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?

An urn contains $5$ balls. Two balls are drawn at random and they are found to be white. The probability that all the balls in the urn are white 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 $\frac{4}{13}$

$A$ computer producing factory has only two plants $T_1$ and $T_2$. Plant $T_1$ produces $20 \%$ and plant $T_2$ produces $80 \%$ of the total computers produced. $7 \%$ of computers produced in the factory turn out to be defective. It is known that $P(\text{defective} | T_1) = 10 P(\text{defective} | T_2)$. $A$ computer produced in the factory is randomly selected and it is not defective. Then the probability that it is produced in plant $T_2$ is

$A$ person goes to office by car, scooter, bus, and train, the probabilities of which are $1/7, 3/7, 2/7,$ and $1/7$ respectively. The probability that he reaches office late if he takes a car, scooter, bus, or train is $2/9, 1/9, 4/9,$ and $1/9$ respectively. Given that he reached the office in time, the probability that he travelled by car is: (in $/7$)

Two digits are selected at random from the digits $1$ through $9$. If their sum is even,then the probability that both are odd is