Here \(P(A)=\frac { 1 }{ 7 } ,P(B)=\frac { 1 }{ 5 } \)
\(P(\overset { \_ }{ A) } =1-\frac { 1 }{ 7 } =\frac { 6 }{ 7 } \)
\(P(\overset { \_ }{ B } )=1-\frac { 1 }{ 5 } =\frac { 4 }{ 5 } \)
Required probability =\(P(A\overset { \_ }{ B } )+P(\overset { \_ }{ A } B)=P(A)P(\overset { \_ }{ B } )+P(\overset { \_ }{ A } )P(B)\)
= \(\frac { 1 }{ 7 } \times \frac { 4 }{ 5 } +\frac { 6 }{ 7 } \times \frac { 1 }{ 5 } \)
\(=\frac { 4 }{ 35 } +\frac { 6 }{ 35 } =\frac { 10 }{ 35 } =\frac { 2 }{ 7 } \)

Let E₁ , E₂ and E₃ be the events that boxes I, II and III are chosen, respectively.
Then P(E₁) = P(E₂) = P(E₃ ) = \(\frac { 1 }{ 3 } \)
Also, let A be the event that ‘the coin drawn is of gold’
Then P(A|E₁) = P(a gold coin from bag I) = \(\frac { 2 }{ 2 } \)= 1,
P(A|E₂) = P(a gold coin from bag II) = 0 and
p(A|E₃) = P(a gold coin from bag III) = \(\frac { 1 }{ 2 } \)
Now, the probability that the other coin in the box is of gold = the probability that gold coin is drawn from the box I.
= P(E₁ |A)
By Bayes' theorem, we know that
P(E₁|A) \(=\frac { P({ E }_{ 1 })P(A|{ E }_{ 1 }) }{ P({ E }_{ 1 })P(A|{ E }_{ 1 })+P({ E }_{ 2 })P(A|{ E }_{ 2 })+P({ E }_{ 3 })P(A|{ E }_{ 3 }) } \)
= \(\frac { \frac { 1 }{ 3 } \times1}{ \frac { 1 }{ 3 } \times1 +\left( \frac { 1 }{ 3 } \right) \times 0 +\frac { 1 }{ 3 } \times \frac { 1 }{ 2 } } =\frac { 2 }{ 3 } \)

(i)\(\text { Given, }P(A)=\frac{1}{2} ; P(B)=\frac{1}{3} ; \)
\(P(\bar{A})=\frac{1}{2} ; P(\bar{B})=\frac{2}{3}\)
(i) P(problem is solved)
= peat least one solves the problem)
\(=1-P(\text { none solves })=1-P(\bar{A} \bar{B}) \)
\(=1-\frac{1}{2} \times \frac{2}{3}=1-\frac{1}{3}=\frac{2}{3} \)
(ii) P(exactly one of them solves the problem)
\(=P(A \bar{B} \text { or } \bar{A} B)=\frac{1}{2} \times \frac{2}{3}+\frac{1}{2} \times \frac{1}{3}=\frac{3}{6}=\frac{1}{2}\)

Let X = Number of defective bulbs out of 4 bulbs drawn
with replacement X = 0, 1, 2, 3, 4
p = Probability of defective bulb \(=\frac{6}{30}=\frac{1}{5}\) 
q = Probability of good bulb = 4/5
\(\begin{array}{c|c|c|c|c|c} \hline X=x_{i} & 0 & 1 & 2 & 3 & 4 \\ \hline P(x)=p_{i} & \left(\frac{4}{5}\right)^{4} & 4 \cdot \frac{1}{5} \cdot \frac{64}{125} & 6 \cdot \frac{1}{25} \cdot \frac{16}{25} & 4 \cdot \frac{1}{125} \cdot \frac{4}{5} & \left(\frac{1}{5}\right)^{4} \\ & =\frac{256}{625} & =\frac{25 \hat{p}}{625} & =\frac{96}{625} & =\frac{16}{625} & =\frac{1}{625} \\ \hline \end{array}\) 
Mean \((\mu)=\Sigma p_{i} x_{i}\) 
\(=0 \times \frac{256}{625}+1 \times \frac{256}{625}+2 \times \frac{96}{625}+3 \times \frac{16}{625}+4 \times \frac{1}{625}\) 
\(=\frac{4}{5}\)

