Let $X$ and $Y$ are two events such that $P(X \cup Y=P)\,(X \cap Y).$
Statement $1:$ $P(X \cap Y' = P)\,(X' \cap Y = 0).$
Statement $2:$ $P(X) + P(Y = 2)\,P\,(X \cap Y)$
Let $X$ and $Y$ are two events such that $P(X \cup Y=P)\,(X \cap Y).$
Statement $1:$ $P(X \cap Y' = P)\,(X' \cap Y = 0).$
Statement $2:$ $P(X) + P(Y = 2)\,P\,(X \cap Y)$
- [AIEEE 2012]
- A
Statement $1$ is false, Statement $2$ is true.
- B
Statement $1$ is true, Statement $2$ is true,Statement $2$ is not a correct explanation of Statement $1$ .
- C
Statement $1$ is true, Statement $2$ is false
- D
Statement $1$ is true, Statement $2$ is true;Statement $2$ is a correct explanation of Statement $1.$
Similar Questions
$A$ and $B$ are two events such that $P(A)=0.54$, $P(B)=0.69$ and $P(A \cap B)=0.35.$ Find $P \left( A \cap B ^{\prime}\right)$ .
$A$ and $B$ are two events such that $P(A)=0.54$, $P(B)=0.69$ and $P(A \cap B)=0.35.$ Find $P \left( A \cap B ^{\prime}\right)$ .
Let $A$ and $B $ be two events such that $P\left( {\overline {A \cup B} } \right) = \frac{1}{6}\;,P\left( {A \cap B} \right) = \frac{1}{4}$ and $P\left( {\bar A} \right) = \frac{1}{4}$ where $\bar A$ stands for the complement of the event $A$. Then the events $A$ and$B$ are
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- [JEE MAIN 2014]
- [AIEEE 2005]
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