x(1) , x(2) , x(3)… . x(50) are fifty re...

`x_(1) , x_(2) , x_(3)`… . `x_(50)` are fifty real numbers such that `x_(r) lt 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 :

`(""^(20) C _(2) xx ""^(30) C _(2))/(""^(50) C _(5))`

`(""^(30)C _(2)xx ""^(19) C_(2))/( ""^(50)C _(5))`

`( ""^(19) C _(2) xx ""^(31) C _(2))/(""^(50) C _(5))`

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BITSAT GUIDE-PROBABILITY -BITSET ARCHIVES

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