$A$ disease affects two-thirds of the population of a country. $A$ test for the disease gives the correct outcome with probability $\frac{2}{3}$. $A$ person $X$ tests positive for the disease. The probability that $X$ has the disease is

$A$ man is known to speak the truth $2$ out of $3$ times. If he throws a die and reports that it is $6$,then the probability that it is actually $5$ is:

$A$ laboratory blood test is $99 \%$ effective in detecting a certain disease when it is in fact,present. However,the test also yields a false positive result for $0.5 \%$ of the healthy persons tested (that is,if a healthy person is tested,then,with probability $0.005,$ the test will imply he has the disease). If $0.1 \%$ of the population actually has the disease,what is the probability that a person has the disease given that his test result is positive?

For $k=1, 2, 3$,the box $B_k$ contains $k$ red balls and $(k+1)$ white balls. Let $P(B_1) = \frac{1}{2}$,$P(B_2) = \frac{1}{3}$,and $P(B_3) = \frac{1}{6}$. $A$ box is selected at random and a ball is drawn from it. If a red ball is drawn,then the probability that it comes from box $B_2$ 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 :

$A$ bag $A$ contains $2$ white and $3$ red balls and bag $B$ contains $4$ white and $5$ red balls. One ball is drawn at random from a randomly chosen bag and is found to be red. The probability that it was drawn from bag $B$ is: