A man throws a die thrice but quits when he gets a number greater than 4.
Let S be the success of getting number > 4 and F be the failure of getting number < 4
\(\therefore\)  The sample space is s = {S, FS, FFS, FFF}
Case 1: E₁:gets a number > 4 on 1^st throw 
\(P({ E }_{ 1 })=\frac { 2 }{ 6 } =\frac { 1 }{ 3 } \)
He wins Rs. 5.
Case 2: E₂: 1^st throw gets number \(\le \)4 and gets number > 4 on 2^nd throw
\(\therefore P({ E }_{ 2 })=\frac { 4 }{ 6 } \times \frac { 2 }{ 6 } =\frac { 2 }{ 3 } \times \frac { 1 }{ 3 } =\frac { 2 }{ 9 } \)
He wins Rs. 5 but loses Rs. 1.
\(\therefore\)  Amount is Rs 4 (Rs. 5 - Rs. 1).
Case 3 : E₃: gets number \(\le \)4 on 1^st & 2^nd throw and gets number > 4 on 3^rd  throw
\(P({ E }_{ 3 })=\frac { 4 }{ 6 } \times \frac { 4 }{ 6 } \times \frac { 2 }{ 6 } \)
\(=\frac { 2 }{ 3 } ​​\times \frac { 2 }{ 3 } \times \frac { 1 }{ 3 } =\frac { 4 }{ 27 } \)
He wins Rs. 5 but loses Rs. 2.
\(\therefore\) Amount is Rs 3 (Rs. 5 - Rs. 2)
Case 4: E₄: gets number  \(\le \)4 on all three throws
\(P\left( { E }_{ 4 } \right) =\frac { 2 }{ 3 } \times \frac { 2 }{ 3 } \times \frac { 2 }{ 3 } =\frac { 8 }{ 27 } \)
He loses - Rs. 3
\(\therefore\) Amount = - Rs. 3
Probability Distribution Table :

Expected value of prize the wins/loses is E(X) = \(\sum { XP(X) } =\frac { 57 }{ 27 } \)
He is expected to win Rs. \(\frac { 57 }{ 27 } i.e.,\) Rs. 2.11.

Let the events be described as below :
A: No change takes place
E₁: Person A gets appointed
E₂: Person B gets appointed
E₃: Person C gets appointed
The chances of selection of A, B and C are in the ratio 1: 2: 4.
Hence,
\(P({ E }_{ 1 })=\frac { 1 }{ 7 } ;P({ E }_{ 2 })=\frac { 2 }{ 7 } ;P({ E }_{ 3 })=\frac { 4 }{ 7 } \)
Probabilities of A, B and C introducing changes to improve profits of company are 0.8, 0.5 and 0.3 respectively.
Hence probability of no changes on appointment of A, B and C are 0.2, 0.5 and 0.7 respectively.
 Hence,
\(P(A|{ E }_{ 1 })=0.2=\frac { 2 }{ 10 } ;P(A/{ E }_{ 2 })=0.5=\frac { 2 }{ 10 } ;\)
\(P(A|{ E }_{ 3 })=0.7=\frac { 7 }{ 10 } \)
Therfore, required probability i.e., \(({ E }_{ 3 }|A)\quad is\quad P({ E }_{ 3 }|A)\)
\(=\frac { P(A|{ E }_{ 3 }).P({ E }_{ 3 }) }{ P(A|{ E }_{ 1 }).P({ E }_{ 1 })+P(A|{ E }_{ 2 }).P({ E }_{ 2 })+P(A|{ E }_{ 3 }).P({ E }_{ 3 }) } \)
\(=\frac { \frac { 4 }{ 7 } .\frac { 7 }{ 10 } }{ \frac { 1 }{ 7 } .\frac { 2 }{ 10 } +\frac { 2 }{ 7 } .\frac { 5 }{ 10 } +\frac { 4 }{ 7 } .\frac { 7 }{ 10 } } =\frac { 7 }{ 10 } \)
\(\therefore\) If no change takes place, the probability that it is due to appointment of C is \(\frac { 7 }{ 10 } \)

Here \(P(A)=\frac { 5 }{ 36 } \)
[(1, 5), (5, 1), (2, 4), (4, 2), (3, 3)]
\(P(\overset { - }{ A } )=1-\frac { 5 }{ 36 } =\frac { 31 }{ 36 } \)
And  \(P(B)=\frac { 6 }{ 36 } =\frac { 1 }{ 6 } \)
[(1, 6), (6, 1), (2, 5), (5, 2), (3, 4), (4, 3)]
\(P(\overset { - }{ B) } =1-\frac { 1 }{ 6 } =\frac { 5 }{ 6 } \)
Reqd.probability = \(P(\overset { \_ }{ A } \overset { \_ }{ B } A)\)
\(=P(\overset { \_ }{ A } )P(\overset { \_ }{ B } )P(A)\)
\(=\frac { 31 }{ 36 } \times \frac { 5 }{ 6 } \times \frac { 5 }{ 36 } =\frac { 775 }{ 7776 } \)

