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

Probability of getting odd number = \(\frac { 3 }{ 6 } =\frac { 1 }{ 2 } \)
Probability of getting even numbers = 1-\(\frac { 1 }{ 2 } =\frac { 1 }{ 2 } \)
Now, probability of getting no odd number when the die is tossed twice = Probability of getting even number when the die is tossed thrice.
 \(\Rightarrow \quad \frac { 1 }{ 2 } \times \frac { 1 }{ 2 } \times \frac { 1 }{ 2 } =\frac { 1 }{ 8 } \)
\(\therefore\)  P(Odd number atleast once) = \(1-\frac { 1 }{ 8 } =\frac { 7 }{ 8 } \)

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)=\(\frac { 1 }{ 3 } \)
\(\therefore P(E/M)=\frac { P(M\cap E) }{ P(M) } =\frac { \frac { 1 }{ 3 } }{ \frac { 2 }{ 3 } } =\frac { 1 }{ 3 } \times \frac { 3 }{ 2 } =\frac { 1 }{ 2 } \)

Given \(P(\bar { A } )=0.6\)
\(\Rightarrow P(A)=1-P(\bar { A } )=1-0.6=0.4\)
P(B) = 0.7 and P(B/A) = 0.4
We know that P(B/A) = \(\frac { P(A\cap B) }{ P(A) } \)
\(\Rightarrow 0.4=\frac { P(A\cap B) }{ 0.4 } \)
\(\Rightarrow P(A\cap B)=0.16\)
Now, P(A/B) = \(\frac { P(A\cap B) }{ P(B) } =\frac { 0.16 }{ 0.7 } =0.2286\)
Also, P(A\(\cup \) B) = P(A) + P(B) - P(A\(\\ \cap \)B)
= 0.4 + 0.7 - 0.16 = 1.1 - 0.16 = 0.94

\(P(\overline { A } \cup \overline { B } )\)=\(\left( \overline { A\cap B } \right) \)(By de Morgan's law)
1-P(A∩B)=1-0.24
=0.76​

Let S be the sample space and A be the event of getting at least two heads.
Therefore, the event Ā denotes, getting at most one head.
n(S) = 2⁹ = 512, n(Ā) = 9C₀ + 9C₁= 1 + 9 = 10
\(P(\bar{A})=\frac{10}{512}=\frac{5}{256} \)
\( P(A)=1-P(\bar{A})=1-\frac{5}{256}=\frac{251}{256}\)

S = {Pack of  52 cards}
\(\therefore\) n(S) = 52
P(ace card or a king card) = P(ace card) + P(king card)
= \(\frac{4}{52}+\frac{4}{52}=\frac{8}{52}=\frac{2}{13}\)

The sample space is S = {HH, HT, TH, TT}
⇒ n(S) = 4
(i) Let A be the event of getting one head and one tail, then
A = {HT, TH}
n(A) = 2
\(\therefore\)  \(P(A)=\frac{n(B)}{n(S)}=\frac{4}{4}=1\)
(ii) Let A be the event of getting atmose two tails, then
\(\therefore\)  B = {HH,HT,TH,TT}
\(\therefore\)  \(P(B)=\frac{n(B)}{n(S)}=\frac{4}{4}=1\)

since the experiment has exactly the three possible mutually exclusive outcomes A, B and C, they must be exhaustive events.
\(\Rightarrow S=A\cup B\cup C\)
Therefore, by axioms of probability
\(P(A)\ge 0,P(B)\ge P(C)\ge 0\) and
\(P(A\cup B\cup C)=P(A)+P(B)+P(C)=P(S)=1\)
Even though P(A) + P(B) + P(C) = 0.421 + 0.527 + 0.042 = 0.990 < 1
therefore, the assignment is not permissible

