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) = \({(0.20)(0.60)\over(0.50)(0.90)+(0.30)(0.70)+(0.20)(0.60)}\)
\(={2\over13}\)

Let A₁, A₂ and A₃ be the events of X, Y and Z becoming managers of the company respectively. Let B be the event that the bonus scheme will be introduced.
We have to find the conditional probability P(A₃/B).
SinceA₁, A₂, and A₃ are mutually exclusive and exhaustive events, applying Bayes’ theorem

We have P(A₃/B) \(={P(A_3)P(B/A_3)\over P(A_1)P(B\ A_1)+P(A_2)P(B/A_2)+P(A_3)P(B/ A_3)}\)
\(P(A_1)={4\over9},P(B/ A_1)=0.3\)
\(P(A_2)={2\over9},P(B/ A_2)=0.5\)
\(P(A_1)={3\over9},P(B/ A_3)=0.4\)
P(A₃/B) \(={P(A_3)P(B/A_3)\over P(A_1)P(B\ A_1)+P(A_2)P(B/A_2)+P(A_3)P(B/ A_3)}\)
P(A₃/B) \(={({3\over9})(0.4)\over ({4\over9})(0.3)+({2\over9})(0.5)+({3\over 9})(0.4)}\)
\(={12\over 34}={6\over 17}\)

Let A be the event of getting award for design of railway bridge and B be the event of getting award for the efficient use of materials
Then, \(P(A)=0.48, P(B)=0.36, P(A \cap B)=0.2\)
(i) P (atleast one of the two awards)
\(=P(A \cup B) \)
\(=P(A)+P(B)-P(A \cap B) \)
\(=0.18+0.36-0.2=0.64\)
(ii) P (will get only one of the award)
\(=P(A \cap \bar{B})+P(\bar{A} \cap B) \)
\(=P(A)-P(A \cap B)+P(B)-P(A \cap B) \)
\(=0.48-0.20+0.36-0.20 \)
\(=0.44\)

Let A, B, and C denote the envelopes and 1, 2, and 3 denote the corresponding letters
The different combination of letters put into the envelopes are shown in the table Let c_i denote the outcomes of the events. Let X be the event of putting the letters into the exactly only one right envelopes Let Y be the event of putting none of the letters into the right envelope 
S = {C₁, C₂, C₃, C₄, C₅, C₆}, n(S) = 6
X = {C₂, C₃, C₆},  n(X) = 3
Y = {C₄, C₅}   n(Y) = 2
P(X) = \(\frac { 3 }{ 6 } =\frac { 1 }{ 2 } \) P(Y) = \(\frac { 2 }{ 6 } =\frac { 1 }{ 3 } \)

Let A₁ be the event that the items are produced by Machine-I, A₂ be the event that items are produced by Machine-II. Let B be the event of drawing a defective item. Now we are asked to find the conditional probability P (A₂/B). Since A₁, A₂ are mutually exclusive and exhaustive events, by Bayes’ theorem,

P(A₂/B) \(={P(A_2)P(B/A_2)\over P(A_1)P(B\ A_1)+P(A_2)P(B/A_2)}\)
We have, P(A₁) = 0.40,P(B/A₁) = 0.04 
P(A₂) = 0.60,P(B/A₂) = 0.05
P(A₂/B) \(={P(A_2)P(B/A_2)\over P(A_1)P(B\ A_1)+P(A_2)P(B/A_2)}\)
P(A₂/B) \(={(0.60)(0.05)\over(0.40)(0.04)+(0.60)(00.05)}={15\over23}\)

Let A_i be the event that the traffic light opens at i th cross, for i = 1, 2, 3, 4.
Let B_i be the event that the traffic light closes at i th cross, for i = 1, 2, 3, 4.
The traffic lights are all independent.
Therefore Ai and B_i are all independent events, for i = 1, 2, 3, 4.
Given that P(A_i) = 0.4, i = 1, 2, 3, 4
P(B_i) = 0.6, i =1, 2, 3, 4
(i) Probability of car crossing the first crossroad without stopping,
P(A₁) = 0.4
(ii) Probability of car crossing first two crossroads without stopping
\(P({ A }_{ 1 }\cap { A }_{ 2 })\)= P(A₁A₂) = (0.4)(0.4) = 0.16
(iii) Probability of car crossing all the crossroads, stopping at third cross
P\(({ A }_{ 1 }\cap { A }_{ 2 }{ B }_{ 3 }\cap { B }_{ 3 }\cap { A }_{ 4 })\) = P(A₁A₂B₃A₄) = (0.4)(0.4)(0.6)(0.4) = 0.0384
(iv) Probability of car crossing all the crossroads, stopping at exactly one of the crossroads is
P(B₁A₂A₃A₄ \(\cup \) A₁B₂A₃A₄ \(\cup \) A₁A₂B₃A₄ \(\cup \) A₂A₃A₃B₄)
= P(B₁A₂A₃A₄)+P(A₁B₂A₃A₄)+P(A₁A₂B₃A₄)+P(A₁A₂A₃B₄)
= 4(0.4)(0.4)(0.6)(0.4) = 4(0.0384) = 0.1536.

