X : Number of kings
'X' takes values 0,1, 2
\(P(X=0)=\frac { ^{ 4 }{ C }_{ 0 }\times ^{ 48 }{ C }_{ 2 } }{ ^{ 52 }{ C }_{ 2 } } =\frac { 48\times 47 }{ 52+1 } =\frac { 188 }{ 221 } \)
\(P(X=1)=\frac { ^{ 4 }{ C }_{ 1 }\times ^{ 48 }{ C }_{ 1 } }{ ^{ 52 }{ C }_{ 2 } } =\frac { 4\times 48\times 2 }{ 52\times 51 } =\frac { 32 }{ 221 } \)
\(P(X=3)=\frac { ^{ 4 }{ C }_{ 2 }\times ^{ 48 }{ C }_{ 0 } }{ ^{ 52 }{ C }_{ 2 } } =\frac { 4\times 3 }{ 52\times 51 } =\frac { 1 }{ 221 } \)
Thus we have:

Mean  \(\mu =\frac { 34 }{ 221 } =\frac { 2 }{ 13 } \)
Standard Deviation,
\(\sigma =\sqrt { \sum { { p }_{ i }{ x }_{ i }^{ 2 } } -{ \mu }^{ 2 } } =\sqrt { \frac { 36 }{ 221 } -{ \left( \frac { 2 }{ 13 } \right) }^{ 2 } } \)
\(=\sqrt { 0.1628-0.0236 } =\sqrt { 0.1392 } =0.373\)
Playing card is not healthy because it wastes precious time and there is no physical activity playing cards.

\(9\over11\)

\(4\over9\)

\(P(S)=0.40 ; P(M)=0.25 ; P(M \cap S)=0.15\)
\(\mathrm{S} \rightarrow Statistics; \mathrm{M} \rightarrow Mathematics \)
\((i) P(S / M)=\frac{P(S \cap M)}{P(M)}=\frac{0.15}{0.25}=\frac{3}{5}\)
\((ii) P(M / S)=\frac{P(S \cap M)}{P(S)}=\frac{0.15}{0.40}=\frac{3}{8}\)

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 the events be as below:
E₁ : Plant 1 is chosen
E₂ : Plant 2 is chosen
and A : Motorcycle is of standard quality
We have :\(P({ E }_{ 1 })=\frac { 70 }{ 100 } P({ E }_{ 2 })=\frac { 30 }{ 100 } \)
\(P({ A/E }_{ 1 })\frac { 80 }{ 100 } P(A/{ E }_{ 2 })=\frac { 90 }{ 100 } \)
By Bayes' Theorem,
(i) \(P({ E }_{ 1 }/A)=\frac { P({ E }_{ 1 })P(A/{ E }_{ 1 }) }{ P({ E }_{ 1 })P(A/{ E }_{ 1 })+P({ E }_{ 2 })P(A/{ E }_{ 2 }) } \)
\(=\frac { \left( \frac { 70 }{ 100 } \right) \left( \frac { 80 }{ 100 } \right) }{ \left( \frac { 70 }{ 100 } \right) \left( \frac { 80 }{ 100 } \right) +\left( \frac { 30 }{ 100 } \right) \left( \frac { 90 }{ 100 } \right) } \)
\(=\frac { 5600 }{ 5600+2700 } =\frac { 56 }{ 83 } \)
(ii) \(P({ E }_{ 2 }/A)=\frac { P({ E }_{ 2 })P(A/{ E }_{ 21 }) }{ P({ E }_{ 1 })P(A/{ E }_{ 1 })+P({ E }_{ 2 })P(A/{ E }_{ 2 }) } \)
\(=\frac { \left( \frac { 30 }{ 100 } \right) \left( \frac { 90 }{ 100 } \right) }{ \left( \frac { 70 }{ 100 } \right) \left( \frac { 80 }{ 100 } \right) +\left( \frac { 30 }{ 100 } \right) \left( \frac { 90 }{ 100 } \right) } \)
\(=\frac { 2700 }{ 2700+5600 } =\frac { 27 }{ 83 } \)
Riding a motorcycle involves higher risk but children ride at very high speed and do stunts causing accidents.

