The probability of solving a question by three students are $\frac{1}{2},\,\,\frac{1}{4},\,\,\frac{1}{6}$ respectively. Probability of question is being solved will be
$\frac{{33}}{{48}}$
$\frac{{35}}{{48}}$
$\frac{{31}}{{48}}$
$\frac{{37}}{{48}}$
If the odds in favour of an event be $3 : 5$, then the probability of non-occurrence of the event is
If ${A_1},\,{A_2},...{A_n}$ are any $n$ events, then
The probabilities of three events $A , B$ and $C$ are given by $P ( A )=0.6, P ( B )=0.4$ and $P ( C )=0.5$ If $P ( A \cup B )=0.8, P ( A \cap C )=0.3, P ( A \cap B \cap$ $C)=0.2, P(B \cap C)=\beta$ and $P(A \cup B \cup C)=\alpha$ where $0.85 \leq \alpha \leq 0.95,$ then $\beta$ lies in the interval
Let $A$,$B$ and $C$ be three events such that $P\left( {A \cap \bar B \cap \bar C} \right) = 0.6$, $P\left( A \right) = 0.8$ and $P\left( {\bar A \cap B \cap C} \right) = 0.1$, then the value of $P$(atleast two among $A$,$B$ and $C$ ) equals
Given $P(A)=\frac{3}{5}$ and $P(B)=\frac{1}{5}$. Find $P(A $ or $B),$ if $A$ and $B$ are mutually exclusive events.