Probability Part 1

Given set $S=\{1,2,3,\dots 8\}$ Choosing 3 numbers from 8 numbers can be done ${}^8C_3$ ways. Choosing 3 numbers from 8 numbers while minimum no is 3 can be done $1 \times {}^5C_2$ ways.$\therefore$ Probability $P(\text{min}=3)=\frac{1 \times {}^5C_2}{{}^8C_3}$ Choosing 3 numbers from 8 numbers while maximum no is 6 can be done $1 \times {}^5C_2$ ways.$\therefore$ Probability $P(\text{max}=6)=\frac{1 \times {}^5C_2}{{}^8C_3}$ Choosing 3 numbers from 8 numbers while minimum number 3 and maximum no is 6 can be done $1 \times {}^2C_1 \times 1$ ways.$\therefore P(\text{min}=3 \cap \text{max}=6)=\frac{1 \times {}^2C_1 \times 1}{{}^8C_3}$The probability that their minimum is 3 , given that their maximum is 6 , is :$P\left(\frac{\text{min}=3}{\text{max}=6}\right)$ $=\frac{P(\text{min}=3 \cap \text{max}=6)}{P(\text{max}=6)}$$=\frac{\frac{{}^2C_1}{{}^8C_3}}{\frac{{}^5C_2}{{}^8C_3}}$$=\frac{{}^2C_1}{{}^5C_2}$$=\frac{1}{5}$