Let E₁ and E₂ be the events of choosing a boy and a girl respectively from the class.
Given that the number of boys to the number of girls = 1: 2
\(\therefore P({ E }_{ 1 })=\frac { 1 }{ 1+2 } =\frac { 1 }{ 3 } \) and P(E₂) = \(\frac { 2 }{ 1+12 } =\frac { 2 }{ 3 } \)
Let A be the event that a student chosen will get first class
Given P(A/E₁) = 0.28 and P(A/E₂) = 0.25
\(\therefore\)  By theorem of total probability, 
P(A) = P(E₁).P(A/E₁) + P(E₂).P(A/E₂)
\(\Rightarrow \quad =\frac { 1 }{ 3 } \times 0.28+\frac { 2 }{ 3 } \times 0.25\)
\(=\frac { 28 }{ 300 } +\frac { 50 }{ 300 } =\frac { 78 }{ 300 } =0.26\) 

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) = \(1-\frac { 1 }{ 8 } =\frac { 7 }{ 8 } \)

\(P(\overline { A } \cup \overline { B } )\)=\(\left( \overline { A\cap B } \right) \)(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
\(P(\bar{A})=\frac{10}{512}=\frac{5}{256} \)
\( P(A)=1-P(\bar{A})=1-\frac{5}{256}=\frac{251}{256}\)

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 \(P(A)=\frac { 4 }{ 7 } \ge 0,\quad P(B)=\frac { 1 }{ 7 } \ge 0,\quad P(C)=\frac { 2 }{ 7 } \ge 0\)
Also \(P(S)=P(A)+P(B)+P(C)=\frac { 4 }{ 7 } +\frac { 1 }{ 7 } +\frac { 2 }{ 7 } =1\)
Therefore the assignment of probability is permissible

\(P(A)=\frac{3}{4}, P(B) =\frac{2}{5}, A \cup B=S \)
\(\Rightarrow P(A \cup B) =1\)
\(P(A \cap B) =P(A)+P(B)-P(A \cup B)\)
\(=\frac{3}{4}+\frac{2}{5}-1\)
\(=\frac{15+8}{20}-1=\frac{23-20}{20}=\frac{3}{20} \)
\(P(A / B) =\frac{P(A \cap B)}{P(B)} \)
\(=\frac{3 / 20}{2 / 5}=\frac{3}{8}\)

Let A₁, A₂, and A₃ be the events that the cars are rented from the agencies X, Y, and Z respectively.
Let G be the event of getting a car in good condition.
We have to find
(i) the total probability of event G that is, P(G)

(ii) find the conditional probability A₃ given G that is, P(A₃ /G)
We have P(A₁) = 0.50,P(G/A₁) = 0.90 
P(A₂) = 0.30, P(G/A₂) = 0.70 
P(A₃) = 0.20, P(G/A₃) =0.60.
(i) Since A₁,A₂, and A₃ are mutually exclusive and exhaustive events and G is an event in S,then the total probability of event G is P(G).
P(G) = P(A₁)P(G/A₁) + P(A₂)P(G/A₂) + P(A₃)P(G/A₃)
P(G) = (0.50)(0.90) + (0.30)(0.70) + (0.20)(0.60)
P(G) = 0.78
(ii) The conditional probability A₃ given G is P(A₃ /G) 
By Bayes’theorem,
P(A₃/G)\(={P(A_3)P(G/A_3)\over P(A_1)P(G/ A_1)+P(A_2)P(G/A_2)+P(A_3)P(G/ A_3)}\)
P(A₃/G) = \({(0.20)(0.60)\over(0.50)(0.90)+(0.30)(0.70)+(0.20)(0.60)}\)
\(={2\over13}\)

Given \(P(A)=0.35\)
\(P(A \text { or } B)=0.85 \text { and } P(A \text { and } B)=0.15\)
\((i) P(A \cup B)=P(A)+P(B)-P(A \cap B)\)
\(0.85 =0.35+P(B)-0.15 \)
\(P(B) =0.85-0.20\)
\(=0.65\)
\((ii) P(\bar{B})=1-P(B)=1-0.65=0.35\)
\((iii) P( only\ A)=P(A)-P(A \cap B)\)
\(=0.35-0.15=0.20\)

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 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, \(P\left(A_1\right)=0.60, P\left(B / A_1\right)=0.02\)
\(P\left(A_2\right)=0.40, P\left(B / A_2\right)=0.04\)
\(P(B)=P\left(A_1\right) \cdot P\left(B / A_1\right)+P\left(A_2\right) \cdot P\left(B / A_2\right)\)
\(=0.60 \times 0.02+0.40 \times 0.04 \)
\(=0.012+0.016 \)
\(=0.028 \)
\(=\frac{.28}{1000}=\frac{7}{250}\)

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)
=\(\frac { 4 }{ 52 } .\frac { 3 }{ 51 } \)
=\(\frac { 1 }{ 121 } \) .

