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, 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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The correct option is C ()/()
As xr<xr+1.This means x1<x2<x3......<x40.
Hence, each of the 19 numbers is less than the number x20.
And each of the 30 numbers x21,x22........x40 is greater than x20.
Out of the five numbers that are randomly picked from the set, when two numbers are picked from the set {x1,x2.....x19} and two others picked from the set {x21,x22........x40} then the number x20 will always be in the middle.
When five numbers are arranged in an order, total number of ways of selecting such five numbers is 19C2×20C2.
And the total number of ways of selecting a set of any five numbers out of 40 is 40C5.
()/()
As xr<xr+1.This means x1<x2<x3......<x40.
Hence, each of the 19 numbers is less than the number x20.
And each of the 30 numbers x21,x22........x40 is greater than x20.
Out of the five numbers that are randomly picked from the set, when two numbers are picked from the set {x1,x2.....x19} and two others picked from the set {x21,x22........x40} then the number x20 will always be in the middle.
When five numbers are arranged in an order, total number of ways of selecting such five numbers is 19C2×20C2.
And the total number of ways of selecting a set of any five numbers out of 40 is 40C5.