mean = 2/3
variance = 5/9

Probability of getting a six = \(\frac { 1 }{ 6 } \)
Probability of not getting a six = \(1-\frac { 1 }{ 6 } =\frac { 5 }{ 6 } \)
P(obtaining the third six in sixth throw)
= {P (obtaining two sixes in the first five throws}\(\times \frac { 1 }{ 6 } \)
= \(^{ 5 }{ C }_{ 2 }{ \left( \frac { 5 }{ 6 } \right) }^{ 3 }{ \left( \frac { 1 }{ 6 } \right) }^{ 2 }\times \frac { 1 }{ 6 } =\frac { 625 }{ 23328 } \)

Given  \(P(A)=\frac { 1 }{ 3 } ,P(\overset { \_ }{ A } )=1-\frac { 1 }{ 3 } =\frac { 2 }{ 3 } \)
\(P(B)=\frac { 2 }{ 7 } ,P(\overset { \_ }{ B } )=1-\frac { 2 }{ 7 } =\frac { 5 }{ 7 } \)
and    \(P(C)=\frac { 3 }{ 8 } ,P(\overset { \_ }{ C } )=1-\frac { 3 }{ 8 } =\frac { 5 }{ 8 } \)
All the three solve the problem simultaneously independently.
We can have case I: A can solve Band C can not solve
Case II : B can solve A and C can not solve
Case III: C can solve A and B can not solve.
P(only one solves) \(P(A\overset { \_ }{ B } \overset { \_ }{ C } )+P(\overset { \_ }{ A } B\overset { - }{ C } )+P(\overset { \_ }{ A } \overset { \_ }{ B } C)\)
= \(P(A)P(\overset { \_ }{ B } )P(\overset { \_ }{ C } )+P(\overset { \_ }{ A } )P(B\overset { - }{ )P(C } )+P(\overset { \_ }{ A } )P(\overset { \_ }{ B } )P(C)\) 
= \(\left( \frac { 1 }{ 3 } \right) \left( \frac { 5 }{ 7 } \right) \left( \frac { 5 }{ 8 } \right) +\left( \frac { 2 }{ 3 } \right) \left( \frac { 2 }{ 7 } \right) \left( \frac { 5 }{ 8 } \right) +\left( \frac { 2 }{ 3 } \right) \left( \frac { 5 }{ 7 } \right) \left( \frac { 3 }{ 8 } \right) \)
= \(\frac { 25 }{ 168 } +\frac { 20 }{ 168 } +\frac { 30 }{ 168 } =\frac { 75 }{ 168 } =\frac { 25 }{ 56 } \)

\(\begin{array}{|c|c|c|c|} \hline \text { Machine } & \boldsymbol{A} & \boldsymbol{B} & \boldsymbol{C} \\ \hline \text { Production } & 60 \% & 30 \% & 10 \% \\ \hline \text { Defective } & 1 \% & 2 \% & 3 \% \\ \hline \end{array}\)
\(P(A)=\frac{60}{100}=\frac{6}{10}, P(B)=\frac{30}{100}=\frac{3}{10}, P(C)=\frac{10}{100}=\frac{1}{10}\)
E: bulb is defective
\(P(E / A)=\frac{1}{100}, P(E / B)=\frac{2}{100}, P(E / C)=\frac{3}{100}\)
Using Bayes' Theorem, probability that defective bulbs was produced by machine A,
\(\begin{array}{r} P(A / E)=\frac{P(A) \cdot P(E / A)}{P(A) \cdot P(E / A)+P(B) \cdot P(E / B)} \\ +P(C) \cdot P(E / C) \end{array}\)
\(=\frac{\frac{6}{10} \times \frac{1}{100}}{\frac{6}{10} \times \frac{1}{100}+\frac{3}{10} \times \frac{2}{100}+\frac{1}{10} \times \frac{3}{100}}=\frac{2}{5}\)

