A / L)+P(\bar{L}) \cdot P(A / \bar{L})\)
\(=\frac{1}{4} \times \frac{2}{7}+\frac{3}{4} \times \frac{1}{7}=\frac{5}{28}\)
\(\text {(ii) } P(L \cap A)=P(L) \cdot P(A / L)\)
\(=\frac{1}{4} \times \frac{2}{7} \)
\(=\frac{1}{14}\)

Let W, W, be the event that the white ball is drawn from bag. 1 and bag 2 respectively. Also let B,B, be the event that the black ball is drawn from bag 1 and 2 respectively.
Then \(P\left(W_1\right)=\frac{5}{8}, \quad P\left(W_2\right)=\frac{4}{10}\)
\(P\left(B_1\right)=\frac{3}{8}, \quad P\left(B_2\right)=\frac{6}{10}\)
(i) P (Both are white \(=P\left(W_1 \cap W_2\right)\)
\(=P\left(W_1\right) \cdot P\left(W_2\right)\)
\(=\frac{5}{8} \times \frac{4}{10}=\frac{1}{4}\)
(ii) P (Both are black) \(=P\left(B_1 \cap B_2\right) \)
\(=P\left(B_1\right) \cdot P\left(B_2\right) \)
\(=\frac{3}{8} \times \frac{6}{10}=\frac{9}{40}\)
(ii) P (One white and one black) \(=P\left(W_1 \cap B_2\right)+P\left(W_2 \cap B_1\right) \)
\(=P\left(W_1\right) \cdot P\left(B_2\right)+P\left(W_2\right) \cdot P\left(B_1\right) \)
\(=\frac{5}{8} \times \frac{6}{10}+\frac{4}{10} \times \frac{3}{8} \)
\(=\frac{3}{8}+\frac{3}{20}=\frac{15+6}{40}=\frac{21}{40} \)

Let A, B, C be the events that the problems solved by 3 students. Then,
\(P(A)=\frac{1}{3}, P(B)=\frac{1}{4}, P(C)=\frac{1}{5}\)
(i) P (Problem is solved) \(=P(A \cup B \cup C)\)
\(=1-P(\overline{A \cup B \cup C})\)
\(=1-P(\bar{A} \cap \bar{B} \cap \bar{C})\)
\(=1-P(\bar{A}) P(\bar{B}) P(\bar{C})\)
\(=1-\frac{2}{3} \times \frac{3}{4} \times \frac{4}{5} \)
\(=1-\frac{2}{5}=\frac{3}{5}\)
(ii) P (exactly one of them will solve)
\(=P(A \bar{B} \bar{C} \cup \bar{A} B \bar{C} \cup \bar{A} \bar{B} C)\)
\(=P(A) \cdot P(\bar{B}) \cdot P(\bar{C})+P(\bar{A}) \cdot P(B) \cdot P(\bar{C}) +P(\bar{A}) \cdot P(\bar{B}) \cdot P(C)\)

\(=\frac{1}{5}+\frac{2}{15}+\frac{1}{10}=\frac{1}{5}\left(1+\frac{2}{3}+\frac{1}{2}\right) \)
\(=\frac{6+4+3}{30}=\frac{13}{30}\)