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:

In an examination, there are $4$ Yes/No type questions. The probability that a student answers a question correctly without guessing is $2/3$. The probability that a student guesses a correct answer is $1/2$. $A$ student writes the examination either by not guessing any of the $4$ questions or by guessing all $4$ questions. The probability that they attempt the exam by guessing all questions is $3/7$. Given that a student answered at least $3$ questions correctly, what is the probability that they answered all questions without guessing?

Bag $1$ contains $4$ white balls and $5$ black balls,and Bag $2$ contains $n$ white balls and $3$ black balls. One ball is drawn randomly from Bag $1$ and transferred to Bag $2$. $A$ ball is then drawn randomly from Bag $2$. If the probability that the ball drawn from Bag $2$ is white is $29/45$,then $n$ is equal to:

$A$ factory has a total of three manufacturing units,$M_1, M_2$,and $M_3$,which produce bulbs independently. The units $M_1, M_2$,and $M_3$ produce bulbs in the proportions of $2: 2: 1$,respectively. It is known that $20\%$ of the bulbs produced in the factory are defective. It is also known that,of all the bulbs produced by $M_1, 15\%$ are defective. Suppose that,if a randomly chosen bulb produced in the factory is found to be defective,the probability that it was produced by $M_2$ is $\frac{2}{5}$. If a bulb is chosen randomly from the bulbs produced by $M_3$,then the probability that it is defective 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

$A$ 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) = $