$A$ and $B$ are two events of a random experiment such that $P(B)=0.4$, $P(A \cap \bar{B})=0.5$, and $P(A \cup B) + P\left(\frac{B}{A \cup \bar{B}}\right) = 1.15$. Then $P(A) = $

Suppose that $5 \%$ of men and $0.25 \%$ of women have gray hair. $A$ person with gray hair is selected at random. If there are an equal number of males and females,then the probability that the selected person is a man is:

Three boxes $B_1$,$B_2$ and $B_3$ contain balls with different colors as follows:

Box	White,Black,Red
$B_1$	$2, 1, 2$
$B_2$	$3, 2, 4$
$B_3$	$4, 3, 2$

$A$ die is thrown. Box $B_1$ is chosen if either $1$ or $2$ turns up. Box $B_2$ is chosen if $3$ or $4$ turns up and box $B_3$ is chosen if $5$ or $6$ turns up. Having chosen a box in this way,a ball is drawn at random from that box. If the ball drawn is found to be Red,then the probability that it is drawn from box $B_2$ is

$A$ man is known to speak the truth $3$ out of $4$ times. He throws a die and reports that it is $6$. Then the probability that it is actually $6$ is:

$A$ letter is known to have come either from $LONDON$ or $CLIFTON$; on the postmark only the two consecutive letters $ON$ are legible. The probability that it came from $LONDON$ is

Suppose that $5 \%$ of men and $0.25 \%$ of women have grey hair. $A$ person with grey hair is selected at random. What is the probability of this person being male? Assume that there are equal numbers of males and females.