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14.Probability
easy
A die is thrown. Let $A$ be the event that the number obtained is greater than $3.$ Let $B$ be the event that the number obtained is less than $5.$ Then $P\left( {A \cup B} \right)$ is
A
$\frac{3}{5}$
B
$0$
C
$1$
D
$\frac{2}{5}$
(AIEEE-2008)
Solution
$A=\{4 ; 5,6\}$ and $B=\{1,2,3,4\}$
$A \cap B=\{4\}$
[by addition theorem of probability] $P(A \cup B)=P(A)+P(B)-P(A \cap B)$
$P(A \cup B)=\frac{3}{6}+\frac{4}{6}-\frac{1}{6}=1$
Standard 11
Mathematics
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