\(P(\bar{A} / B) =\frac{P(\bar{A} \cap B)}{P(B)} \)
\(=\frac{P(B)-P(A \cap B)}{P(B)} \)
\(=\frac{0.5-0.2}{0.5}=\frac{0.3}{0.5}=\frac{3}{5}\)

P(card will be 6 or smaller)
= \(\frac{5+5+5+5}{52}=\frac{20}{52}=\frac{5}{13}\)   [∵ 5 cards which are 6 or smaller from each variety]

If A and B are mutually exclusive, then
\(P(A \cap B)=0\)
But if A and B are independent, then
\(P(A \cap B)=P(A) \cdot P(B)\)
So if A and B are non empty, then they are not mutually exclusive and independent simultaneously.

When'a die is thrown twice, then the sanmple space
\(S=\{(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),\\ (2,1),(2,2),(2,3),(2,4),(2,5),(2,6), \\ (3,1),(3,2),(3,3),(3,4),(3,5),(3,6), \\ (4,1),(4,2),(4,3),(4,4),(4,5),(4,6), \\ (5,1),(5,2),(5,3),(5,4),(5,5),(5,6), \\ (6,1),(6,2),(6,3),(6,4),(6,5),(6,6)\}\)
n(S) = 36
A be the event, first die shows 5 and
B be the event, second die shows 5 then,
\(A =\{(5,1),(5,2),(5,3),(5,4),(5,5),(5,6)\}\)
\(n(A) =6\)
\(B =\{(1,5),(2,5),(3,5),(4,5),(5,5),(6,5)\} \)
\(n(B) =6 \)
\(A \cap B =\{(5,5)\}, \quad n(A \cap B)=1\)
Now, \(P(A \cup B) =P(A)+P(B)-P(A \cap B) \)
\(=\frac{n(A)}{n(S)}+\frac{n(B)}{n(S)}-\frac{n(A \cap B)}{n(S)} \)
\(=\frac{6}{36}+\frac{6}{36}-\frac{1}{36} \)
\(=\frac{11}{36}\)

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 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) \(=P\left(W_1 \cap B_2\right)+P\left(W_2 \cap B_1\right) \)
\(=P\left(W_1\right) \cdot P\left(B_2\right)+P\left(W_2\right) \cdot P\left(B_1\right) \)
\(=\frac{5}{8} \times \frac{6}{10}+\frac{4}{10} \times \frac{3}{8} \)
\(=\frac{3}{8}+\frac{3}{20}=\frac{15+6}{40}=\frac{21}{40} \)

\(\mathrm{P}(\mathrm{A} \cup \mathrm{B}) =\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A} \cap \mathrm{B}) \)
\(0.6 =0.35+\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A}) \cdot \mathrm{P}(\mathrm{B}) \)
\(0.25 =\mathrm{P}(\mathrm{B})-0.35 \mathrm{P}(\mathrm{B}) \)
\(=\mathrm{P}(\mathrm{B})(1-0.35) \)
\(=\mathrm{P}(\mathrm{B})(0.65) \)
\(\mathrm{P}(\mathrm{B}) =\frac{0.25}{0.65}=\frac{5}{13} \)

\(n(S) =2^{14}=16 \)
\(A =\left\{\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}|,| \begin{array}{ll} 0 & 1 \\ 1 & 0 \end{array}|,| \begin{array}{ll} 1 & 1 \\ 0 & 1 \end{array}|,| \begin{array}{ll} 1 & 0 \\ 1 & 1 \end{array}|,| \begin{array}{ll} 0 & 1 \\ 1 & 1 \end{array}|,| \begin{array}{ll} 1 & 1 \\ 1 & 0 \end{array} \mid\right\} \)
\(n(A) =6 \)
\(P(A) =\frac{6}{16}=\frac{3}{8} \)

Defective = 4
Not Defective = 8
\(\mathrm{n}(\mathrm{S})=12 \mathrm{C}_{2}=\frac{12 \times 11}{1 \times 2}=66\)
\(\mathrm{P} \text { (one defective) }=\frac{4 \mathrm{C}_{1} \times 8 \mathrm{C}_{1}}{12 \mathrm{C}_{2}}=\frac{32}{66}\)
\(\mathrm{P} \text { (two defective) }=\frac{4 \mathrm{C}_{2}}{12 \mathrm{C}_{2}}=\frac{6}{66}\)
\(P \text { (atleast one defective) }=\frac{32}{66}+\frac{6}{66}=\frac{19}{33}\)

Let I be the event of getting State Government service and C be the event of getting Central Government job.
Given that P(I) = 0.12, P(C) = 0.25, and \(P(I \cap C)\) = 0.07
(i) P( at least one of the two jobs) \(=P(I \text { or } C)=P(I \cup C)\)
\(=P(I)+P(C)-P(I \cap C) \)
\(=0.12+0.25-0.07=0.30\)
(ii) P(only one of the two jobs) = P [only I or only C]
=\(P\left( I\cap \overline { C } \right) +P\left( \overline { I } \cup C \right) \)
\(=\{0.12-0.07\}+\{0.25-0.07\}\)
\(=0.23 \)

Let S be the sample space and A be the event of taking 2 hundred rupee note.
THerefore, n(S) = 12c₂ = 66, n(A) = 4c₂ = 6 and \(\left( \overline { A } \right) \) = 66 - 6 = 60
Therefore, odds in favour of A is 6 : 60
That is, odds in favour of A is 1 : 10, and P(A) = \(\frac { 1 }{ 11 } \)