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(

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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

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