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?

There are three bags $B_1, B_2$ and $B_3$. The bag $B_1$ contains $5$ red and $5$ green balls,$B_2$ contains $3$ red and $5$ green balls,and $B_3$ contains $5$ red and $3$ green balls. Bags $B_1, B_2$ and $B_3$ have probabilities $\frac{3}{10}, \frac{3}{10}$ and $\frac{4}{10}$ respectively of being chosen. $A$ bag is selected at random and a ball is chosen at random from the bag. Then which of the following options is/are correct?
$(1)$ Probability that the selected bag is $B_3$ and the chosen ball is green equals $\frac{3}{20}$
$(2)$ Probability that the chosen ball is green equals $\frac{39}{80}$
$(3)$ Probability that the chosen ball is green,given that the selected bag is $B_3$,equals $\frac{3}{8}$
$(4)$ Probability that the selected bag is $B_3$,given that the chosen ball is green,equals $\frac{4}{13}$

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

There are three coins. One is a two-headed coin (having heads on both faces),another is a biased coin that comes up heads $75 \%$ of the time,and the third is an unbiased coin. One of the three coins is chosen at random and tossed; it shows heads. What is the probability that it was the two-headed coin?

$A$ man is known to speak the truth $3$ out of $4$ times. He throws a die and reports that it is a six. The probability that it is actually a six,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) = $