Total cards are 12
A : number drawn is more than 3, i.e. 4, 5, 6, ..., 12.
B : getting an even number, i.e. 2, 4, 6, 8, 10, 12.
\(A \cap B: 4,6,8,10,12 \)
\(P(B / A)=\frac{P(A \cap B)}{P(A)}=\frac{5 / 12}{9 / 12}=\frac{5}{9} . \)

Total number of ways of selecting two numbers = ⁶C₂ = 15
Values of x (larger of the two) can be 2, 3, 4, 5, 6
\(P(x=2)={1\over 15}\)
\(P(x=3)={2\over 15}\)
\(P(x=4)={3\over 15}\)
\(P(x=5)={4\over 15}\)
and \(P(x=6)={5\over 15}\)
Distribution can be written as

Mean \(=\sum xP(x)={70\over 15}={14\over 3}\)

Let C₁, C₂, C₃, C₄ be the events that the lost card is of heart, spades, diamond or club respectively.
Obviously P(C₁) = P(C₂) = P(C₃) = P(C₄)
\(={13\over 52}={1\over 4}\)
Let S be the event of drawing two cards of heart from the remaining 51 cards.We wish to find \(P\left(C_1\over S\right)\)
Now \(P\left(C_1\over S\right)\) is the probability of drawing two heart cards from 51 cards given that one heart card is lost
\(={^{12}C_2\over ^{51}C_2}={12\times11\over 1\times2}\times{1\times2\over 51\times50}={22\over 425}\)
\(P\left( S\over C_3\right)=P\left( S\over C_3\right)=P\left( S\over C_4\right)={^{13}C_2\over ^{51}C_2}\)
\(={13\times12\over 1\times2}\times{1\times2\over 51\times50}={26\over 425}\)
By Bayes' Theorem
\(P\left(C_1\over S\right)={P(C_1).P\left(S\over C_1\right)\over \sum P(C_1).P\left(S\over C_1\right)}\)
\(={{1\over4}\times{22\over 425}\over{1\over 4}\times{22\over 425}+{1\over 4}\times{26\over 425}+{1\over4}\times{26\over 425}}\)
\(={22\over 22+26+26+26}\)
\(={11\over 50}\)

Total number of oranges = 25
number of good oranges = 20
number of bad oranges = 5
Probability of getting a bad orange,
\(P(B)=\frac { 5 }{ 25 } =\frac { 1 }{ 5 } \)
Now  \(p=\frac { 1 }{ 5 } \)
\(\\ q=1-p=\frac { 4 }{ 5 } \)
Let X be the random variable of "Number of bad oranges".
\(\Rightarrow X=0,1,2,3,4\)
\(P(X=0)={ n }_{ { C }_{ r } }{ \left( p \right) }^{ n-r }{ \left( q \right) }^{ r }\)
\(P(X=0)={ 4 }_{ { C }_{ 0 } }{ \left( \frac { 4 }{ 5 } \right) }^{ 4 }=\frac { 256 }{ 625 } \)
\(P(X=1)=4\times \frac { 64 }{ 125 } \times \frac { 1 }{ 5 } =\frac { 256 }{ 625 } \)
\(P(X=2)={ 4 }_{ { C }_{ 2 } }{ \left( \frac { 4 }{ 5 } \right) }^{ 2 }{ \left( \frac { 1 }{ 5 } \right) }^{ 2 }\)
\(=6\times \frac { 16 }{ 625 } =\frac { 96 }{ 625 } \)
\(P(X=3)={ 4 }_{ { C }_{ 3 } }\left( \frac { 4 }{ 5 } \right) { \left( \frac { 1 }{ 5 } \right) }^{ 3 }\)
\(=4\times \frac { 4 }{ 625 } =\frac { 16 }{ 625 } \)
\(P(X=4)={ 4 }_{ { C }_{ 4 } }{ \left( \frac { 4 }{ 5 } \right) }^{ 0 }{ \left( \frac { 1 }{ 5 } \right) }^{ 4 }\)
\(=\frac { 1 }{ 625 } \)
\(\therefore\)   Probability distribution is

