$A$ 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) = $

In an entrance test,there are multiple-choice questions. There are four possible answers to each question,of which one is correct. The probability that a student knows the answer to a question is $9/10$. If he gets the correct answer to a question,then the probability that he was guessing 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:

$A$ shopkeeper buys a particular type of electric bulbs from three manufacturers $M_1, M_2$,and $M_3$. He buys $25 \%$ of his requirement from $M_1$,$45 \%$ from $M_2$,and $30 \%$ from $M_3$. Based on past experience,he found that $2 \%$ of type $M_3$ bulbs are defective,whereas only $1 \%$ of type $M_1$ and type $M_2$ bulbs are defective. If a bulb chosen by him at random is defective,then the probability that it was of type $M_3$ is

Bag $A$ contains $6$ Green and $8$ Red balls and bag $B$ contains $9$ Green and $5$ Red balls. $A$ card is drawn at random from a well-shuffled pack of $52$ playing cards. If it is a spade,two balls are drawn at random from bag $A$,otherwise two balls are drawn at random from bag $B$. If the two balls drawn are found to be of the same colour,then the probability that they are drawn from bag $A$ 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: