Bag $I$ contains $3$ red,$4$ black,and $3$ white balls. Bag $II$ contains $2$ red,$5$ black,and $2$ white balls. One ball is transferred from Bag $I$ to Bag $II$,and then a ball is drawn from Bag $II$. The ball drawn is found to be black. What is the probability that the transferred ball was red?

In a manufacturing company,three machines $A$,$B$ and $C$ respectively produce $20 \%$,$30 \%$ and $50 \%$ of the total product. The defective products from $A$,$B$ and $C$ are respectively $5 \%$,$3 \%$ and $2 \%$. If an article produced by the company is selected at random and is found to be defective,then the probability that it is produced by machine $B$ is

In an entrance test,there are multiple-choice questions. There are four possible answers to each question,of which one is correct. The probability that a student knows the answer to a question is $9/10$. If he gets the correct answer to a question,then the probability that he was guessing is:

Two cards are drawn at random from a pack of $52$ playing cards. If both the cards drawn are found to be black in colour,then the probability that at least one of them is a face card is

There are three bags $B_1, B_2$ and $B_3$. The bag $B_1$ contains $5$ red and $5$ green balls,$B_2$ contains $3$ red and $5$ green balls,and $B_3$ contains $5$ red and $3$ green balls. Bags $B_1, B_2$ and $B_3$ have probabilities $\frac{3}{10}, \frac{3}{10}$ and $\frac{4}{10}$ respectively of being chosen. $A$ bag is selected at random and a ball is chosen at random from the bag. Then which of the following options is/are correct?
$(1)$ Probability that the selected bag is $B_3$ and the chosen ball is green equals $\frac{3}{20}$
$(2)$ Probability that the chosen ball is green equals $\frac{39}{80}$
$(3)$ Probability that the chosen ball is green,given that the selected bag is $B_3$,equals $\frac{3}{8}$
$(4)$ Probability that the selected bag is $B_3$,given that the chosen ball is green,equals $\frac{4}{13}$

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...