Bag $I$ contains $3$ red and $4$ black balls,while Bag $II$ contains $5$ red and $6$ black balls. One ball is drawn at random from one of the bags and it is found to be red. Find the probability that it was drawn from Bag $II$.

$A$ bag contains $8$ balls,whose colors are either white or black. $4$ balls are drawn at random without replacement and it was found that $2$ balls are white and $2$ balls are black. The probability that the bag originally contained an equal number of white and black balls is:

$A$ bag contains $5$ balls of unknown colours. There are equal chances that out of these five balls,there may be $0, 1, 2, 3, 4,$ or $5$ red balls. $A$ ball is taken out from the bag at random and is found to be red. The probability that it is the only red ball in the bag is

The probability of men getting a certain disease is $\frac{1}{2}$ and that of women getting the same disease is $\frac{1}{5}$. The blood test that identifies the disease gives the correct result with probability $\frac{4}{5}$. Suppose a person is chosen at random from a group of $30$ males and $20$ females,and the blood test of that person is found to be positive. What is the probability that the chosen person is a man?

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 $\frac{4}{13}$

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