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UPSC NDA 2 2012 Mathematics Paper

Directions: For the next two (02) questions that follow:

The dice each numbered from 1 to 6 are thrown together.

Let

A

and

B

be two events given by

A: even number on the first die

B: number on the second die is greater than 4

Question : 119 of 120

Marks: +1, -0

What is

P(A \cup B)

Solution:

A : even numbers

=\{2,4,6\}

B

: number

> 4=\{5,6\} A \cap B=\{6\}

P(A \cap B)=\frac{1}{6}

P(A)=\frac{3}{6}=\frac{1}{2}

P(B)=\frac{2}{6}=\frac{1}{3}

P(A \cup B)=P(A)+P(B)-P(A \cap B)

=\frac{1}{2}+\frac{1}{3}-\frac{1}{6}=\frac{3+2-1}{6}

=\frac{4}{6}=\frac{2}{3}

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