$A$ 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:

On every evening,a student either watches $TV$ or reads a book. The probability of watching $TV$ is $\frac{4}{5}$. If he watches $TV$,the probability that he will fall asleep is $\frac{3}{4}$ and it is $\frac{1}{4}$ when he reads a book. If the student is found to be asleep on an evening,the probability that he watched the $TV$ is

$A$ bag contains $19$ unbiased coins and one coin with head on both sides. One coin is drawn at random and tossed,and a head turns up. If the probability that the drawn coin was unbiased is $\frac{m}{n}$,where $\operatorname{gcd}(m, n) = 1$,then $n^2 - m^2$ is equal to:

$A$ box contains $4$ black,$2$ white,and $6$ red balls. Another box contains $3$ black and $5$ white balls. An unbiased die is thrown. If $1$ or $2$ appears on the die,a ball is drawn from the first box; otherwise,a ball is drawn from the second box. If the drawn ball is black,what is the probability that $2$ appeared on the die?

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 $2$.

$A$ 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