Let E_I = Letter has come from CALCUITA
E₂ = Letter has come from TATANAGAR
and E = Two consecutive letters (i.e. alphabets) TA are visible on envelope
\(\therefore P\left(E_{1}\right)=\frac{1}{2}, P\left(E_{2}\right)=\frac{1}{2}, P\left(\frac{E}{E_{1}}\right)=\frac{n\left(E \cap E_{1}\right)}{n\left(E_{1}\right)}=\frac{1}{7}\) 
[\(\therefore\) pairs of consecutive letters are CA, AL, LC, CU, UT, TT, TA]
and \(P\left(\frac{E}{E_{2}}\right)=\frac{n\left(E \cap E_{2}\right)}{n\left(E_{2}\right)}=\frac{2}{8}\) 
[\(\therefore\)8 pairs of consecutive letters are TA, AT, TA;AN, NA, AG, GA, AR]
\(\therefore P\left(\frac{E_{2}}{E}\right)=\frac{P\left(E_{2}\right) \cdot P\left(\frac{E}{E_{2}}\right)}{P\left(E_{1}\right) \cdot P\left(\frac{E}{E_{1}}\right)+P\left(E_{2}\right) \cdot P\left(\frac{E}{E_{2}}\right)}\) 
[using Baye's theorem]
\(=\frac{\frac{1}{2} \times \frac{2}{8}}{\frac{1}{2} \times \frac{1}{7}+\frac{1}{2} \times \frac{2}{8}}=\frac{\frac{2}{16}}{\frac{8+14}{2 \times 7 \times 8}}=\frac{2 \times 7}{22}=\frac{7}{11}\) 
Hence, the probability that the letter TA came from TATANAGAR is \(\frac{7}{11}\).

\(2\over7\)

Let the events be:
E₁ : The girl gets 1, 2, 3 and 4 on the dice
E₂ : The girl gets 5 or 6 on the dice.
A : Exactly one head shows up.
\(P({ E }_{ 1 })=\frac { 4 }{ 6 } =\frac { 2 }{ 3 } ,P({ E }_{ 2 })=\frac { 2 }{ 6 } =\frac { 1 }{ 3 } \)
and \(P(A/{ E }_{ 1 })=\frac { 1 }{ 2 } ,P(A/{ E }_{ 2 })=\frac { 3 }{ 8 } \)
By Bayes' Theorem,
\(P({ E }_{ 1 }/A)=\frac { P({ E }_{ 1 })P(A/{ E }_{ 1 }) }{ P({ E }_{ 1 })P(A/{ E }_{ 1 })+P({ E }_{ 2 })P(A/{ E }_{ 2 }) } \)
\(=\frac { \frac { 2 }{ 3 } \times \frac { 1 }{ 2 } }{ \frac { 2 }{ 3 } \times \frac { 1 }{ 2 } +\frac { 1 }{ 3 } \times \frac { 3 }{ 8 } } =\frac { \frac { 1 }{ 3 } }{ \frac { 1 }{ 3 } +\frac { 1 }{ 8 } } =\frac { 8 }{ 11 } \)

Let the events be as below:
E₁ : 2 red balls are transfered from Bag I to Bag II
E₂ : 2 black balls are transferd frm Bag I to Bag II
E₃ : 1 red and 1 black balls are transferd from Bag I to Bag
and A : 1 red ball is drawn from Bag II
\(P({ E }_{ 1 })=\frac { ^{ 3 }{ C }_{ 2 } }{ ^{ 7 }{ C }_{ 2 } } =\frac { 3\times 2 }{ 7\times 6 } =\frac { 1 }{ 7 } \)
\(P({ E }_{ 1 })=\frac { ^{ 4 }{ C }_{ 2 } }{ ^{ 7 }{ C }_{ 2 } } =\frac { 4\times 3 }{ 7\times 6 } =\frac { 2 }{ 7 } \)
\(P({ E }_{ 3 })=\frac { ^{ 3 }{ C }_{ 1 }\times ^{ 4 }{ C }_{ 1 } }{ ^{ 7 }{ C }_{ 2 } } =\frac { 3\times 4 }{ \frac { 7\times 6 }{ 1\times 2 } } =\frac { 3\times 4 }{ 7\times 3 } =\frac { 4 }{ 7 } \)
and \(P(A/{ E }_{ 1 })=\frac { 6 }{ 11 } ,P(A/{ E }_{ 2 })=\frac { 4 }{ 11 } ,P(A/{ E }_{ 3 })=\frac { 5 }{ 11 } \)
By Bayes' Theorem
\(P({ E }_{ 2 }/A)=\frac { P({ E }_{ 2 })P(A/{ E }_{ 2 }) }{ P({ E }_{ 1 })P(A/{ E }_{ 1 })+P({ E }_{ 2 })P(A/{ E }_{ 2 })+P({ E }_{ 3 })P(A/{ E }_{ 3 }) } \)
\(=\frac { \left( \frac { 2 }{ 7 } \right) \left( \frac { 4 }{ 11 } \right) }{ \left( \frac { 1 }{ 7 } \right) \left( \frac { 6 }{ 11 } \right) +\left( \frac { 2 }{ 7 } \right) \left( \frac { 4 }{ 11 } \right) +\left( \frac { 4 }{ 7 } \right) \left( \frac { 5 }{ 11 } \right) } \)
\(=\frac { 8 }{ 6+8+20 } =\frac { 8 }{ 34 } =\frac { 4 }{ 17 } \)

