Let $A$ and $B$ be two events such that $P(A \cap B) = \frac{1}{6}$, $P(A \cup B) = \frac{31}{45}$, and $P(\bar{B}) = \frac{7}{10}$, then:

The probability of obtaining an even prime number on each die,when a pair of dice is rolled,is . . . . . . .

$A$ random experiment has five outcomes $w_1, w_2, w_3, w_4$ and $w_5$. The probabilities of the occurrence of the outcomes $w_1, w_2, w_3, w_4$ and $w_5$ are respectively $\frac{1}{6}, a, b, c$ and $\frac{1}{12}$ such that $12a + 12b - 1 = 0$. If $p(w_4) = c$ and the outcomes are equally likely for $w_2, w_3, w_4$,find the probability of the outcome $w_3$. Given the condition $12a + 12b = 1$,and assuming $a=b=c$ for the remaining outcomes,find $p(w_3)$.

One card is drawn from a well-shuffled deck of $52$ cards. If each outcome is equally likely,calculate the probability that the card will be a diamond but not an ace.

If $A, B$ and $C$ are mutually exclusive and exhaustive events of a random experiment such that $P(B) = \frac{3}{2} P(A)$ and $P(C) = \frac{1}{2} P(B)$, then $P(A \cup C)$ equals to

When a die is thrown twice,what is the probability of getting $4, 5,$ or $6$ on the first throw and $1, 2, 3,$ or $4$ on the second throw?