\(\therefore Mean=\sum { XP(X) } \)
\(=\frac { 500 }{ 625 } =\frac { 20 }{ 25 } =\frac { 4 }{ 5 } \)
\(Var.\quad (X)=E({ X }^{ 2 })-[E(X){ ] }^{ 2 }\)
\(=\frac { 800 }{ 625 } -\frac { 16 }{ 25 } \)
\(Var.(X)=\frac { 32 }{ 25 } -\frac { 16 }{ 25 } =\frac { 16 }{ 25 } \)

Let X is the number of red balls
So, X can take value 0, 1, 2, 3, 4
\(\therefore p=P(red\quad ball)=\frac { 6 }{ 9 } =\frac { 2 }{ 3 } \)
\(\therefore \ q=1-p=\frac { 1 }{ 3 } \)
Case 1: When X = 0, all white balls
\(P(X=0)={ 4 }_{ { C }_{ 0 } }{ \left( \frac { 1 }{ 3 } \right) }^{ 4-0 }{ \left( \frac { 2 }{ 3 } \right) }^{ 0 }=\frac { 1 }{ 81 } \)
Case 2: When X = 1, 3 white and 1 red balls
\(P(X=1)={ 4 }_{ { C }_{ 1 } }{ \left( \frac { 1 }{ 3 } \right) }^{ 4-1 }{ \left( \frac { 2 }{ 3 } \right) }^{ 1 }\)
\(=4\times \frac { 1 }{ 27 } \times \frac { 2 }{ 3 } =\frac { 8 }{ 81 } \)
Case 3: When X = 2, 2 white and 2 red balls
\(P(x=2)={ 4 }_{ { C }_{ 2 } }{ \left( \frac { 1 }{ 3 } \right) }^{ 4-2 }{ \left( \frac { 2 }{ 3 } \right) }^{ 2 }\)
\(=6\times \frac { 1 }{ 9 } \times \frac { 4 }{ 9 } =\frac { 24 }{ 81 } \)
Case 4: When X = 3, 1 white and 3 red balls
\(P(X=3)={ 4 }_{ { C }_{ 3 } }{ \left( \frac { 1 }{ 3 } \right) }^{ 4-3 }{ \left( \frac { 2 }{ 3 } \right) }^{ 3 }\)
\(=4\times \frac { 1 }{ 3 } \times \frac { 8 }{ 27 } =\frac { 32 }{ 81 } \)
Case 5: When X = 4, 4 red balls
P(X = 4) = \({ 4 }_{ { C }_{ 4 } }{ \left( \frac { 1 }{ 3 } \right) }^{ 4-4 }{ \left( \frac { 2 }{ 3 } \right) }^{ 4 }\)
\(=1\times 1\times \frac { 16 }{ 81 } =\frac { 16 }{ 81 } \)
\(\therefore \) the probability distribution of number of red ball is:

\(Mean=\sum { XP(X)=\frac { 8 }{ 3 } } \)
\(Var\ (X)=\sum { { X }^{ 2 }P(X)-\{ \sum { XP(X){ \} }^{ 2 } } } \)
\(=8-{ \left( \frac { 8 }{ 3 } \right) }^{ 2 }=8-\frac { 64 }{ 9 } =\frac { 8 }{ 9 } =0.88\)

