In a certain recruitment test with multiple-choice questions,there are four options for 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 a weak student gets the correct answer,what is the probability that they were guessing?

$A$ 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 $.....$ .

In a factory which manufactures bolts,machines $A, B$ and $C$ manufacture respectively $25\%, 35\%$ and $40\%$ of the bolts. Of their outputs,$5, 4$ and $2$ percent are respectively defective bolts. $A$ bolt is drawn at random from the product and is found to be defective. What is the probability that it is manufactured by the machine $B$?

Let $n_1$ and $n_2$ be the number of red and black balls,respectively,in box $I$. Let $n_3$ and $n_4$ be the number of red and black balls,respectively,in box $II$.
$1.$ One of the two boxes,box $I$ and box $II$,was selected at random and a ball was drawn randomly out of this box. The ball was found to be red. If the probability that this red ball was drawn from box $II$ is $\frac{1}{3}$,then the correct option$(s)$ with the possible values of $n_1, n_2, n_3$ and $n_4$ is(are):
$(A)$ $n_1=3, n_2=3, n_3=5, n_4=15$
$(B)$ $n_1=3, n_2=6, n_3=10, n_4=50$
$(C)$ $n_1=8, n_2=6, n_3=5, n_4=20$
$(D)$ $n_1=6, n_2=12, n_3=5, n_4=20$
$2.$ $A$ ball is drawn at random from box $I$ and transferred to box $II$. If the probability of drawing a red ball from box $I$,after this transfer,is $\frac{1}{3}$,then the correct option$(s)$ with the possible values of $n_1$ and $n_2$ is(are):
$(A)$ $n_1=4, n_2=6$
$(B)$ $n_1=2, n_2=3$
$(C)$ $n_1=10, n_2=20$
$(D)$ $n_1=3, n_2=6$
Give the answer for question $1$ and $2$.

There are $2$ bags each containing $3$ white and $5$ black balls and $4$ bags each containing $6$ white and $4$ black balls. If a ball drawn randomly from a bag is found to be black,then the probability that this ball is from the first set of bags is