Given two mutually exclusive events $A$ and $B$ such that $P(A) = 0.45$ and $P(B) = 0.35,$ then $P (A$ or $B ) =$
$0.1$
$0.25$
$0.15$
$0.8$
(d) $P(A \cup B) = P(A) + P(B) = 0.45 + 0.35 = 0.8.$
The probabilities that $A$ and $B$ will die within a year are $p$ and $q$ respectively, then the probability that only one of them will be alive at the end of the year is
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
Suppose that $A, B, C$ are events such that $P\,(A) = P\,(B) = P\,(C) = \frac{1}{4},\,P\,(AB) = P\,(CB) = 0,\,P\,(AC) = \frac{1}{8},$ then $P\,(A + B) = $
Two events $A$ and $B$ will be independent, if
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.
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