Sample space ={B₁B₂, B₁G₂, G₁B₂, G₁G₂}, B₁ and G₁ are the older boy and girl respectively.
Let   E₁ = both the children are boys;
E₂ = one of the children are boys;
E₃ = the older child is a boy
Then, (a) P(E₁/E₂) = \(P\left( \frac { { E }_{ 1 }\cap { E }_{ 2 } }{ { E }_{ 2 } } \right) =\frac { \frac { 1 }{ 4 } }{ \frac { 3 }{ 4 } } =\frac { 1 }{ 3 } \)
(b) P(E₁/E₃) = \(P\left( \frac { { E }_{ 1 }\cap { E }_{ 3 } }{ { E }_{ 3 } } \right) =\frac { \frac { 1 }{ 4 } }{ \frac { 2 }{ 4 } } =\frac { 1 }{ 2 } \)

We know that, \(A \cup B\) denotes the occurrence of at least one of A and B and \(A \cap B\) denotes the occurrence of both A and B, simultaneously.
Then, \(P(A \cup B)=0.6 \text { and } P(A \cap B)=0.3\) 
\(\because P(A \cup B)=P(A)+P(B)-P(A \cap B)\)
\(\therefore \quad 0.6=P(A)+P(B)-0.3 \Rightarrow P(A)+P(B)=0.9\) 
\(\Rightarrow\{[1-P(\bar{A})]+[1-P(\bar{B})]\}=0.9\) 
\([\because P(A)=1-P(\bar{A}) \text { and } P(B)=1-P(\bar{B})]\) 
\(\Rightarrow P(\bar{A})+P(\bar{B})=2-0.9=1.1\) 

\(64 \% , as \mathrm{P} (same fact )=P(A B or \bar{A} \bar{B}) \)
\(=\frac{70}{100} \times \frac{85}{100}+\frac{30}{100} \times \frac{15}{100} \)
\(=\frac{5950+450}{10000}=\frac{6400}{10000}=64 \%\)

Let A_T : Event that A speaks truth
and B_T : Event that B speaks truth.
Given, \(\begin{aligned} P\left(A_T\right)=\frac{80}{100}=\frac{4}{5} \end{aligned}\)
\(\begin{aligned} P\left(B_T\right)=\frac{90}{100}=\frac{9}{10} \end{aligned}\)
P(agree) = P(both speaking truth or both telling lie)
\(\begin{aligned} =P\left(A_T B_T \text { or } \bar{A}_T \bar{B}_T\right) \end{aligned}\)
\(\begin{aligned} =P\left(A_T\right) P\left(B_T\right) \text { or } P\left(\bar{A}_T\right) P\left(\bar{B}_T\right) \end{aligned}\)
\(\begin{aligned} =\left(\frac{4}{5}\right)\left(\frac{9}{10}\right)+\left(\frac{1}{5}\right)\left(\frac{1}{10}\right) \end{aligned}\)
\(\begin{aligned} =\frac{36+1}{50}=\frac{37}{50}=\frac{74}{100}=74 \% \end{aligned}\)

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).  

Let, E₁: Two white balls are transferred
E₂: Two red balls are transferred
E₃: One red and one white ball are transferred.
 A: The ball drawn from the bag II is red.
\(P({ E }_{ 1 })=\frac { { 4 }_{ C_{ 2 } } }{ { 9 }_{ { c }_{ 2 } } } =\frac { 4\times 3 }{ 9\times 8 } =\frac { 1 }{ 6 } \)
\({ P({ E } }_{ 2 })=\frac { { 5 }_{ C_{ 2 } } }{ { 9 }_{ C_{ 2 } } } =\frac { 4\times 3 }{ 9\times 8 } =\frac { 5 }{ 18 } \)
\(P(E_{ 3 })=\frac { { 5 }_{ { C }_{ 1 } }\times { 4 }_{ C_{ 1 } } }{ { 9 }_{ C_{ 2 } } } =\frac { 4\times 5\times 2 }{ 9\times 8 } =\frac { 5 }{ 9 } \)
\(P(A/{ E }_{ 1 })=\frac { 3 }{ 8 } ,P(A/{ E }_{ 2 })=\frac { 5 }{ 8 } ,\)
\(P(A/{ E }_{ 3 })=\frac { 4 }{ 8 } \)
The required probability, P(E₃/A), by Bayes' Theorem
\(=\frac { P({ E }_{ 3 }).P(A/{ E }_{ 3 }) }{ P({ E }_{ 1 }).P(A/{ E }_{ 1 })+P({ E }_{ 2 }).P(A/{ E }_{ 2 })+P({ E }_{ 3 }).P(A/{ E }_{ 3 }) } \)
\(=\frac { \frac { 5 }{ 9 } \times \frac { 4 }{ 8 } }{ \frac { 1 }{ 6 } \times \frac { 3 }{ 8 } +\frac { 5 }{ 18 } \times \frac { 5 }{ 8 } +\frac { 5 }{ 9 } \times \frac { 4 }{ 8 } } \)
\(=\frac { 20 }{ 37 } \) 

