There are $3$ bags,each containing $5$ white balls and $3$ black balls. Also,there are $2$ bags,each containing $2$ white balls and $4$ black balls. $A$ white ball is drawn at random. Find the probability that this white ball is from a bag of the first group.

Two coins are available, one fair and the other two-headed. Choose a coin and toss it once; assume that the unbiased coin is chosen with probability $\frac{3}{4}$. Given that the outcome is head, the probability that the two-headed coin was chosen, is

Three boxes $B_1$,$B_2$ and $B_3$ contain balls with different colors as follows:

Box	White,Black,Red
$B_1$	$2, 1, 2$
$B_2$	$3, 2, 4$
$B_3$	$4, 3, 2$

$A$ die is thrown. Box $B_1$ is chosen if either $1$ or $2$ turns up. Box $B_2$ is chosen if $3$ or $4$ turns up and box $B_3$ is chosen if $5$ or $6$ turns up. Having chosen a box in this way,a ball is drawn at random from that box. If the ball drawn is found to be Red,then the probability that it is drawn from box $B_2$ 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 $\frac{4}{13}$

$A$ diagnostic test has a probability of $0.95$ of giving a positive result when applied to a person suffering from a certain disease and a probability of $0.10$ of giving a positive result when given to a non-sufferer. It is estimated that $0.5 \%$ of the population are suffering from the disease. If this test is administered to a person from this population about whom there is no information relating to the incidence of this disease and the test gives a positive result, then the probability that the person is a sufferer is:

Let $U_1$ and $U_2$ be two urns such that $U_1$ contains $3$ white and $2$ red balls,and $U_2$ contains only $1$ white ball. $A$ fair coin is tossed. If head appears,then $1$ ball is drawn at random from $U_1$ and put into $U_2$. However,if tail appears,then $2$ balls are drawn at random from $U_1$ and put into $U_2$. Now $1$ ball is drawn at random from $U_2$.
$1.$ The probability of the drawn ball from $U_2$ being white is
$(A)$ $\frac{13}{30}$ $(B)$ $\frac{23}{30}$ $(C)$ $\frac{19}{30}$ $(D)$ $\frac{11}{30}$
$2.$ Given that the drawn ball from $U_2$ is white,the probability that head appeared on the coin is
$(A)$ $\frac{17}{23}$ $(B)$ $\frac{11}{23}$ $(C)$ $\frac{15}{23}$ $(D)$ $\frac{12}{23}$
Give the answer for question $1$ and $2.$