$A$ 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) = $

$A$ company has two plants $A$ and $B$ to manufacture motorcycles. $60 \%$ of motorcycles are manufactured at plant $A$ and the remaining are manufactured at plant $B$. $80 \%$ of the motorcycles manufactured at plant $A$ are rated of standard quality,while $90 \%$ of the motorcycles manufactured at plant $B$ are rated of standard quality. $A$ motorcycle picked up randomly from the total production is found to be of standard quality. If $p$ is the probability that it was manufactured at plant $B$,then $126 p$ is

$A$ bag $P$ contains $3$ blue and $5$ red balls. Another bag $Q$ contains $4$ blue and $6$ red balls. $A$ ball is drawn at random from one of the bags and is found to be red. The probability that it is from bag $Q$ is

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:

Given three identical bags each containing $10$ balls,whose colours are as follows:

	Red	Blue	Green
Bag $I$	$3$	$2$	$5$
Bag $II$	$4$	$3$	$3$
Bag $III$	$5$	$1$	$4$

$A$ person chooses a bag at random and takes out a ball. If the ball is Red,the probability that it is from bag $I$ is $p$,and if the ball is Green,the probability that it is from bag $III$ is $q$,then the value of $\left(\frac{1}{p}+\frac{1}{q}\right)$ is:

Two coins $A$ and $B$ are kept in an urn. When coin $A$ is flipped,the probability of getting a head is $1/4$,while for coin $B$ it is $3/4$. One coin is randomly chosen from this bag,tossed twice,and it falls heads on both occasions. The probability that it is coin $A$ is: