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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