Fill in the blanks in following table :
$P(A)$ | $P(B)$ | $P(A \cap B)$ | $P (A \cup B)$ |
$0.35$ | ........... | $0.25$ | $0.6$ |
Here, $P(A)=0.35$, $P(A \cap B)=0.25$, $P(A \cup B)=0.6$
We know that $P (A \cup B)= P ( B )+ P ( B )- P (A \cap B)$
$\therefore $ $0.6=0.35+ P ( B )-0.25$
$\Rightarrow $ $P ( B )=0.6-0.35+0.25$
$\Rightarrow $ $P ( B )=0.5$
Given two independent events $A$ and $B$ such that $P(A) $ $=0.3, \,P(B)=0.6$ Find $P(A$ and $B)$.
If $P(A \cup B) = 0.8$ and $P(A \cap B) = 0.3,$ then $P(\bar A) + P(\bar B) = $
The chance of an event happening is the square of the chance of a second event but the odds against the first are the cube of the odds against the second. The chances of the events are
If $A$ and $B$ are events such that $P(A \cup B) = 3/4,$ $P(A \cap B) = 1/4,$ $P(\bar A) = 2/3,$ then $P(\bar A \cap B)$ is
Let $\mathrm{E}$ and $\mathrm{F}$ be events with $\mathrm{P}(\mathrm{E})=\frac{3}{5}, \mathrm{P}(\mathrm{F})$ $=\frac{3}{10}$ and $\mathrm{P}(\mathrm{E} \cap \mathrm{F})=\frac{1}{5} .$ Are $\mathrm{E}$ and $\mathrm{F}$ independent ?