Probability of an event cannot be negative. Here P(C) =-0.06
∴  the above set of events are not possible.
P(A) =\(\frac{1}{3}\), P(B)=\(\frac{1}{6}\), P(C) =\(\frac{2}{9}\), P(D)=\(\frac{5}{18}\)

Since A and B are mutually exclusive, P(A\(\cap \)B) = 0
Now, P(AUB) = P(A) + P(B) - P(A\(\cap \)B)
\(\Rightarrow \frac { 3 }{ 5 } =\frac { 1 }{ 2 } +p-0\)
\(\Rightarrow p=\frac { 3 }{ 5 } -\frac { 1 }{ 2 } =\frac { 6-5 }{ 10 } =\frac { 1 }{ 10 } \)
\(\therefore\) \(p=\frac { 1 }{ 10 } \)

Given P(A) = \(\frac { 1 }{ 2 } \), P(B) = \(\frac { 7 }{ 12 } \) and P(\(\bar { A } \cup \bar { B } \)) = \(\frac { 1 }{ 4 } \).
Now, \(P(\bar { A } \cup \bar { B } )=P(\overline { A\cap B } )=1-P(A\cap B)\)
\(\Rightarrow \frac { 1 }{ 4 } =1-P(A\cap B)\quad \Rightarrow P(A\cap B)=1-\frac { 1 }{ 4 } =\frac { 3 }{ 4 } \)
Now P(A) \(\times\) P(B) = \(\frac { 1 }{ 2 } \times \frac { 7 }{ 12 } =\frac { 7 }{ 24 } \)
\(\therefore P(A\cap B)\neq P(A)\times P(B)\)
Thus, A and B are not independent.

Let m be the event the selected student passes in mathematics and E be the event that the selected student passes in English.
\(\therefore P(M)=\frac { 2 }{ 3 } \) and P(M\(\cap \)E)=\(\frac { 1 }{ 3 } \)
\(\therefore P(E/M)=\frac { P(M\cap E) }{ P(M) } =\frac { \frac { 1 }{ 3 } }{ \frac { 2 }{ 3 } } =\frac { 1 }{ 3 } \times \frac { 3 }{ 2 } =\frac { 1 }{ 2 } \)

\(P(B)=0.5, P(A \cup B)=0.7, P(A \cap B)=0.2\)
\(\text {WKT } P(A \cup B)=P(A)+P(B)-P(A \cap B)\)
\(0.7=P(A)+0.5-0.2\)
\(0.7=P(A)+0.3\)
\(\therefore P(A)=0.4 \text {. }\)
To prove that A and B are independent
\(\text { i.e, } P(A \cap B)=P(A) \cdot P(B) \text { is true. }\)
\(P(A \cap B) =0.2 \)
\(P(A) \cdot P(B) =0.4 \times 0.5=0.20\)
\(\text {From (1) and (2), } P(A \cap B)=P(A) \cdot P(B)=0.2\)
Hence A and B are independent.