\(P(A)=0.3, \quad P(B)=0.6, \quad P(A \cap B)=0.18 \)
\(P(A \cup B)=P(A)+P(B)-P(A \cap B) \)
\(=0.3+0.6-0.18 =0.72 \)
\(P(\bar{A} \cup \bar{B})=1-P(A \cap B) \)
\(=1-0.72=0.28 \)

\(P\left(A_{1}\right)=0.6, \quad P\left(A_{2}\right)=0.4\)
\(A_{1}=\text { Event of selecting a girl }\)
\(\mathrm{B}=\text { Event of student taller than } 1.8 \mathrm{~m}\)
\(A_{2}=\text { Event of selecting a boy }\)
\(P\left(B / A_{1}\right)=\frac{1}{100}, P\left(B / A_{2}\right)=\frac{4}{100}\)
To find \(=\frac{P\left(A_{1}\right) \cdot P\left(B / A_{1}\right)}{P\left(A_{1}\right) \cdot P\left(B / A_{1}\right)+P\left(A_{2}\right) \cdot P\left(B / A_{2}\right)} \)
\(=\frac{0.6 \times \frac{1}{100}}{0.6 \times \frac{1}{100}+0.4 \times \frac{4}{100}} \)
\(=\frac{6}{\frac{6}{1000}+\frac{16}{1000}} \)
\(=\frac{6}{22}=\frac{3}{11} \)

\(\mathrm{P}(\mathrm{A} \cup \mathrm{B}) =\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A} \cap \mathrm{B}) \)
\(0.6 =0.35+\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A}) \cdot \mathrm{P}(\mathrm{B}) \)
\(0.25 =\mathrm{P}(\mathrm{B})-0.35 \mathrm{P}(\mathrm{B}) \)
\(=\mathrm{P}(\mathrm{B})(1-0.35) \)
\(=\mathrm{P}(\mathrm{B})(0.65) \)
\(\mathrm{P}(\mathrm{B}) =\frac{0.25}{0.65}=\frac{5}{13} \)

6G, 2W 7B two are selected
\(n(S) =15 C_{2}=\frac{15 \times 14}{1 \times 2}=105 \)
\(A =1 G 1 W \text { or } 1 W 1 B \text { or } 1 B 1 G \)
\(n(A) =6 C_{1} \times 2 C_{1}+2 C_{1} \times 7 C_{1}+7 C_{1} \times 6 C_{1} \)
\(=12+14+42=68 \)
\(P(A) =\frac{68}{105} \)

Required Probability ={P(WBWBW)+ P(BWBWB)}
\(=\left(\frac{5}{8} \times \frac{3}{7} \times \frac{4}{6} \times \frac{2}{5} \times \frac{3}{4}\right) +\left(\frac{3}{8} \times \frac{5}{7} \times \frac{2}{6} \times \frac{4}{5} \times \frac{1}{4}\right) \)
\(=\frac{3}{56}+\frac{1}{56}=\frac{4}{56}=\frac{1}{14} \)

Number of divisible by 3 = 3, 6, 9,12,15, 18
Number of divisible by 4 = 4, 8, 12, 16,20
n(A) = 10, n(S) = 20
\(P(A) =\frac{10}{20} =\frac{1}{2}\)