The probability that the product of the outcomes when three dice are rolled simultaneously is divisible by $4$ is equal to

If a $3$-digit number is randomly chosen,what is the probability that either the number itself or some permutation of the number (which is a $3$-digit number) is divisible by $4$ and $5$?

Given two independent events,if the probability that exactly one of them occurs is $\frac{26}{49}$ and the probability that none of them occurs is $\frac{15}{49}$,then the probability of the more probable of the two events is (in $/7$)

$A$ and $B$ are independent events. The probability that both $A$ and $B$ occur is $\frac{1}{20}$ and the probability that neither of them occurs is $\frac{3}{5}$. The probability of occurrence of $A$ is

Let $A, B$ and $C$ be three events such that the probability that exactly one of $A$ and $B$ occurs is $(1-k)$,the probability that exactly one of $B$ and $C$ occurs is $(1-2k)$,the probability that exactly one of $C$ and $A$ occurs is $(1-k)$ and the probability that all $A, B$ and $C$ occur simultaneously is $k^2$,where $0 < k < 1$. Then the probability that at least one of $A, B$ and $C$ occurs is:

Cards are drawn one by one without replacement from a pack of $52$ cards. The probability that $10$ cards will precede the first ace is