$A$ box contains $n$ coins,$m$ of which are fair and the rest are biased. When a biased coin is tossed,the probability of getting a head is twice as likely as tail. $A$ coin is drawn from the box at random and is tossed twice. It is found that the first time it shows head and the second time it shows tail. Then,the probability that the coin drawn is fair 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:

Let $U_1$ and $U_2$ be two urns such that $U_1$ contains $3$ white and $2$ red balls,and $U_2$ contains only $1$ white ball. $A$ fair coin is tossed. If it shows head,$1$ ball is drawn at random from $U_1$ and transferred to $U_2$. If it shows tail,$2$ balls are drawn at random from $U_1$ and transferred to $U_2$. Now,$1$ ball is drawn at random from $U_2$. Given that the ball drawn from $U_2$ is white,what is the probability that the coin showed head?

Three companies $C_1, C_2, C_3$ produce car tyres. $A$ car manufacturing company buys $40 \%$ of its requirement from $C_1, 35 \%$ from $C_2$ and $25 \%$ from $C_3$. The company knows that $2 \%$ of the tyres supplied by $C_1, 3 \%$ by $C_2$ and $4 \%$ by $C_3$ are defective. If a tyre chosen at random from the consignment received is found defective,then the probability that it was supplied by $C_2$ is:

$A$ factory has two machines $A$ and $B$. Past record shows that machine $A$ produced $60 \%$ of the items of output and machine $B$ produced $40 \%$ of the items. Further,$2 \%$ of the items produced by machine $A$ and $1 \%$ produced by machine $B$ were defective. All the items are put into one stockpile and then one item is chosen at random from this and is found to be defective. What is the probability that it was produced by machine $B$?

Let $H_1, H_2, \ldots, H_{n}$ be mutually exclusive and exhaustive events with $P(H_i) > 0, i = 1, 2, \ldots, n$. Let $E$ be any other event with $0 < P(E) < 1$.
$STATEMENT-1$: $P(H_i \mid E) > P(E \mid H_i) \cdot P(H_i)$ for $i = 1, 2, \ldots, n$.
$STATEMENT-2$: $\sum_{i=1}^{n} P(H_i) = 1$.