Two events $A$ and $B$ will be independent, if
$A$ and $B$ are mutually exclusive
$P\left(A^{\prime} B^{\prime}\right)=[1-P(A)][1-P(B)]$
$P(A)=P(B)$
$P(A)+P(B)=1$
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
$P(A \cup B) = P(A \cap B)$ if and only if the relation between $P(A)$ and $P(B)$ is
Consider an experiment of tossing a coin repeatedly until the outcomes of two consecutive tosses are same. If the probability of a random toss resulting in head is $\frac{1}{3}$, then the probability that the experiment stops with head is.
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
$A$ and $B$ are events such that $P(A)=0.42$, $P(B)=0.48$ and $P(A$ and $B)=0.16 .$ Determine $P ($ not $B).$