For an event, odds against is $6 : 5$. The probability that event does not occur, is
$\frac{5}{6}$
$\frac{6}{{11}}$
$\frac{5}{{11}}$
$\frac{1}{6}$
(b) Required probability $ = \frac{6}{{6 + 5}} = \frac{6}{{11}}.$
A die is thrown. Let $A$ be the event that the number obtained is greater than $3.$ Let $B$ be the event that the number obtained is less than $5.$ Then $P\left( {A \cup B} \right)$ is
$A$ 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 events for which $P(A) = x$, $P(B) = y,$$P(A \cap B) = z,$ then $P(\bar A \cap B)$ equals
If $A$ and $B$ are two events of a random experiment, $P\,(A) = 0.25$, $P\,(B) = 0.5$ and $P\,(A \cap B) = 0.15,$ then $P\,(A \cap \bar B) = $
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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