JEE MAIN - Mathematics (2012 - No. 7)
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 :
$${3 \over 8}$$
$${1 \over 5}$$
$${1 \over 4}$$
$${2 \over 5}$$
Explanation
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 :
$$P\left( {{{\min = 3} \over {\max = 6}}} \right)$$
= $${{P\left( {\min = 3 \cap \max = 6} \right)} \over {P\left( {\max = 6} \right)}}$$
= $${{{{{}^2{C_1}} \over {{}^8{C_3}}}} \over {{{{}^5{C_2}} \over {{}^8{C_3}}}}}$$
= $${{{}^2{C_1}} \over {{}^5{C_2}}}$$
= $${{1 \over 5}}$$
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 :
$$P\left( {{{\min = 3} \over {\max = 6}}} \right)$$
= $${{P\left( {\min = 3 \cap \max = 6} \right)} \over {P\left( {\max = 6} \right)}}$$
= $${{{{{}^2{C_1}} \over {{}^8{C_3}}}} \over {{{{}^5{C_2}} \over {{}^8{C_3}}}}}$$
= $${{{}^2{C_1}} \over {{}^5{C_2}}}$$
= $${{1 \over 5}}$$
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