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UPSC NDA I 2014 Mathematics Paper

For the next four (04) items that follow:

Number X is randomly selected from the set of odd numbers and Y is randomly selected from the set of even numbers of the set

{1, 2, 3, 4, 5, 6, 7}.

Let Z = (X + Y).

[NDA I 2014]

Question : 114 of 120

Marks: +1, -0

Solution:

The given set = {1,2,3,4,5,6,7}

Odd set = {1,3,5,7}

Even set = {2,4,6}

Number of ways selecting odd number(Y) and even number(X) is

{}^4C_1 \times {}^3C_1

Where

Z = X \times Y

= 1 \times 2

= 3 \times 2

= 5 \times 2

= 7 \times 2

Therefore, four valid cases exist.

Probability

= \frac{4}{{}^4C_1 \times {}^3C_1} = \frac{1}{3}

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