Let the events be:
E₁ = Choosing 1^st coin
E₂ = Choosing 2^nd coin
E₃ = Choosing 3^rd coin
A: Getting Heads
\(P({ E }_{ 1 })=P({ E }_{ 2 })=P(E_{ 3 })=\frac { 1 }{ 3 } \)
\(P(A/{ E }_{ 1 })=\frac { 40 }{ 100 } ,P(A/{ E }_{ 2 })=\frac { 75 }{ 100 } ,\)
\(P(A/{ E }_{ 3 })=\frac { 1 }{ 2 } \)
\(P({ E }_{ 1 }/A)\)
\(=\frac { P({ E }_{ 1 })P(A/{ E }_{ 1 }) }{ P({ E }_{ 1 })P(A/{ E }_{ 1 })+P({ E }_{ 2 })P(A/{ E }_{ 2 })+P({ E }_{ 3 })P(A/{ E }_{ 3 }) } \)
\(=\frac { \frac { 1 }{ 3 } .\frac { 40 }{ 100 } }{ \frac { 1 }{ 3 } .\frac { 40 }{ 100 } +\frac { 1 }{ 3 } .\frac { 75 }{ 100 } +\frac { 1 }{ 3 } .\frac { 1 }{ 2 } } =\frac { 8 }{ 33 } \)

Two dice are thrown together, so number of outcomes in the sample space is \(36 \Rightarrow n(S)=36\) 
E = Total of4 = {(2, 2),(3, 1),(1, 3)}
\(\Rightarrow n(E)=3\) 
F = Total of 9 or more
\(=\left\{\begin{array}{l} (3,6),(6,3),(4,5),(5,4),(4,6), \\ (6,4),(5,5),(5,6),(6,5),(6,6) \end{array}\right\}\) 
\(\Rightarrow n(F)=10\) 
and G = Total divisible by 5
= {(1, 4), (4, 1), (2, 3), (3, 2), (4, 6), (6, 4), (5, 5)
\(\Rightarrow n(G)=7\) 
Here,\((E \cap F)=\phi \text { and }(E \cap G)=\phi\) 
\(\Rightarrow n(F \cap G)=3 \text { and }(E \cap F \cap G)=\phi\) 
\(\therefore P(E)=\frac{n(E)}{n(S)}=\frac{3}{36}=\frac{1}{12}\) 
\(P(F)=\frac{n(F)}{n(S)}=\frac{10}{36}=\frac{5}{18}\) 
and \(P(G)=\frac{n(G)}{n(S)}=\frac{7}{36}\) 
\(P(F \cap G)=\frac{3}{36}=\frac{1}{12}\) 
Here, we see that \(P(F \cap G) \neq P(F) \cdot P(G)\) 
[since only F and G have common events, so only F and G are used here] 
Hence, there is no pair which is independent.

