One bag contains $5$ white and $4$ black balls. Another bag contains $7$ white and $9$ black balls. A ball is transferred from the first bag to the second and then a ball is drawn from second. The probability that the ball is white, is
$\frac{8}{{17}}$
$\frac{{40}}{{153}}$
$\frac{5}{9}$
$\frac{4}{9}$
The probabilities of three mutually exclusive events are $\frac{2}{3} , \frac{1}{4}$ and $\frac{1}{6}$. The statement is
$A$ and $B$ are two independent events. The probability that both $A$ and $B$ occur is $\frac{1}{6}$ and the probability that neither of them occurs is $\frac{1}{3}$. Then the probability of the two events are respectively
For the three events $A, B$ and $C, P$ (exactly one of the events $A$ or $B$ occurs) = $P$ (exactly one of the events $B$ or $C$ occurs)= $P$ (exactly one of the events $C$ or $A$ occurs)= $p$ and $P$ (all the three events occur simultaneously) $ = {p^2},$ where $0 < p < 1/2$. Then the probability of at least one of the three events $A, B$ and $C$ occurring is
A die marked $1,\,2,\,3$ in red and $4,\,5,\,6$ in green is tossed. Let $A$ be the event, $'$ the number is even,$'$ and $B$ be the event, 'the number is red'. Are $A$ and $B$ independent?
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