Let A :Two drawn balls are white
E₁ : All the balls are white
E₂ : Three balls are white
E₃ : Two balls are white 
Since, E₁, E₂, and E₃ are mutually exclusive and exhaustive events.
\(\therefore \quad P\left(E_1\right)=P\left(E_2\right)=P\left(E_3\right)=\frac{1}{3}\)
Now, \(P\left(\frac{A}{E_1}\right)=\frac{{ }^4 C_2}{{ }^4 C_2}=1, P\left(\frac{A}{E_2}\right)=\frac{{ }^3 C_2}{{ }^4 C_2}=\frac{3}{6}=\frac{1}{2}\)
and \(P\left(\frac{A}{E_3}\right)=\frac{{ }^2 C_2}{{ }^4 C_2}=\frac{1}{6}\)
\(\therefore\) Probability that all balls in the bag are white
\(P\left(\frac{E_1}{A}\right)=\frac{P\left(E_1\right) \cdot P\left(\frac{A}{E_1}\right)}{\left[P\left(E_1\right) \cdot P\left(\frac{A}{E_1}\right)+P\left(E_2\right) \cdot P\left(\frac{A}{E_2}\right)\right.}\)\(\left.+P\left(E_3\right) \cdot P\left(\frac{A}{E_3}\right)\right]\)
\(=\frac{\frac{1}{3} \times 1}{\frac{1}{3} \times 1+\frac{1}{3} \times \frac{1}{2}+\frac{1}{3} \times \frac{1}{6}}\)\(=\frac{1}{1+\frac{1}{2}+\frac{1}{6}}=\frac{6}{10}=0.6\)

Let the events E₁ and E₂ be the events when lost card is a diamond and not a diamond respectively.
\(P({ E }_{ 1 })=\frac { 13 }{ 52 } =\frac { 1 }{ 4 } \)and
\(P({ E }_{ 2 })=\frac { 39 }{ 52 } =\frac { 3 }{ 4 } \)
Let A be the event:
"two cards drawn from the remaining pack are diamonds"
\(P(A/{ E }_{ 1 })=\frac { 12\times 11 }{ 51\times 50 } \)
\(P(A/{ E }_{ 2 })=\frac { 13\times 12 }{ 51\times 50 } \)
By Bayes' Theorem
\(P({ E }_{ 1 }/A)=\frac { P({ E }_{ 1 })P(A/{ E }_{ 1 }) }{ P({ E }_{ 1 })P(A/{ E }_{ 1 })+P({ E }_{ 2 })P(A/{ E }_{ 2 }) } \)
\(=\frac { \left( \frac { 1 }{ 4 } \right) \left( \frac { 12\times 11 }{ 51\times 50 } \right) }{ \left( \frac { 1 }{ 4 } \right) \left( \frac { 12\times 11 }{ 51\times 50 } \right) +\left( \frac { 3 }{ 4 } \right) \left( \frac { 13\times 12 }{ 51\times 50 } \right) } \)
\(=\frac { 12\times 11 }{ 12\times 11+3\times 13\times 12 } \)
\(=\frac { 132 }{ 132+468 } =\frac { 132 }{ 600 } =\frac { 11 }{ 50 } \)