Given, A and B are two independent events with
\(P(\bar{A} \cap B)=\frac{2}{15} \text { and } P(A \cap \bar{B})=\frac{1}{6}\) 
Now, \(P(\bar{A} \cap B)=\frac{2}{15} \Rightarrow P(B) \cdot P(\bar{A})=\frac{2}{15}\) 
\(\Rightarrow P(B) \cdot[1-P(A)]=\frac{2}{15}\) 
\(\Rightarrow P(B)-P(A) \cdot P(B)=\frac{2}{15}\) 
and \(P(A \cap \bar{B})=\frac{1}{6} \Rightarrow P(A) \cdot P(\bar{B})=\frac{1}{6}\) 
\(\Rightarrow P(A) \cdot[1-P(B)]=\frac{1}{6}\) 
\(\Rightarrow P(A)-P(A) P(B)=\frac{1}{6}\) 
On subtracting Eq. (i) from Eq. (ii), we get
\(P(A)-P(B)=\frac{1}{6}-\frac{2}{15}=\frac{5-4}{30}=\frac{1}{30}\) 
\(\Rightarrow P(A)=\frac{1}{30}+P(B)\) 
Now, on substituting the value of ~A) in Eq. (i), we get
\(P(B)-\left[\frac{1}{30}+P(B)\right] \cdot P(B)=\frac{2}{15}\) 
Let P(B) = x, then
\(x-\left(\frac{1}{30}+x\right) x=\frac{2}{15}\) 
\( \Rightarrow 30 x-(1+30 x) x=4 \Rightarrow 30 x-x-30 x^{2}=4 \)
\(\Rightarrow 30 x^{2}-29 x+4=0 \Rightarrow(6 x-1)(5 x-4)=0\)
\(x=\frac{1}{c} \text { or } \frac{4}{c} \Rightarrow P(B)=\frac{1}{5} \text { or } \frac{4}{-}[\because x=P(B)] \) 
Now, if \(\text { if } P(B)=\frac{1}{6}, \text { then } P(A)=\frac{1}{5}\) 
and if \(P(B)=\frac{4}{5}, \text { then } P(A)=\frac{5}{6}\)

Let E₁ = Event that 5 or 6 is shown on die
and E₂ = Event that 1, 2, 3 or 4 is shown on die
Here, n(E₁) = 2andn(E₂) = 4
Also, n(S) = 6
\(\therefore P\left(E_{1}\right)=\frac{2}{6}=\frac{1}{3}\) 
and \(P\left(E_{2}\right)=\frac{4}{6}=\frac{2}{3}\) 
Let E = The event that exactly one head show up.
\(\therefore P\left(\frac{E}{E_{1}}\right)=P\) (exactly one head show up when coin is tossed thrice)
\(=P\{H T T, T H T, T T H\}=\frac{3}{8}\) 
\(\left[\because \text { total number of outcomes }=2^{3}=8\right]\)
\(P\left(\frac{E}{E_{2}}\right)=P\) (head shows up when coin is tossed once) \(=\frac{1}{2}\) 
The probability that the girl threw 1, 2, 3 or 4 with the die, if she obtained exactly one head, is given by
\(P\left(\frac{E_{2}}{E}\right)=\frac{P\left(E_{2}\right) \cdot P\left(\frac{E}{E_{2}}\right)}{P\left(E_{1}\right) \cdot P\left(\frac{E}{E_{1}}\right)+P\left(E_{2}\right) \cdot P\left(\frac{E}{E_{2}}\right)}\) 
[by Baye's theorem]
\(=\frac{\frac{2}{3} \times \frac{1}{2}}{\frac{1}{3} \times \frac{3}{8}+\frac{2}{3} \times \frac{1}{2}}=\frac{\frac{1}{3}}{\frac{1}{8}+\frac{1}{3}}=\frac{8}{8+3}=\frac{8}{11}\)