Let X represent the random variable,
then X = 0, 1, 2, 3
\(p=\frac { 1 }{ 6 } ,q=1-p\)
\(q=\frac { 5 }{ 6 } \)
\(P(X=0)=P(r=0)={ 3 }_{ { C }_{ 0 } }{ \left( \frac { 1 }{ 6 } \right) }^{ 0 }{ \left( \frac { 5 }{ 6 } \right) }^{ 3 }=\frac { 125 }{ 216 } \)
\(P(X=1)=P(r=1)={ 3 }_{ { C }_{ 1 } }{ \left( \frac { 1 }{ 6 } \right) }^{ 1 }{ \left( \frac { 5 }{ 6 } \right) }^{ 2 }=\frac { 75 }{ 216 } \)
\(P(X=2)=P(r=2)={ 3 }_{ { C }_{ 2 } }{ \left( \frac { 1 }{ 6 } \right) }^{ 2 }{ \left( \frac { 5 }{ 6 } \right) }^{ 1 }=\frac { 15 }{ 216 } \)
\(P(X=3)=P(r=3)={ 3 }_{ { C }_{ 3 } }{ \left( \frac { 1 }{ 6 } \right) }^{ 3 }{ \left( \frac { 5 }{ 6 } \right) }^{ 0 }=\frac { 1 }{ 216 } \)

\(Mean=\sum { { x }_{ i }{ p }_{ i }=\frac { 1 }{ 2 } } \)
\(Variance(X)=\sum { { ({ x }_{ i }) }^{ 2 }{ p }_{ i }-{ (\sum { { x }_{ i }{ p }_{ i } } ) }^{ 2 } } \)
\(=\frac { 2 }{ 3 } -{ \left( \frac { 1 }{ 2 } \right) }^{ 2 }\)
\(=\frac { 2 }{ 3 } -\frac { 1 }{ 4 } =\frac { 5 }{ 12 } \)
Standard Deviation
\(\sqrt { Variance(X) } =\frac { \sqrt { 15 } }{ 6 } \)

Let E₁: Event selecting bag with 4 red & 4 black balls
E₂: Event selecting bag with 2 red & 6 black balls
A: Event selecting 2 red balls without replacement
Then, P(E₁) = P(E₂) = \(\frac { 1 }{ 2 } \)
\(P(A/{ E }_{ 1 })=\frac { 4_{ C_{ 2 } } }{ { 8 }_{ C_{ 2 } } } =\frac { 3 }{ 14 } ,\)
\(P(A/{ E }_{ 2 })=\frac { { 2 }_{ { C }_{ 2 } } }{ { 8 }_{ { C }_{ 2 } } } =\frac { 1 }{ 28 } .\)
\(P\left( \frac { { E }_{ 1 } }{ A } \right) =\frac { P({ E }_{ 1 }).P(A/{ E }_{ 1 }) }{ P({ E }_{ 1 }).P(A/{ E }_{ 1 })+P({ E }_{ 2 }).P(A/{ E }_{ 2 }) } \)
\(=\frac { \frac { 1 }{ 2 } .\frac { 3 }{ 14 } }{ \frac { 1 }{ 2 } .\frac { 3 }{ 14 } +\frac { 1 }{ 2 } .\frac { 1 }{ 28 } } =\frac { 6 }{ 7 } \)

Let E₁: urn has 2 white balls
E₂: urn has 3 white balls
E₃: urn has 4 white balls
A: 2 balls drawn are white
P(E₁) = P(E₂) = P(E₃) = \(\frac { 1 }{ 3 } \)
P(A/E₁)=\(\frac { { 2 }_{ { C }_{ 2 } } }{ { 4 }_{ { C }_{ 2 } } } =\frac { 1 }{ 6, } \)
\(P(A/{ E }_{ 2 })=\frac { { 2 }_{ { C }_{ 2 } } }{ { 4 }_{ C_{ 2 } } } =\frac { 1 }{ 6 } ,\)
\(P(A/{ E }_{ 2 })=\frac { { 3 }_{ { C }_{ 2 } } }{ { 4 }_{ C_{ 2 } } } =\frac { 1 }{ 2 } ,\)
\(P(A/{ E }_{ 3 })=\frac { { 4 }_{ { C }_{ 2 } } }{ { 4 }_{ { C }_{ 2 } } } =1\)
\(\therefore P({ E }_{ 3 }/A)\)
\(=\frac { { P(E }_{ 3 }).P(A/{ E }_{ 3 }) }{ P({ E }_{ 1 })P(A/{ E }_{ 1 })+P({ E }_{ 2 })P(A/{ E }_{ 2 })+P({ E }_{ 3 })P(A/{ E }_{ 3 }) } \)
\(=\frac { \frac { 1 }{ 3 } .1 }{ \frac { 1 }{ 3 } .\frac { 1 }{ 6 } +\frac { 1 }{ 3 } .\frac { 1 }{ 2 } +\frac { 1 }{ 3 } .1 } \)
\(=\frac { 6 }{ 10 } =0.6\)

