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 1

The correct option is B

$\frac{{}^{30}{C}_2\times {}^{19}{C}_2}{{}^{50}{C}_5}$

Explanation for the correct option

Step 1: Note the given data

Total number of terms$=50$

$x_r<x_{r+1},r=1,2,3,....50$. The real numbers are arranged in ascending order.

Since $x_{20}$ is the middle term, the number of cases for the first and second places will be $19$ and for the fourth and fifth places, the number of cases will be $30$.

Step 2: Find the probability of the event

The number of ways to $n$ terms in $r$place is ${}^{n}C_r$.

Let $S$ be the event of arranging $50$ numbers differently taking $5$ numbers at a time.

$\therefore n\left(S\right)={{}^{50}C}_5$

The number of ways numbers can be arranged in the first two places${{}^{19}C}_2$.

The number of ways numbers can be arranged in the last two places${{}^{30}C}_2$.

Let $E$be the event that $x_{20}$is the middle number and the numbers are arranged in ascending order.

$\therefore P\left(E\right)=\frac{{{}^{19}C}_2\times {{}^{30}C}_2}{{{}^{50}C}_5}$

Hence the correct option is B.