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}}.$
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 ?
Given that the events $A$ and $B$ are such that $P(A)=\frac{1}{2}, P(A \cup B)=\frac{3}{5}$ and $\mathrm{P}(\mathrm{B})=p .$ Find $p$ if they are mutually exclusive.
If $A$ and $B$ are arbitrary events, then
Three persons $P, Q$ and $R$ independently try to hit a target . If the probabilities of their hitting the target are $\frac{3}{4},\frac{1}{2}$ and $\frac{5}{8}$ respectively, then the probability that the target is hit by $P$ or $Q$ but not by $R$ is
If $P(B) = \frac{3}{4}$, $P(A \cap B \cap \bar C) = \frac{1}{3}{\rm{ }}$ and $P(\bar A \cap B \cap \bar C) = \frac{1}{3},$ then $P(B \cap C)$ is
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