If $A$ and $B$ are two events such that $P\,(A \cup B) = P\,(A \cap B),$ then the true relation is
If $A$ and $B$ are two events such that $P\,(A \cup B) = P\,(A \cap B),$ then the true relation is
- [IIT 1998]
- A
$P\,(A) + P\,(B) = 0$
- B
$P\,(A) + P\,(B) = P\,(A)\,P\,\left( {\frac{B}{A}} \right)$
- C
$P\,(A) + P\,(B) = 2\,P\,(A)\,P\,\left( {\frac{B}{A}} \right)$
- D
None of these
Similar Questions
$A$ and $B$ are events such that $P(A)=0.42$, $P(B)=0.48$ and $P(A$ and $B)=0.16 .$ Determine $P ($ not $B).$
$A$ and $B$ are events such that $P(A)=0.42$, $P(B)=0.48$ and $P(A$ and $B)=0.16 .$ Determine $P ($ not $B).$
If $P\,({A_1} \cup {A_2}) = 1 - P(A_1^c)\,P(A_2^c)$ where $c$ stands for complement, then the events ${A_1}$ and ${A_2}$ are
If $P\,({A_1} \cup {A_2}) = 1 - P(A_1^c)\,P(A_2^c)$ where $c$ stands for complement, then the events ${A_1}$ and ${A_2}$ are
If $A$ and $B$ are two independent events, then $P\,(A + B) = $
If $A$ and $B$ are two independent events, then $P\,(A + B) = $
Events $\mathrm{A}$ and $\mathrm{B}$ are such that $\mathrm{P}(\mathrm{A})=\frac{1}{2}, \mathrm{P}(\mathrm{B})=\frac{7}{12}$ and $\mathrm{P}$ $($ not $ \mathrm{A}$ or not $\mathrm{B})=\frac{1}{4} .$ State whether $\mathrm{A}$ and $\mathrm{B}$ are independent?
Events $\mathrm{A}$ and $\mathrm{B}$ are such that $\mathrm{P}(\mathrm{A})=\frac{1}{2}, \mathrm{P}(\mathrm{B})=\frac{7}{12}$ and $\mathrm{P}$ $($ not $ \mathrm{A}$ or not $\mathrm{B})=\frac{1}{4} .$ State whether $\mathrm{A}$ and $\mathrm{B}$ are independent?
In a hostel, $60 \%$ of the students read Hindi newspaper, $40 \%$ read English newspaper and $20 \%$ read both Hindi and English newspapers. A student is selected at random. If she reads Hindi newspaper, find the probability that she reads English newspaper.
In a hostel, $60 \%$ of the students read Hindi newspaper, $40 \%$ read English newspaper and $20 \%$ read both Hindi and English newspapers. A student is selected at random. If she reads Hindi newspaper, find the probability that she reads English newspaper.