If probability that exactly one of events $A, B, C$ occurs, is $0.6$ and probability that none of $A, B, C$ occur is $0.2$ then probability that atleast two of $A, B, C$ occur is

A dice is weighted so that the probability of different faces to turn up is as given

If $P\left(\frac{A}{B}\right)=p_1$ and $P\left(\frac{B}{C}\right)=p_2$ and $P\left(\frac{C}{A}\right)=p_3$, then the values of $p_1, p_2, p_3$, respectively are Take the events $A, B \& C$ as $A=\left\{1,2,3\right\}, B=\left\{2,3,5\right\}$ and $C=\left\{2,4,6\right\}$.