Since the experiment has exactly the three possible mutually exclusive outcomes A, B and C, they must be exhaustive events.
\(\Rightarrow S=A\cup B\cup C\)
Therefore, by axioms of probability
\(P(A)\ge 0,P(B)\ge P(C)\ge 0\) and
\(P(A\cup B\cup C)=P(A)+P(B)+P(C)=P(S)=1\)
Given that \(P(A)=\frac { 2 }{ 5 } \ge ,\quad P(B)=\frac { 1 }{ 5 } \ge 0,\quad P(C)=\frac { 3 }{ 5 } \ge 0\)
But \(P(S)=P(A)+P(B)+P(C)=\frac { 2 }{ 5 } +\frac { 1 }{ 5 } +\frac { 6 }{ 5 } >1\)
Therefore the assignment is not permissible

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)
\(=\frac{P(B/I)P(I)}{P(B)}\)
\(=\frac{\frac{6}{10}\times \frac{1}{2}}{\frac{11}{20}}=\frac{6}{20}/\frac{11}{20}=\frac{6}{11}\)

P\( \left( A\cap \overline { B } \right) \) = P(A) - P(A∩B)
= 0.52 - 0.24 = 0.28
P\( \left( A\cap \overline { B } \right) \) = 0.28

Let A, B, C be the three possible events of selection.
A is 3 bowlers 8 others
B is 4 bowlers 7 others
C is 5 bowlers 6 others
\(P \quad(A t \text { loast } 3 \text { bowlers })=P(A \cup B \cup C) \)
\(=P(A)+P(B)+P(C)\)
\(=\frac{5 C_3 \times 11 C_8}{16 C_{11}}+\frac{5 C_1 \times 11 C_7}{16 C_{11}}+\frac{5 C_5 \times 11 C_6}{16 C_{11}}\)
\(=\frac{5 C_2 \times 11 C_3+5 C_1 \times 11 C_4+1 \times 11 C_5}{16 C_5}\)
\(=\frac{\frac{5 \times 4}{1 \times 2} \times \frac{11 \times 10 \times 9}{1 \times 2 \times 3}+\frac{5 \times 11 \times 10 \times 9 \times 8}{1 \times 2 \times 3 \times 4}}{+\frac{11 \times 10 \times 9 \times 8 \times 7}{1 \times 2 \times 3 \times 4 \times 5}} \)
\(=\frac{16 \times 15 \times 14 \times 13 \times 12}{1 \times 2 \times 3 \times 4 \times 5}\)
\(P(A \cup B \cup C) =\frac{\frac{11 \times 10 \times 9}{2 \times 3}\left[\frac{5 \times 4}{1 \times 2}+\frac{5 \times 8}{4}+\frac{8 \times 7}{4 \times 5}\right]}{2 \times 14 \times 13 \times 12}\)
\(=\frac{11 \times 5 \times 3 \times\left[10+10+\frac{14}{5}\right]}{2 \times 14 \times 13 \times 12} \)
\(=\frac{11 \times 5 \times 3 \times 114}{2 \times 14 \times 13 \times 12 \times 5}\)
\(=\frac{11 \times 57}{14 \times 13 \times 4} \)
\(=\frac{627}{728} \)

Let A be the event of getting 3 red balls, and B be the evenl of getting onered and 2 black balls, then
\((i) P(A)=\frac{7 C_3}{11 C_3}=\frac{7 \times 6 \times 5}{11 \times 10 \times 9}=\frac{7}{33} \)
\((ii) P(B)=\frac{7 C_1 \times 4 C_2}{11 C_3}=\frac{7 \times \frac{A^2 \times \not p}{1 \times \not 2}}{\frac{11 \times 10^5 \times \phi^2}{1 \times 2 \times} \times \not p}=\frac{14}{55}\)

(i) Let A be the event ofgetting one mango and one apple, Then
\(P(A)=\frac{5 C_1 \times 4 C_1}{9 C_2}=\frac{5}{9}\)

(ii) Let B and C be the events of getting both are mango and both are an apple respectively then
\(P(\text { Bor } C) =P(B)+P(C) \)
\(=\frac{5 C_2}{9 C_2}+\frac{4 C_2}{9 C_2}=\frac{5 \times 4}{9 \times 8}+\frac{4 \times 3}{9 \times 8} \)
\(=\frac{8}{18}=\frac{4}{9}\)