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

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

\(P(A / \bar{B}) =\frac{P(A \cap \bar{B})}{P(\bar{B})}\)
\(=\frac{P(A)-P(A \cap B)}{1-P(B)} \)
\(=\frac{0.6-0.2}{1-0.5}=\frac{0.4}{0.5}=\frac{4}{5}\)

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
\(P(A\cup B\cup C)=P(A)+P(B)+P(C)=P(S)=1\)
The assignment is permissible because
\(P(A)=\frac { 1 }{ \sqrt { 3 } } \ge 0,\quad P(B)-1-\frac { 1 }{ \sqrt { 3 } } \ge 0,\quad P(C)-0\ge 0\)
\(P(S)=P(A)+P(B)+P(C)=\frac { 1 }{ \sqrt { 3 } } +1-\frac { 1 }{ \sqrt { 3 } } +0-1\)

Given A and B are independent
\( \Rightarrow P(A \cap B) =P(A) P(B) \)
\(P(A \cup B) =0.6, P(A)=0.2\)
\(P(A \cup B) =P(A)+P(B)-P(A \cap B) \)
\(=P(A)+P(B)-P(A) \cdot P(B) \)
\(0 \cdot 6 =0 \cdot 2+P(B)(1-0 \cdot 2) \)
\(\frac{0 \cdot 4}{0 \cdot 8} =P(B) \)
\(\text { i.e., } P(B) =\frac{1}{2}=0.5\)

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 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 A be the event of drawing a Jack in the first draw,
B be the event of drawing a Jack in the second draw.
Case (i)
Card is replaced
n(A) = 4  (Jack)
n(B) = 4  (Jack)
and n(S) = 52 (Total)
Clearly the event A will not affect the probability of the occurrence of event B and therefore A and B are independent.
P(A\(\cap \) B) = P(A). P(B)
P(A) = \(\frac{4}{52}\) , P(B) = \(\frac{4}{52}\)
P(A\(\cap \) B) = P(A).P(B)
=\(\frac{4}{52}\).\(\frac{4}{52}\)
=\(\frac{1}{169}\) .
Case (ii)
Card is not replaced
In the first draw, there are 4 Jacks and 52 cards in total. Since the Jack, drawn at the first draw is not replaced, in the second draw there are only 3 Jacks and 51 cards in total. Therefore the first event A affects the probability of the occurrence of the second event B.
Thus A and B are not independent. That is, they are dependent events.
Therefore, P( A \(\cap \) B ) = P(A). P(B)
P(A) = \(\frac{4}{52}\)
P(B/A) = \(\frac{3}{51}\)
P( A \(\cap \) B ) = P(A).P(B/A)
=\(\frac { 4 }{ 52 } .\frac { 3 }{ 51 } \)
=\(\frac { 1 }{ 121 } \) .

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)
\(=P(A \bar{B} \bar{C} \cup \bar{A} B \bar{C} \cup \bar{A} \bar{B} C)\)
\(=P(A) \cdot P(\bar{B}) \cdot P(\bar{C})+P(\bar{A}) \cdot P(B) \cdot P(\bar{C}) +P(\bar{A}) \cdot P(\bar{B}) \cdot P(C)\)

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

\(m \leq 5\)
\(2 x^{2}+2 m x+m+1>0\)
For real roots
\(4 m^{2}-8(m+1)>0 \)
\(4 m^{2} - 8 m-8>0 \)
\(m^{2}-2 m-2>0 \)
\(\text {Then } m \text { can be } 5,4,3\)
\(n(S)=5 ,n(A)=3 ,P(A)=\frac{3}{5} \)

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

Required Probability ={P(WBWBW)+ P(BWBWB)}
\(=\left(\frac{5}{8} \times \frac{3}{7} \times \frac{4}{6} \times \frac{2}{5} \times \frac{3}{4}\right) +\left(\frac{3}{8} \times \frac{5}{7} \times \frac{2}{6} \times \frac{4}{5} \times \frac{1}{4}\right) \)
\(=\frac{3}{56}+\frac{1}{56}=\frac{4}{56}=\frac{1}{14} \)

P(A or B but not C)
\(= P(A) P(B) P(\bar{C})+P(A) P(\bar{B}) P(\bar{C})+P(\bar{A}) P(B) P(\bar{C}) \)
\(= \frac{3}{4} \times \frac{1}{2} \times \frac{3}{8}+\frac{3}{4} \times \frac{1}{2} \times \frac{3}{8}+\frac{1}{4} \times \frac{1}{2} \times \frac{3}{8} \)
\(= \frac{9+9+3}{64}=\frac{21}{64} \)

Consider the following events.
A: getting a white ball in I draw
B: getting a black ball in II draw
Required probability = P(getting white ball in I draw and getting a black ball in II draw)
= P(\(A\cap B\))
= P(A).P(B/A)   \(\left[ \because P(B/A)=\frac { P(A\cap B) }{ P(A) } \right] \)

n(S) = 52
There are 4 kings and 4 queens in a pack of cards
\(\therefore\) n(E) = 8
There are 4 queens and 4 aces in a pack of cards
\(\therefore\) n(F) = 8
\(\therefore P(E)=\frac { 8 }{ 52 } =\frac { 2 }{ 13 } \) and \(P(F)=\frac { 8 }{ 52 } =\frac { 2 }{ 13 } \) and \(P(E\cap F)=\frac { 4 }{ 52 } =\frac { 1 }{ 13 } \)
\(P(E/F)=\frac { P(E\cap F) }{ P(F) } =\frac { \frac { 1 }{ 13 } }{ \frac { 2 }{ 13 } } =\frac { 1 }{ 13 } \times \frac { 13 }{ 2 } =\frac { 1 }{ 2 } \)

\(\mathrm{P}(\mathrm{A})>0 ; P(B)>0 ; P(C)>0 ; P(D)>0 \text { and } P(A)+P(B)+P(C)+\mathrm{P}(\mathrm{D}) \)
\(=0.22+0.38+0.16+0.34 \)
\(=1.1 \neq 1\)
\(\therefore\) The assignment of probability are not permissible.

Given P(A) = 0.5 P(B) = 0.3
A and B are mutually cxclusive events \(P(A \cap B)=0\)
\((i) P(A \cup B)=P(A)+P(B)=0.5+0.3=0.8\)
\((ii) P(A \cap \bar{B})=P(A)-P(A \cap B)=0.5-0=0.5\)
\(\text { (iii) } P{\bar{A} \cap B}=P(B)-P(A \cap B)=0.3-0=0.3\)