$A$ bag contains $2n+1$ coins. It is known that $n$ of these coins have a head on both sides, whereas the remaining $n+1$ coins are fair. $A$ coin is picked up at random from the bag and tossed. If the probability that the toss results in a head is $\frac{31}{42}$, then $n$ is equal to

$A$ 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?

$A$ boy speaks the truth in $3$ out of $5$ times. If he throws a die and reports that the number appeared on it is $5$, then the probability that it is actually $5$ is:

The urns $A, B$ and $C$ contain $4$ red,$6$ black; $5$ red,$5$ black and $\lambda$ red,$4$ black balls respectively. One of the urns is selected at random and a ball is drawn. If the ball drawn is red and the probability that it is drawn from urn $C$ is $0.4$,then find the square of the length of the side of the largest equilateral triangle inscribed in the parabola $y^2 = \lambda x$ with one vertex at the vertex of the parabola.

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:

There are two boxes, each containing $10$ balls. In each box, some are black and the rest are white. $A$ ball is drawn at random from one of the boxes and it is found to be black. If the probability that the black ball drawn is from the second box is $\frac{1}{5}$, then the number of black balls in the first box is: