Let the events be As:
A : At least one girl is chosen
B : At least 2 girls are chosen
We requireP(B/A)
\(P(\overset { - }{ A } )\)= P(No girl is chosen)
\(=\frac { ^{ 8 }{ C }_{ 4 } }{ ^{ 12 }{ C }_{ 4 } } =\frac { 70 }{ 495 } =\frac { 14 }{ 99 } \)
\(P(A)=1-P(\overset { - }{ A } )=1-\frac { 14 }{ 99 } =\frac { 85 }{ 99 } \)
And \(P(A\cap B)\)= P(2 boys and 2 girls) = \(\frac { ^{ 8 }{ C }_{ 2 }.^{ 4 }{ C }_{ 2 } }{ ^{ 12 }{ C }_{ 4 } } \)
\(=\frac { 6\times 28 }{ 495 } =\frac { 56 }{ 165 } \)
\(P(B/A)=\frac { P(A\cap B) }{ P(A) } =\frac { 56 }{ 165 } \times \frac { 99 }{ 85 } =\frac { 168 }{ 425 } \)

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 = Bag I is selected
E ₂ = Bag II is selected
and A = Black ball is drawn
Then,\(P\left(E_{1}\right)=P\left(E_{2}\right)=\frac{1}{2}, P\left(\frac{A}{E_{1}}\right)=\frac{3}{5}, P\left(\frac{A}{E_{2}}\right)=\frac{2}{6}\) 
\(\therefore\) Required probability
\(P(A)=P\left(E_{1}\right) \times P\left(\frac{A}{E_{1}}\right)+P\left(E_{2}\right) \times P\left(\frac{A}{E_{2}}\right)\) 
= \(\frac{7}{15}\) 

Since, X = Number of four seen
On tossing two dice, X = 0, 1, 2
Also, \(P(4)=\frac{1}{10} \text { and } P(\operatorname{not} 4)=\frac{9}{10}\) 
So, \(P(X=0)=P(\operatorname{not} 4) \cdot P(\operatorname{not} 4)=\frac{9}{10} \cdot \frac{9}{10}=\frac{81}{100}\) 
\(P(X=1)=P(\operatorname{not} 4) \cdot P(4)+P(4) \cdot P(\operatorname{not} 4)\) 
\(=\frac{9}{10} \cdot \frac{1}{10}+\frac{1}{10} \cdot \frac{9}{10}=\frac{18}{100}\) 
\(P(X=2)=P(4) \cdot P(4)=\frac{1}{10} \cdot \frac{1}{10}=\frac{1}{100}\) 
Thus, we get the following table
\(\begin{array}{|c|c|c|c|} \hline X & 0 & 1 & 2 \\ \hline P(X) & 81 / 100 & 18 / 100 & 1 / 100 \\ \hline \end{array}\) 
\(\begin{array}{c|c|c|c} \hline X P(X) & 0 & 18 / 100 & 2 / 100 \\ \hline X^{2} P(X) & 0 & 18 / 100 & 4 / 100 \\ \hline \end{array}\) 
\(\therefore \operatorname{Var}(X)=E\left(X^{2}\right)-[E(X)]^{2}=\Sigma X^{2} P(X)-\left[\sum X P(X)\right]^{2}\) 
\(=\left(0+\frac{18}{100}+\frac{4}{100}\right)-\left(0+\frac{18}{100}+\frac{2}{100}\right)^{2}\) 
\(=\frac{22}{100}-\left(\frac{20}{100}\right)^{2}=\frac{11}{50}-\frac{1}{25}\) 
\(=\frac{11-2}{50}=\frac{9}{50}=\frac{18}{100}=0.18\)

\(\begin{array}{c|l|l|l|l} \hline x & 0 & 500 & 2000 & 3000 \\ \hline P(X) & \frac{9995}{10000} & \frac{3}{10000} & \frac{1}{10000} & \frac{1}{10000} \\ \hline P_{i} X_{1} & 0 & \frac{1500}{10000} & \frac{2000}{10000} & \frac{3000}{10000} \\ \hline \end{array}\) 
\(\because E(X)=\Sigma X P(X)\) 
= 0.65