Let the events Be:
E₁ : coin is two headed
E₂ : coin is biased(heads 75%)
E₃ : Coin is biased(tails 40%)
and A : coin shows up head
\(P({ E }_{ 1 })=P({ E }_{ 2 })=P({ E }_{ 3 })=\frac { 1 }{ 3 } \)
\(P(A/{ E }_{ 1 })=1,P(A/{ E }_{ 2 })=\frac { 75 }{ 100 } =\frac { 3 }{ 4 } ,\)
By Bayes' theorem
\(P({ E }_{ 1 }​/A)=\frac { P({ E }_{ 1 })P(A/{ E }_{ 1 }) }{ P({ E }_{ 1 })P(A/{ E }_{ 1 })+P({ E }_{ 2 })P(A/{ E }_{ 2 })+P({ E }_{ 3 })P(A/{ E }_{ 3 }) } \)
\(=\frac { \frac { 1 }{ 3 } \times 1 }{ \frac { 1 }{ 3 } \times 1+\frac { 1 }{ 3 } \times \frac { 3 }{ 4 } +\frac { 1 }{ 3 } \times \frac { 3 }{ 5 } } \)
\(=\frac{4}{4+3+2}=\frac{4}{9}\)

Let E₁ and E₂ denote the events that first and second group will win. Then,
P(E₁) = 0.6 and P(E₂) = 0.4
Let E be the event of introducing the new product.
Then, \(P\left(\frac{E}{E_1}\right)=0.7 \text { and } P\left(\frac{E}{E_2}\right)=0.3\)
Now, we have to find the probability that new product is introduced by second event.
\(\therefore P\left(\frac{E_2}{E}\right)=\frac{P\left(E_2\right) P\left(\frac{E}{E_2}\right)}{P\left(E_1\right) P\left(\frac{E}{E_1}\right)+P\left(E_2\right) P\left(\frac{E}{E_2}\right)}\)
\(=\frac{0.4 \times 0.3}{0.6 \times 0.7+0.4 \times 0.3}=\frac{0.12}{0.42+0.12}=\frac{0.12}{0.54}\)
= 0.22

Out of 9 students 4 can be selected in ⁹C₄ ways favourable
cases of selecting 2 boys and 2 girls out of 4 boys and 5 girls is \({ }^{4} C_{2} \times{ }^{5} C_{2}\)
probability of selecting 2 boys and 2 girls out of 9
\(=\frac{{ }^{4} C_{2} \times{ }^{5} C_{2}}{{ }^{9} C_{4}}=\frac{6 \times 10 \times 24}{9 \times 8 \times 7 \times 6}=\frac{10}{21}\)

Bag A. : 6 red + 5 blue; Bag B : 5 red + 8 blue
let a red ball be drawn from bag A
\(P\left(\mathrm{red}_{A}\right)=\frac{6}{11}\)
Number of balls in bag B: (5+1) red + 8 blue
\(P\left(\text { blue }_{B}\right)=\frac{8}{14}\)
therefore probability of drawing a blue ball from bag B, when a red ball is transferred from bag A to bag B is
\(=P\left(\text { red }_{A}\right) \cdot P\left(\text { blue }_{B}\right) \)
\(=\frac{6}{11} \times \frac{8}{14} \)
let blue ball is drawn from bag A
\(P\left(\text { blue }_{A}\right)=\frac{5}{11}\)
Number of balls in bag B : 5 red + (8 + I) blue
\(P\left(\text { blue }_{B}\right)=\frac{9}{14}\)
Probability of drawing a blue ball from bag B when a blue ball is transferred from bag A to bag B is
\(=P\left(\text { blue }_{A}\right) \cdot P\left(\text { blue }_{B}\right)\)
\(=\frac{5}{11} \times \frac{9}{14} \)
Hence, probability of drawing a blue ball from bag B when a ball is transferred from bag A to bag B = \(\frac{6}{11} \times \frac{8}{14}+\frac{5}{11} \times \frac{9}{14}=\frac{48+45}{154}=\frac{93}{154}\)

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
\(P({ E }_{ 1 })=\frac { 13 }{ 52 } =\frac { 1 }{ 4 } ,\quad P({ E }_{ 2 })=1-\frac { 1 }{ 4 } =\frac { 3 }{ 4 } \)
\(P(A/{ E }_{ 1 })=\frac { { 12 }_{ C_{ 3 } } }{ { 51 }_{ { C }_{ 3 } } } ,\quad P(A/{ E }_{ 2 })=\frac { 13_{ { C }_{ 3 } } }{ { 51 }_{ { C }_{ 3 } } } \)
\(P({ E }_{ 1 }/A)=\frac { \frac { 1 }{ 4 } .\frac { { 12 }_{ C_{ 3 } } }{ { 51 }_{ { C }_{ 3 } } } }{ \frac { 1 }{ 4 } .\frac { { 12 }_{ C_{ 3 } } }{ { 51 }_{ { C }_{ 3 } } } +\frac { 3 }{ 4 } .\frac { { 12 }_{ C_{ 3 } } }{ { 51 }_{ { C }_{ 3 } } } } \)
\(=\frac { 10 }{ 49 } \)