Management Aptitude Test Dec 2013 Paper

Question : 98 of 200

Marks: +1, -0

Out of 20 consecutive positive integers, two are chosen at random. The probability that their sum is odd is

$10/19$
$1/20$
$19/20$
$9/19$

Solution:

(a ) Total number of integers =20
Number of ways in which 2 integers can be selected =

^20C_2

=190
For sum to be odd one integer must be odd and other even.
∴ Number of ways in which 1 odd and 1 even integer can be selected = 10 x 10 = 100
∴ P (sum is odd) =

100/190=10/19

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