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 $n=$

Events $E_{1}$ and $E_{2}$ form a partition of the sample space $S$. $A$ is any event such that $P(E_{1}) = P(E_{2}) = \frac{1}{2}$,$P(E_{2} | A) = \frac{1}{2}$,and $P(A | E_{2}) = \frac{2}{3}$. Then $P(E_{1} | A)$ is:

In a factory which manufactures bolts,machines $A, B$ and $C$ manufacture respectively $25\%, 35\%$ and $40\%$ of the bolts. Of their outputs,$5, 4$ and $2$ percent are respectively defective bolts. $A$ bolt is drawn at random from the product and is found to be defective. What is the probability that it is manufactured by the machine $B$?

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 $2.$

$A$ bag contains $6$ balls. If three balls are drawn at a time and all of them are found to be green,then the probability that exactly $5$ of the balls in the bag are green is:

Let $H_1, H_2, \ldots, H_{n}$ be mutually exclusive and exhaustive events with $P(H_i) > 0, i = 1, 2, \ldots, n$. Let $E$ be any other event with $0 < P(E) < 1$.
$STATEMENT-1$: $P(H_i \mid E) > P(E \mid H_i) \cdot P(H_i)$ for $i = 1, 2, \ldots, n$.
$STATEMENT-2$: $\sum_{i=1}^{n} P(H_i) = 1$.