Three numbers are chosen at random without replacement from \[\left\{ {1,2,3,..8} \right\}.\] The probability that their minimum is 3, given that their maximum is 6, is :
Solutions · 1
Official solution
Given set S = \[\left\{ {1,2,3,..8} \right\}\]
Choosing 3 numbers from 8 numbers can be done 8C3 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 :
\[P\left( {{{\min = 3} \over {\max = 6}}} \right)\]
= \[{{P\left( {\min = 3 \cap \max = 6} \right)} \over {P\left( {\max = 6} \right)}}\]
= \[{{{{{}^2{C_1}} \over {{}^8{C_3}}}} \over {{{{}^5{C_2}} \over {{}^8{C_3}}}}}\]
= \[{{{}^2{C_1}} \over {{}^5{C_2}}}\]
= \[{{1 \over 5}}\]