Let us detine the events as
E₁ : Students reside in a hostel
E₂ : Students are day scholars
A : Students get A grade
Then,
P(E₁) = Probability that student reside in a hostel
\(=60 \%=\frac{60}{100}\)
and P(E₂) = Probability that students are day scholars = 1 - \(\frac{60}{100}=\frac{40}{100}\)
Also, P(A/E₁) = Probability that hostelers get A grade
\(=30 \%=\frac{30}{100}\)
and P(A/E₂)= Probability that students having day scholars get A grade
\(=20 \%=\frac{20}{100}\)
\(\therefore\) The probability that the selecting student is a hosteler having A grade,
\(P\left(E_1 / A\right)=\frac{P\left(E_1\right) \cdot P\left(A / E_1\right)}{P\left(E_1\right) \cdot P\left(A / E_1\right)+P\left(E_2\right) \cdot P\left(A / E_2\right)}\)
                                 [by Baye's theorem]
\(\begin{aligned} =\frac{\frac{60}{100} \times \frac{30}{100}}{\left(\frac{60}{100} \times \frac{30}{100}\right)+\left(\frac{40}{100} \times \frac{20}{100}\right)} \end{aligned}\)
\(\begin{aligned} =\frac{1800}{1800+800}=\frac{1800}{2600}=\frac{18}{26}=\frac{9}{13} \end{aligned}\)

As there are 2 red balls, so maximum ’ red balls can be 2.

Here, sample space = {1, 2, 3,4,5, 6}, A ⋂ B = {5}, B ⋂ C = {2, 5}, A ⋂ C = {1, 5}, A ⋂ B ⋂ C = {5} and {A U B} ⋂ C = {1, 2, 5}
Also, \(P(A)=\frac{2}{6}, P(B)=\frac{3}{6}, P(C)=\frac{3}{6}\)
\(P(A \cap B)=\frac{1}{6}, P(B \cap C)=\frac{2}{6}, P(A \cap C)=\frac{2}{6} \)
\(P(A \cap B \cap C)=\frac{1}{6} \text { and } P((A \cup B) \cap C)=\frac{3}{6}\)
\((i) \ (\mathbf{b}): P(A \mid B)=\frac{P(A \cap B)}{P(B)}=\frac{1 / 6}{3 / 6}=\frac{1}{3}\)
\((ii) \ (a): P(B \mid C)=\frac{P(B \cap C)}{P(C)}=\frac{2 / 6}{3 / 6}=\frac{2}{3}\)
\((iii) \ (\mathrm{d}): P(A \cap B \mid C)=\frac{P(A \cap B \cap C)}{P(C)}=\frac{1 / 6}{3 / 6}=\frac{1}{3}\)
\((iv) \ (c): P(A \mid C)=\frac{P(A \cap C)}{P(C)}=\frac{2 / 6}{3 / 6}=\frac{2}{3}\)
\( (v) \ (\mathbf{d}): P(A \cup B \mid C)=\frac{P((A \cup B) \cap C)}{P(C)}=\frac{3 / 6}{3 / 6}=1\)

(i) (b): Required probability = \(\frac{{ }^{12} C_{2}}{{ }^{51} C_{2}}\)
\(=\frac{12 \times 11}{51 \times 50}=\frac{22}{425}\)
(ii) (a): Required probability = \(\frac{{ }^{13} C_{2}}{{ }^{51} C_{2}}=\frac{13 \times 12}{51 \times 50}=\frac{26}{425}\)
(iii) (b): Clearly, P(A l E₁) = \(\frac{{ }^{12} C_{2}}{{ }^{51} C_{2}}=\frac{22}{425}\)
\(P\left(A \mid E_{2}\right)=\frac{{ }^{13} C_{2}}{{ }^{51} C_{2}}=\frac{26}{425} \)
\(P\left(A \mid E_{3}\right)=P\left(A \mid E_{4}\right)=\frac{26}{425}\)
\(\therefore \sum_{i=1}^{4} P\left(A \mid E_{i}\right)=\frac{22}{425}+\frac{26}{425}+\frac{26}{425}+\frac{26}{425}=\frac{100}{425}=0.24\)
(iv) (b): P(getting both king) = \(\frac{{ }^{4} C_{2}}{{ }^{52} C_{2}}=\frac{4 \times 3}{52 \times 51}=\frac{1}{221}\)
(v) (a): P(drawing a king) = \(\frac{4}{52}=\frac{1}{13}\)
\( \therefore \quad P(\text { not drawing a king })=1-\frac{1}{13}=\frac{12}{13} \)
\(\therefore \quad \text { Required probability }=\frac{12}{13} \times \frac{12}{13}=\frac{144}{169}\)