$A$ student answers a multiple choice question with $5$ alternatives,of which exactly $1$ is correct. The probability that he knows the correct answer is $p$,where $0 < p < 1$. If he does not know the correct answer,he randomly ticks $1$ answer. Given that he has answered the question correctly,the probability that he did not tick the answer randomly is:

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 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:

Bag $I$ contains $3$ red and $4$ black balls and Bag $II$ contains $4$ red and $5$ black balls. One ball is transferred from Bag $I$ to Bag $II$ and then a ball is drawn from Bag $II$. The ball so drawn is found to be red in colour. Find the probability that the transferred ball is black.

$A$ bag contains $6$ balls. If $4$ balls are drawn at a time and all of them are found to be red,then the probability that exactly $5$ of the balls in the bag are red is

Let $n_1$ and $n_2$ be the number of red and black balls,respectively,in box $I$. Let $n_3$ and $n_4$ be the number of red and black balls,respectively,in box $II$.
$1.$ One of the two boxes,box $I$ and box $II$,was selected at random and a ball was drawn randomly out of this box. The ball was found to be red. If the probability that this red ball was drawn from box $II$ is $\frac{1}{3}$,then the correct option$(s)$ with the possible values of $n_1, n_2, n_3$ and $n_4$ is(are):
$(A)$ $n_1=3, n_2=3, n_3=5, n_4=15$
$(B)$ $n_1=3, n_2=6, n_3=10, n_4=50$
$(C)$ $n_1=8, n_2=6, n_3=5, n_4=20$
$(D)$ $n_1=6, n_2=12, n_3=5, n_4=20$
$2.$ $A$ ball is drawn at random from box $I$ and transferred to box $II$. If the probability of drawing a red ball from box $I$,after this transfer,is $\frac{1}{3}$,then the correct option$(s)$ with the possible values of $n_1$ and $n_2$ is(are):
$(A)$ $n_1=4, n_2=6$
$(B)$ $n_1=2, n_2=3$
$(C)$ $n_1=10, n_2=20$
$(D)$ $n_1=3, n_2=6$
Give the answer for question $1$ and $2$.