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:

$A$ pack of playing cards was found to contain only $51$ cards. If the first $13$ cards which are examined are all red,then the probability that the missing card is black,is

Two balls are selected at random one by one without replacement from a bag containing $4$ white and $6$ black balls. If the probability that the first selected ball is black,given that the second selected ball is also black,is $\frac{m}{n}$,where $\operatorname{gcd}(m, n) = 1$,then $m + n$ is equal to :

In an entrance test, there are multiple-choice questions. Each question has four possible answers, only one of which is correct. The probability that a student knows the answer to a question is $90\%$. If the student gets the correct answer to a question, what is the probability that they were guessing?

Bag $A$ contains $2$ white,$1$ black and $3$ red balls and bag $B$ contains $3$ black,$2$ red and $n$ white balls. One bag is chosen at random and $2$ balls drawn from it at random,are found to be $1$ red and $1$ black. If the probability that both balls come from Bag $A$ is $\frac{6}{11}$,then $n$ is equal to

$A$ box $B_1$ contains $3$ blue balls and $6$ red balls. Another box $B_2$ contains $8$ blue balls and $n$ red balls $(n \in N)$. $A$ ball selected at random from a box is found to be red. If $p$ is the probability that this red ball drawn is from box $B_2$,then