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14.Probability
easy
$A$ and $B$ are two events such that $P(A)=0.54$, $P(B)=0.69$ and $P(A \cap B)=0.35.$ Find $P \left( A ^{\prime} \cap B ^{\prime}\right)$.
A
$0.12$
B
$0.12$
C
$0.12$
D
$0.12$
Solution
It is given that $P ( A )=0.54$, $P ( B )=0.69$, $P (A \cap B)=0.35$
$A^{\prime} \cap B^{\prime}=(A \cup B)^{\prime}$ [by De Morgan's law]
$\therefore P \left(A^{\prime} \cap B^{\prime}\right)$ $= P (A \cup B)^{\prime}=1- P (A \cup B)=1-0.88=0.12$
Standard 11
Mathematics