\(P(A / B)=\frac{P(A \cap B)}{P(B)}=\frac{P(A)}{P(B)}[\because A \subset B \Rightarrow A \cap B=A]\)

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

\(n(S) =2^{14}=16 \)
\(A =\left\{\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}|,| \begin{array}{ll} 0 & 1 \\ 1 & 0 \end{array}|,| \begin{array}{ll} 1 & 1 \\ 0 & 1 \end{array}|,| \begin{array}{ll} 1 & 0 \\ 1 & 1 \end{array}|,| \begin{array}{ll} 0 & 1 \\ 1 & 1 \end{array}|,| \begin{array}{ll} 1 & 1 \\ 1 & 0 \end{array} \mid\right\} \)
\(n(A) =6 \)
\(P(A) =\frac{6}{16}=\frac{3}{8} \)

Assistant    Statistics
\(\begin{array}{ll} A=2 & S=3 \\ S=3 & T=3 \\ T=2 & A=1 \\ I=1 & I=2 \\ N=1 & C=1 \end{array}\)
\(\mathrm{n}(\mathrm{S})=9 \mathrm{C}_{1} \times 10 \mathrm{C}_{1}=90\)
One letter in I word and another letter (same) in II word the combinations are
\(\begin{array}{ll} \mathrm{AA} & \mathrm{A} \\ \mathrm{SSS} & \mathrm{SSS} \\ \mathrm{TT} & \mathrm{TTT} \\ \mathrm{I} & \mathrm{II} \end{array}\)
No. of doublets are
AA, AA, SS, SS, SS, SS, SS, SS, S$ SS, SS, TT TT,
TT, TT, TT, TT, II, II
n(A) = 19
\(P(A) =\frac{n(A)}{n(S)} \)
\(=\frac{19}{90} \)

Fifty rupee note = 3
Hundred rupee note = 4
500 rupee note = 6
Totsl = 13
\(\mathrm{n}(\mathrm{S})=13 \mathrm{C}_{2}=78\)
\(\text {No. of ways the event to occur }=4\)
\(n(A)=4 C_{2}=6\)
\(\text {No. of ways the event not to occur }=72\)
The odds of event are a:b
\(6: 72 \Rightarrow 1: 12\)

Defective = 4
Not Defective = 8
\(\mathrm{n}(\mathrm{S})=12 \mathrm{C}_{2}=\frac{12 \times 11}{1 \times 2}=66\)
\(\mathrm{P} \text { (one defective) }=\frac{4 \mathrm{C}_{1} \times 8 \mathrm{C}_{1}}{12 \mathrm{C}_{2}}=\frac{32}{66}\)
\(\mathrm{P} \text { (two defective) }=\frac{4 \mathrm{C}_{2}}{12 \mathrm{C}_{2}}=\frac{6}{66}\)
\(P \text { (atleast one defective) }=\frac{32}{66}+\frac{6}{66}=\frac{19}{33}\)

\(P(\bar{A})=\frac{1}{4}, P(A) =\frac{3}{4}, P(A \cap B)=\frac{1}{4} \)
\(P(A \cup B) =\frac{3}{4}+P(B)-\frac{1}{4} \)
\(=\frac{1}{2}+P(B) \)
\(\frac{5}{6} =\frac{1}{2}+P(B) \)
\(P(B) =\frac{5}{6}-\frac{1}{2}=\frac{2}{6}=\frac{1}{3} \)
\(P(\bar{A})=\frac{1}{4}, P(A)=\frac{3}{4}, P(A \cap B)=\frac{1}{4} \)
\(P(A \cap B)=P(A) \cdot P(B)=\frac{3}{4} \times \frac{1}{3}=\frac{1}{4}=P(\text { Any }) \)
A and B are independent but not equally likely.

P(A or B but not C)
\(= P(A) P(B) P(\bar{C})+P(A) P(\bar{B}) P(\bar{C})+P(\bar{A}) P(B) P(\bar{C}) \)
\(= \frac{3}{4} \times \frac{1}{2} \times \frac{3}{8}+\frac{3}{4} \times \frac{1}{2} \times \frac{3}{8}+\frac{1}{4} \times \frac{1}{2} \times \frac{3}{8} \)
\(= \frac{9+9+3}{64}=\frac{21}{64} \)

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
\(P(A) =\frac{10}{20} =\frac{1}{2}\)