If odds against solving a question by three students are $2 : 1 ,  5:2$ and $5:3$ respectively, then probability that the question is solved only by one student is

The probability that $A$ speaks truth is $\frac{4}{5}$, while this probability for $B$ is $\frac{3}{4}$. The probability that they contradict each other when asked to speak on a fact

$A , B, C$ try to hit a target simultaneously but independently. Their respective probabilities of hitting targets are $\frac{3}{4},\frac{1}{2},\frac{5}{8}$. The probability that the target is hit by $A$ or $B$ but not by $C$ is

Twelve tickets are numbered $1$ to $12$. One ticket is drawn at random, then the probability of the number to be divisible by $2$ or $3$, is

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

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