If $A$ and $B$ are any two events, then $P(A \cup B) = $
$P(A) + P(B)$
$P(A) + P(B) + P(A \cap B)$
$P(A) + P(B) - P(A \cap B)$
$P(A)\,\,.\,\,P(B)$
(c) It is a fundamental concept.
The probability that $A$ speaks truth is $\frac{4}{5}$, while this probability for $B$ is $\frac{3}{4}$. The probability that they contradict each other when asked to speak on a fact
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 two events such that $P(A)=0.54$, $P(B)=0.69$ and $P(A \cap B)=0.35.$ Find $P \left( B \cap A ^{\prime}\right)$.
Let $A$ and $B$ be two events such that $P\overline {(A \cup B)} = \frac{1}{6},P(A \cap B) = \frac{1}{4}$ and $P(\bar A) = \frac{1}{4},$ where $\bar A$ stands for complement of event $A$. Then events $A$ and $B$ are
Check whether the following probabilities $P(A)$ and $P(B)$ are consistently defined $P ( A )=0.5$, $ P ( B )=0.4$, $P ( A \cap B )=0.8$
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