QJEEM12Ma64a6: JEE Main 2012 Mathematics - Class 12 JEEM12Ma64a6 Solution: Three numbers are chosen at random without replacement from The …

Given set S = $\left\{ {1,2,3,..8} \right\}$

Choosing 3 numbers from 8 numbers can be done ${{}^8{C_3}}$ ways.

Choosing 3 numbers from 8 numbers while minimum no is 3 can be done $1 \times {}^5{C_2}$ ways.

$\therefore$ Probablity P(min = 3) = ${{1 \times {}^5{C_2}} \over {{}^8{C_3}}}\,$

Choosing 3 numbers from 8 numbers while maximum no is 6 can be done $1 \times {}^5{C_2}$ ways.

$\therefore$ Probablity P(max = 6) = ${{1 \times {}^5{C_2}} \over {{}^8{C_3}}}\,$

Choosing 3 numbers from 8 numbers while minimum number 3 and maximum no is 6 can be done $1 \times {}^2{C_1} \times 1$ ways.

$\therefore$ $P\left( {\min = 3 \cap \max = 6} \right)$ = ${{1 \times {}^2{C_1} \times 1} \over {{}^8{C_3}}}$

The probability that their minimum is $3,$ given that their maximum is $6,$ is :
$P\left( {{{\min = 3} \over {\max = 6}}} \right)$

= ${{P\left( {\min = 3 \cap \max = 6} \right)} \over {P\left( {\max = 6} \right)}}$

= ${{{{{}^2{C_1}} \over {{}^8{C_3}}}} \over {{{{}^5{C_2}} \over {{}^8{C_3}}}}}$

= ${{{}^2{C_1}} \over {{}^5{C_2}}}$

= ${{1 \over 5}}$