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 those 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 1

The correct option is B $\frac{{}^{30}{C}_2{\times }^{19}{C}_2}{{}^{50}{C}_5}$

$n\left(S\right){=}^{50}{C}_5,n\left(E\right){=}^{19}{C}_2{\times }^{30}{C}_2$
$\therefore P\left(E\right)=\frac{{}^{30}{C}_2{\times }^{19}{C}_2}{{}^{50}{C}_5}$
By selecting 2 from first 19
and selecting 2 from last 30