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 A₁ and A₂ be the events of job done by engineer-1 and engineer-2 of the company respectively. Let B be the event that the error occurs in the work.
We have to find the conditional probability
P(A₁/B) and P(A₂/B) to compare their errors in their work.
From the given information, we have
P(A₁) = 0.60,P(B/A₁) = 0.03
P(A₂) = 0.40, P(B/A₂) = 0.04 
A₁and A₂ are mutually exclusive and exhaustive events.
Applying Bayes’ theorem,

P(A₁/B)\(={P(A_1)P(B/A_1)\over P(A_1)P(B\ A_1)+P(A_2)P(B/A_2)}\)
\(={(0.60)(0.03)\over(0.60)(0.03)+(0.40)(00.04)}\)
\(P(A_1/B)={9\over 17}\)
P(A₂/B)\(={P(A_2)P(B/A_2)\over P(A_1)P(B\ A_1)+P(A_2)P(B/A_2)}\)
\(={(0.40)(0.04)\over(0.60)(0.03)+(0.40)(00.04)}\)
\(P(A_2/B)={8\over 17}\)
Since P (A₁/B)>P(A₂/B), the chance of error done by engineer-1 is greater than the chance of error done by engineer-2. Therefore one may guess that the serious error would have been be done by engineer-1.

Let A, B, C be the event of winning FIDE cup respectively by X,Y, and Z this year.
Given that X is thrice as likely to win as Y.
A : B :: 3 : 1        (1)
Y is twice as likely as to win Z
B : C :: 2 : 1     (2)
From (1) and (2)
A : B : C :: 6 : 2 : 1
A = 6k, B = 2k, C = k, where k is proportional constant.
Probability to win the cup by X is \(P(A)=\frac { 6k }{ 9k } =\frac { 2 }{ 3 } \)
Probability to win the cup by Y is \(P(B)=\frac { 2k }{ 9k } =\frac { 2 }{ 9 } \) and
Probability to win the cup by Z is \(P(C)=\frac { k }{ 9k } =\frac { 1 }{ 9 } \)

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 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.
We have to find the total probability of event B. That is, P(B).

Clearly A₁ and A₂ are mutually exclusive and exhaustive events.
Therefore, P(B) = P(A₁).P(B\A₁) + P(A₂).P(B\A₂)
We have, P(A₁) = 0.40, P(B/A₁) = 0.04
P(A₂) = 0.60, P(B/A₂) = 0.05 
P(B) = P(A₁) . P(B/A₂) + P(A₂).P(B/A₂)
= (0.40)(0.40) + (0.60)(0.05)
= 0.046.

Let be the event of speaks the truth, be the event of speaks the truth
∴ \(\bar { A } \) is the event of X not speaking the truth and \(\bar { B } \) is the event of Y not speaking the truth.
Let C be the event that they will contradict each other.
Given that
P(A) = 0.70 ⇒ P(\(\bar { A } \)) = 1 - P(A) = 0.30
P(B) = 0.90 ⇒ P(\(\bar { B } \)) = 1 - P(B) = 0.10
C = (A speaks truth and B does not speak truth or B speaks truth and A does not speak truth)
C =\(\left[ (A\cap \bar { B } )\cup (\bar { A } \cap B) \right] \) (see figure)
since \((A\cap \bar { B } )\) and \((\bar { A } \cap B)\) are mutually exclusively,
P(C) = \((A\cap \bar { B } )+(\bar { A } \cap B)\)
= P(A)P(\(\bar { B } \)) + P(\(\bar { A } \))P(B)
( Since A, B are independent event A, \(\bar { B } \) are also independent events
= (0.70) (0.10) + (0.30) (0.90)
= 0.070 + 0.270 = 0.34
P(C) = 0.34
 

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
\(n(C)=37(28+9) \)
\(P(C)=\frac{n(C)}{n(S)}=\frac{37}{256}\)

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]
=\(P\left( I\cap \overline { C } \right) +P\left( \overline { I } \cup C \right) \)
\(=\{0.12-0.07\}+\{0.25-0.07\}\)
\(=0.23 \)