Let H₁, H₂, H₃ and H₄ be the events of hitting the plane by the anti-aircraft gun in the first second, third and fourth shot respectively.
Let H be the event that anti-aircraft gun hits the plane. Therefore \(\bar{H}\) is the event that the plane is not shot down. Given that
P(H₁) = 0.2 ⇒ P(\({ \bar { H } }_{ 1 }\)) = 1 - P(H₁) = 0.8
P(H₂) = 0.4 ⇒  P(\({ \bar { H } }_{ }\)) = 1-P(H₂) = 0.6
P(H₃) = 0.2  ⇒  P(\({ \bar { H } }_{ 3 }\)) = 1-P(H₃) = 0.8
P(H₄) = 0.1  ⇒  P(\({ \bar { H } }_{ 4 }\)) = 1-P(H₄) = 0.9
The probability that the gun hits the plane is
P(H) = 1 - P\((\bar { H } )\) = 1 - \((\overline { { H }_{ 1 }\cup { H }_{ 2 }\cup { H }_{ 3 }\cup { H }_{ 4 } } )\)
= 1 - P\(\left( { \bar { H } }_{ 1 }\cap { \bar { H } }_{ 2 }{ \bar { H } }_{ 3 }\cap { \bar { H } }_{ 4 } \right) \)
= 1 - P\(P(\bar { { H }_{ 1 } } )P(\bar { { H }_{ 2 } } )P(\bar { { H }_{ 3 } } )P(\bar { { H }_{ 4 } } )\)
= 1 - (0.8)(0.6)(0.8)(0.9) = 1 - 0.3456
P(H) = 0.6544

Let H be the event that the husband is watching television.
Let A₁ be the event that the wife is watching television in the prime time.
Let A₂ be the event that the wife is not watching television in the prime time.
Then. \(P\left(A_1\right)=\frac{60}{100}=0.6, P\left(A_2\right)=\frac{40}{100}=0.4\)
\(P\left(H / A_1\right)=\frac{40}{100}=0.4, P\left(A / A_2\right)=\frac{30}{100}=0.3\)
(i) P (husband is watching the television during the prime time)
\(P(H)=P\left(H / A_1\right) \cdot P\left(A_1\right)+P\left(H / A_2\right) \cdot P\left(A_2\right)\)
\(=0.6 \times 0.4+0.4 \times 0.3=0.4(0.6+0.3)\)
\(=0.4 \times 0.9=0.36=\frac{30}{100}=\frac{9}{25}\)
(ii) P (husband is watching the television, the wife is also watching the television)
\(P\left(A_1 / H\right)=\frac{P\left(A_1\right) \cdot P\left(H / A_1\right)}{P\left(A_1\right) \cdot P\left(H / A_1\right)+P\left(A_2\right) \cdot P\left(H / A_2\right)}\)
\(=\frac{0.6 \times 0.4}{0.36}=\frac{6}{9}=\frac{2}{3}\)

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

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 L be the' event of select a leap year
Then, \(P(L)=\frac{1}{4}, P(\bar{L})=\frac{3}{4}\)
If A is the year containing 53 Sundays
Then, \(P(A / L)=\frac{2}{7} \text { and } P(A / \bar{L})=\frac{1}{7}\)
\(\text {(i) } P(A)=P(L \cap A)+P(\bar{L} \cap A)\)
\(=P(L) \cdot P(A / L)+P(\bar{L}) \cdot P(A / \bar{L})\)
\(=\frac{1}{4} \times \frac{2}{7}+\frac{3}{4} \times \frac{1}{7}=\frac{5}{28}\)
\(\text {(ii) } P(L \cap A)=P(L) \cdot P(A / L)\)
\(=\frac{1}{4} \times \frac{2}{7} \)
\(=\frac{1}{14}\)