Let, E₁: Event that lost card is a spade
E₂: Event that lost card is a not spade
A: Event that three spades are drawn without replacement from 51 cards
\(P({ E }_{ 1 })=\frac { 13 }{ 52 } =\frac { 1 }{ 4 } ,\quad P({ E }_{ 2 })=1-\frac { 1 }{ 4 } =\frac { 3 }{ 4 } \)
\(P(A/{ E }_{ 1 })=\frac { { 12 }_{ C_{ 3 } } }{ { 51 }_{ { C }_{ 3 } } } ,\quad P(A/{ E }_{ 2 })=\frac { 13_{ { C }_{ 3 } } }{ { 51 }_{ { C }_{ 3 } } } \)
\(P({ E }_{ 1 }/A)=\frac { \frac { 1 }{ 4 } .\frac { { 12 }_{ C_{ 3 } } }{ { 51 }_{ { C }_{ 3 } } } }{ \frac { 1 }{ 4 } .\frac { { 12 }_{ C_{ 3 } } }{ { 51 }_{ { C }_{ 3 } } } +\frac { 3 }{ 4 } .\frac { { 12 }_{ C_{ 3 } } }{ { 51 }_{ { C }_{ 3 } } } } \)
\(=\frac { 10 }{ 49 } \)

E₁ = Event that 1 occur;
E₂ = Event that 1 does not occur;
A = Event that the man reports that 1 occur.
\(P({ E }_{ 1 })=\frac { 1 }{ 6 } ;P({ E }_{ 2 })=\frac { 5 }{ 6 } ;\)
\(P(A/{ E }_{ 1 })=\frac { 3 }{ 5 } ;P(A/{ E }_{ 2 })=\frac { 2 }{ 5 } ;\)
\(P({ E }_{ 1 }/A)=\frac { \frac { 1 }{ 6 } .\frac { 3 }{ 5 } }{ \frac { 1 }{ 6 } .\frac { 3 }{ 5 } +\frac { 5 }{ 6 } .\frac { 2 }{ 5 } } =\frac { 3 }{ 13 } \)

Let the events defined are :
E₁: Person chosen is a cyclist
E_2: Person chosen is a scooter driver
E₃: Person chosen is a motorbike driver
A:Person meets with an accident
P(E₁_= 1/6, P(E₂) = 1/3, P(E₃) = 1/2
P(A/E₁) = 0.01, P(A/E₂) = 0.03,
P(A/E₃) = 0.15
\(P{ (E }_{ 2 }/A)=\frac { P({ E }_{ 2 }).P(A/{ E }_{ 2 }) }{ P({ E }_{ 1 }).P(A/{ E }_{ 1 })+P({ E }_{ 2 }).P(A/{ E }_{ 2 })+P({ E }_{ 3 }).P(A/{ E }_{ 3) } } \)
\(=\frac { \frac { 1 }{ 3 } \times 0.03 }{ \frac { 1 }{ 6 } \times 0.01+\frac { 1 }{ 3 } \times 0.03+\frac { 1 }{ 2 } \times 0.15 } \)
\(=\frac { 0.01 }{ 0.086 } \)
= 0.11627
Suggestion: Cycle should be promoted as it is:
(i) Good for health
(ii) Pollution free
(iii) Saves energy (no petrol).