x1,x2,x3,...,x40 are forty real numbers such that xr<xr+1 for r = 1, 2, 3... 39. Five numbers out of these are picked up at random. The probability that the five numbers have x20 as the middle number is:

x1,x2,x3,...,x40 are forty real numbers such that xr<xr+1 for r = 1, 2

1.	x1,x2,x3,...,x40 are forty real numbers such that xr<xr+1 for r = 1, 2, 3... 39. Five numbers out of these are picked up at random. The probability that the five numbers have x20 as the middle number is:
Answer» $x_{1}$ , $x_{2}$ , $x_{3}$ ,..., $x_{40}$ 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:

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x1,x2,x3,...,x40 are forty real numbers such that xr&lt;xr+1 for r = 1, 2, 3... 39. Five numbers out of these are picked up at random. The probability that the five numbers have x20 as the middle number is:

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x1,x2,x3,...,x40 are forty real numbers such that xr<xr+1 for r = 1, 2, 3... 39. Five numbers out of these are picked up at random. The probability that the five numbers have x20 as the middle number is: