Let $A, B,$ and $C$ be three mutually independent events. Consider the two statements $S_1$ and $S_2$:
$S_1 : A$ and $B \cup C$ are independent.
$S_2 : A$ and $B \cap C$ are independent.
Then:

The probabilities of having a defective toy in three cartons $A, B, C$ are $\frac{1}{3}, \frac{1}{4}, \frac{2}{5}$ respectively. If a carton is selected at random and a toy drawn randomly from it is found to be defective,then the probability that it is drawn from carton $B$ is

Suppose that a bag $A$ contains $n$ red and $2$ black balls and another bag $B$ contains $2$ red and $n$ black balls. One of the two bags is selected at random and two balls are drawn from it at a time. When it is known that the two balls drawn are red,if the probability that those two balls drawn are from bag $A$ is $\frac{6}{7}$,then $n=$

$A$ laboratory blood test is $99 \%$ effective in detecting a certain disease when it is in fact,present. However,the test also yields a false positive result for $0.5 \%$ of the healthy persons tested (that is,if a healthy person is tested,then,with probability $0.005,$ the test will imply he has the disease). If $0.1 \%$ of the population actually has the disease,what is the probability that a person has the disease given that his test result is positive?

If $E_1$ and $E_2$ are equally likely, mutually exclusive and exhaustive events and $P(A|E_1) = 0.2$, $P(A|E_2) = 0.3$, then $P(E_1|A)$ is equal to...

$A$ bag contains $10$ balls out of which $k$ are red and $(10-k)$ are black, where $0 \le k \le 10$. If three balls are drawn at random without replacement and all of them are found to be black, then the probability that the bag contains $1$ red and $9$ black balls is: