Question

x${}_1$,x${}_2$,x${}_3$,...,x${}_{40}$ are forty real numbers such that x${}_r$$<$x${}_{r+1}$for r = 1, 2, 3... 39. Five numbers out of these are picked up at random. The probability that the five numbers have x${}_{20}$ as the middle number is:

• A. 19/55
• B. (¹⁹C₂ײ⁰C₂)/( ⁴⁰C₄)
• C. (¹⁹C₂ײ⁰C₂)/( ⁴⁰C₅)
• D. (¹⁹C₃ײ⁰C₂)/( ⁴⁰C₄)
• E. None of these

Answer 1

The correct option is C

(¹⁹C₂ײ⁰C₂)/( ⁴⁰C₅)

As x${}_r$$<$x${}_{r+1}$.This means x${}_1$$<$x${}_2$$<$x${}_3$......$<$x${}_{40}$.

Hence, each of the 19 numbers is less than the number x${}_{20}$.

And each of the 30 numbers x${}_{21}$,x${}_{22}$........x${}_{40}$ is greater than x${}_{20}$.

Out of the five numbers that are randomly picked from the set, when two numbers are picked from the set {x${}_1$,x${}_2$.....x${}_{19}$} and two others picked from the set {x${}_{21}$,x${}_{22}$........x${}_{40}$} then the number x${}_{20}$ will always be in the middle.

When five numbers are arranged in an order, total number of ways of selecting such five numbers is ${}^{19}$C${}_2$$\times$${}^{20}$C${}_2$.

And the total number of ways of selecting a set of any five numbers out of 40 is ${}^{40}$C${}_5$.