Question
$${x_1},\,{x_2},\,{x_3},.....,{x_{50}}$$ are fifty real numbers such that $${x_r} < {x_{r + 1}}$$ for $$r = 1,\,2,\,3,.....,49.$$ Five numbers out of these are picked up at random. The probability that the five numbers have $${x_{20}}$$ as the middle number is :
A.
$$\frac{{{}^{20}{C_2} \times {}^{30}{C_2}}}{{{}^{50}{C_5}}}$$
B.
$$\frac{{{}^{30}{C_2} \times {}^{19}{C_2}}}{{{}^{50}{C_5}}}$$
C.
$$\frac{{{}^{19}{C_2} \times {}^{31}{C_3}}}{{{}^{50}{C_5}}}$$
D.
none of these
Answer :
$$\frac{{{}^{30}{C_2} \times {}^{19}{C_2}}}{{{}^{50}{C_5}}}$$
Solution :
$$\eqalign{ & n\left( S \right) = {}^{50}{C_5},\,n\left( E \right) = {}^{30}{C_2} \times {}^{19}{C_2} \cr & \therefore \,\,P\left( E \right) = \frac{{{}^{30}{C_2} \times {}^{19}{C_2}}}{{{}^{50}{C_5}}}. \cr} $$
$$\eqalign{ & n\left( S \right) = {}^{50}{C_5},\,n\left( E \right) = {}^{30}{C_2} \times {}^{19}{C_2} \cr & \therefore \,\,P\left( E \right) = \frac{{{}^{30}{C_2} \times {}^{19}{C_2}}}{{{}^{50}{C_5}}}. \cr} $$