\(\text {(i) } P(A)=0.4 \ \&\ P(A \cup B)=0.7\)
A and B are mutually exclusive then
\(P(A \cup B) =P(A)+P(B) \)
\(0.7 =0.4+P(B) \)
\(0.3 =P(B)\)
\(P(B) =0.3\)
(ii) A and B are independent then,
\(P(A \cup B) =P(A)+P(B)-P(A \cap B) \)
\(\therefore 0.7 =P(A)+P(B)-P(A) \cdot P(B) \)
\(0.7 =0.4+P(B)[1-0.4]\)
\(P(B) =\frac{0.7-0.4}{1-0.4}=\frac{0.3}{0.6}=\frac{1}{2}=0.5\)
\(\text { (iii) } P(A / B)=0.4\)
\(\frac{P(A \cap B)}{P(B)}=0.4\)
\(\Rightarrow \frac{P(A)+P(B)-P(A \cup B)}{P(B)}=0.4\)
\(\frac{0.4+P(B)-0.7}{P(B)}=0.4\)
\(P(B)-0.3=0.4 P(B)\)
\((1-0.4) P(B)=0.3\)
\(P(B)=\frac{0.3}{0.6}=\frac{1}{2}=0.5\)
\(\text {(iv) } P(B / A)=0.5\)
\(\frac{P(A \cap B)}{P(A)}=0.5\)
\(P(A)+P(B)-P(A \cup B)=0.5 P(A)\)
\(0.4+P(B)-0.7=0.5 \times 0.4\)
\(P(B)=0.2+0.3=0.5\)

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

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

Probabilrty of getting atleast 8 heads
\(=P(8)+P(9)+P(10) \)
\(=10 C_{8}\left(\frac{1}{2}\right)^{10}+10 C_{9}\left(\frac{1}{2}\right)^{10}+10 C_{10}\left(\frac{1}{2}\right)^{10} \)
\(=\frac{45+10+1}{1024}=\frac{56}{1024}=\frac{7}{128} \)

\(P\left(A_{1}\right)=0.6, \quad P\left(A_{2}\right)=0.4\)
\(A_{1}=\text { Event of selecting a girl }\)
\(\mathrm{B}=\text { Event of student taller than } 1.8 \mathrm{~m}\)
\(A_{2}=\text { Event of selecting a boy }\)
\(P\left(B / A_{1}\right)=\frac{1}{100}, P\left(B / A_{2}\right)=\frac{4}{100}\)
To find \(=\frac{P\left(A_{1}\right) \cdot P\left(B / A_{1}\right)}{P\left(A_{1}\right) \cdot P\left(B / A_{1}\right)+P\left(A_{2}\right) \cdot P\left(B / A_{2}\right)} \)
\(=\frac{0.6 \times \frac{1}{100}}{0.6 \times \frac{1}{100}+0.4 \times \frac{4}{100}} \)
\(=\frac{6}{\frac{6}{1000}+\frac{16}{1000}} \)
\(=\frac{6}{22}=\frac{3}{11} \)

\(n(S)=n(A \times A) \quad=16\)
\(\text {To find } n(A)\)
\(\text {Let } a=4, b=1 \text { then } a^{2}-4 b \text { is }\)
A = 1,b = 1 then 16 - 4 > 0
b = 2. then 16 - 8 > 0
b=3 then 16-L2>0
b = 4 then 16 - 16 = 0
Let Q = 3,b = 1 thg,l, 9-4 > 0
b = 2 then 9 - 8 > 0
Let a = 2,b = l then 4 - 4 > 0
n(A).= 7,
\(P(A)=\frac{n(A)}{n(S)}=\frac{7}{16}\)

\(P(A \cap \bar{B})=P(A)-P(A \cap B) \)
\(\frac{1}{2} =\frac{7}{10}-P(A \cap B) \)
\(P(A \cap B) =\frac{7}{10}-\frac{1}{2}=\frac{1}{5} \)

The equation is not valid for 1 to 9. 30 to 49
(i.e) for 9 + 20 = 29 numbers
Valid for 10 to 29 and 50 to 100 = 71 numbers
n(A) = 71, n(S) = 100
\(\mathrm{P}(\mathrm{A})=\frac{71}{100}=0.71\)

\(P(A / B)=\frac{P(A \cap B)}{P(B)}=\frac{P(A)}{P(B)}[\because A \subset B \Rightarrow A \cap B=A]\)

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

Fifty rupee note = 3
Hundred rupee note = 4
500 rupee note = 6
Totsl = 13
\(\mathrm{n}(\mathrm{S})=13 \mathrm{C}_{2}=78\)
\(\text {No. of ways the event to occur }=4\)
\(n(A)=4 C_{2}=6\)
\(\text {No. of ways the event not to occur }=72\)
The odds of event are a:b
\(6: 72 \Rightarrow 1: 12\)

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

\( P(A)>0 ; P(B)>0 ; P(C)>0 ; P(D)>0 \) and
\( P(A)+P(B)+P(C)+P(D) =0.15+0.30+0.43+0.12=1.0 \)
\(\therefore\) The assignment of probability are permissible.