In a certain recruitment test with multiple choice questions,there are four options to answer each question,out of which only one is correct. An intelligent student knows $90 \%$ of the correct answers,while a weak student knows only $20 \%$ of the correct answers. If an intelligent student gets the correct answer for a question,what is the probability that he was guessing it?

$A$ person is known to speak truth in $3$ out of $5$ times. If he throws a die and reports that it is six,then the probability that it is actually six,is

$A$ 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 $.....$ .

$A$ man speaks truth $2$ out of $3$ times. He picks one of the natural numbers in the set $S=\{1, 2, 3, 4, 5, 6, 7\}$ and reports that it is even. The probability that it is actually even is

$A$ manufacturing company of bulbs has $3$ units $A, B$ and $C$ which produce $25 \%$,$35 \%$ and $40 \%$ of the bulbs respectively. Out of the bulbs produced by $A, B, C$ units,$5 \%, 4 \%$ and $2 \%$ are defective respectively. If a bulb is chosen at random and found to be defective,then the probability that it is produced by unit $B$ is

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 $\frac{4}{13}$