There are four machines and it is known that exactly two of them are faulty. They are tested one by one,in a random order,until both the faulty machines are identified. The probability that only two tests are needed is:

If $A$ and $B$ are independent events of a random experiment such that $P(A \cap B) = \frac{1}{6}$ and $P(\bar{A} \cap \bar{B}) = \frac{1}{3}$,then $P(A)$ is equal to (Here,$\bar{E}$ is the complement of the event $E$)

In a random experiment,a fair die is rolled until two fours are obtained in succession. The probability that the experiment will end in the fifth throw of the die is equal to

$A$ random variable $X$ has the following probability distribution:
| $X=x$ | $1$ | $2$ | $3$ | $4$ | $5$ | $6$ | $7$ | $8$ |
| :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- |
| $P(X=x)$ | $0.15$ | $0.23$ | $0.12$ | $0.20$ | $0.08$ | $0.10$ | $0.05$ | $0.07$ |
For the events $E = \{X \text{ is a prime number}\}$ and $F = \{X < 5\}$,find $P(E \cup F)$.

An unbiased coin is tossed. If the result is a head,a pair of unbiased dice is rolled and the sum of the numbers on the two faces is noted. If the result is a tail,a card from a well-shuffled pack of eleven cards numbered $2, 3, 4, \dots, 12$ is picked and the number on the card is noted. The probability that the noted number is either $7$ or $8$ is:

$A$ and $B$ are independent events. If $P(A \cup B)=0.5$ and $P(A)=0.2$,then $P(B) = $ . . . . . . . (in $/8$)