\(\text {(i) } P(A)=0.4 \ \&\ P(A \cup B)=0.7\)
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 B & G are the representation of boys & girls respectively.
Let F be the student taking first grade.
Then \(P(B)=\frac{2}{3}, P(G)=\frac{1}{3}\)
\(P(F / B)=0.70, P(F / G)=0.85\)
\(Now, P(F \cap B or F \cap G)=P(F \cap B)+P(F \cap G)\)
\(=P(B) \cdot P(F / B)+P(G) .P(F / G)\)
\(=\frac{2}{3} \times 0.7+\frac{1}{3} \times 0.85\)
\(=\frac{1.4+0.85}{3} \)
\( =\frac{2.25}{3}=\frac{225}{300} \)
\(=\frac{3}{4}=0.75\)

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

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

Let A be the event that a student chosen randomly studies in class XII and B be the event that the randomly chosen student is a girl.
\(\therefore\) We have to find P(A/B)
Clearly, n(A\(\cap \)B) = 10% of 430 = 430\(\times\) \(\frac { 10 }{ 100 } =43\)
\(\Rightarrow \quad P(A\cap B)=\frac { 43 }{ 1000 } \)  [ n(S) = 1000]
and n(B) = 430
\(\Rightarrow\)\(P(B)=\frac { 430 }{ 1000 } \)
\(\therefore P(A/B)=\frac { P(A\cap B) }{ P(B) } =\frac { \frac { 43 }{ 1000 } }{ \frac { 430 }{ 1000 } } =\frac { 43 }{ 1000 } \times \frac { 1000 }{ 430 } \)
P(A/B) = \(\frac { 1 }{ 10 } \)

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
\((i) P(A)=\frac{7 C_3}{11 C_3}=\frac{7 \times 6 \times 5}{11 \times 10 \times 9}=\frac{7}{33} \)
\((ii) P(B)=\frac{7 C_1 \times 4 C_2}{11 C_3}=\frac{7 \times \frac{A^2 \times \not p}{1 \times \not 2}}{\frac{11 \times 10^5 \times \phi^2}{1 \times 2 \times} \times \not p}=\frac{14}{55}\)

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

M	A	T
5	4	9

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

\(P(A)>0 ; P(B)>0 ;P(C)<0\)
\(\therefore\) The assignment of probability are not permissible.

Notice that three coins are tossed simultaneously = one coin is tossed three times.
The sample space S = { H,T } \(\times\) {H,T} \(\times\) {H,T}
S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}, n(S) = 8
Let A be the event of getting exactly one head, B be the event of getting atleast one head and C be the event of getting at most one head
A = {HTT, THT, TTH}; n(A) = 3
B = {HTT, THT, TTH, HHT, HTH, THH, HHH} n(B) = 7
C = {TTT, HTT, THT, TTH}; n(C) = 4
Therefore the required probabilities are
i) P(A) = \(\frac { n(A) }{ n(S) } =\frac { 3 }{ 8 } \)
ii) P(B) = \(\frac { n(B) }{ n(S) } =\frac { 7 }{ 8 } \)
iii)P(C) \(​​=\frac{n(C)}{n(S)}=\frac{4}{8}=\frac{1}{2}\)

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

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

Sample space S such that n{S) = 2ⁿ = 256
(i) If A be the event of getting exactly two tails, then
\(n(A)=28\)
\(P(A)=\frac{n(A)}{n(S)}=\frac{28}{256}=\frac{7}{64}\)
i) If B be the event of getting atleast two tails, then
n(B) = 247
\(P(B)=\frac{n(B)}{n(S)}=\frac{247}{256}\)

(iii) It C be the event of atmost two tails, then
\(n(C)=37(28+9) \)
\(P(C)=\frac{n(C)}{n(S)}=\frac{37}{256}\)

P\((A\cup B)\) = P(A) + P(B) - P(\(A\cap B\))
P\((A\cup B)\) = P(A) + P(B) - P(A)P(B) (since A and B are independent)
That is, 0.9 = 0.4 + P(B) − (0.4) P(B)
0.9 − 0.4 = (1− 0.4) P(B)
Therefore, P(B) = \(\frac{5}{6}\).