A) \[\frac{14}{29}\]
B) \[\frac{20}{39}\]
C) \[\frac{1}{2}\]
D) None of these
Correct Answer: B
Solution :
Key Idea In out of \[40\] consecutive numbers, \[20\] are odd and \[20\] are even numbers. Now, the sum of two numbers is odd only when one is odd and other is even. \[\therefore \]Required probability\[=\frac{^{20}{{C}_{1}}{{\cdot }^{20}}{{C}_{1}}}{^{40}{{C}_{2}}}\] \[=\frac{20\times 20}{\frac{40\times 39}{2\times 1}}=\frac{20\times 20}{20\times 39}\] \[=\frac{20}{39}\]
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