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CBSE Class 12 Math 2020 Delhi Set 1 Solved Paper

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Question : 10 of 36
Marks: +1, -0
A die is thrown once. Let AA be the event that the number obtained is greater than 3. Let BB be the event that the number obtained is less than 5 . Then P(A∪B)P(A \cup B) is
Solution:     👈: Video Solution
P(A)=36P(A)=\frac{3}{6}
P(B)=46P(B)=\frac{4}{6}
and
P(A∩B)=16P(A \cap B)=\frac{1}{6}
We know,
P(A∪B)=P(A)+P(B)−P(A∩B)P(A \cup B)=P(A)+P(B)-P(A \cap B)
=36+46−16=\frac{3}{6}+\frac{4}{6}-\frac{1}{6}
=66=1=\frac{6}{6}=1
∴\therefore Option (d) is correct.
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