$A$ pack of cards contains $4$ aces,$4$ kings,$4$ queens,and $4$ jacks. Two cards are drawn at random. The probability that at least one of these is an ace is:

$12$ balls are distributed among $3$ boxes. The probability that the first box will contain exactly $3$ balls is:

Three numbers are selected at random from the set ${1, 2, 3, \dots, 8}$ without replacement. Given that the minimum of the selected numbers is $3$ and the maximum is $6$, what is the probability that the third number is $4$ or $5$?

Consider the $6 \times 6$ square grid in the figure. Let $A_1, A_2, \ldots, A_{49}$ be the points of intersection (dots in the picture) in some order. We say that $A_i$ and $A_j$ are friends if they are adjacent along a row or along a column. Assume that each point $A_i$ has an equal chance of being chosen.
$(1)$ Let $p_i$ be the probability that a randomly chosen point has $i$ many friends,$i=0, 1, 2, 3, 4$. Let $X$ be a random variable such that for $i=0, 1, 2, 3, 4$,the probability $P(X=i)=p_i$. Then the value of $7 E(X)$ is
$(2)$ Two distinct points are chosen randomly out of the points $A_1, A_2, \ldots, A_{49}$. Let $p$ be the probability that they are friends. Then the value of $7 p$ is

One die has two faces marked $1$,two faces marked $2$,one face marked $3$ and one face marked $4$. Another die has one face marked $1$,two faces marked $2$,two faces marked $3$ and one face marked $4$. The probability of getting the sum of numbers to be $4$ or $5$,when both the dice are thrown together,is

$P$ speaks truth in $70\%$ of the cases and $Q$ in $80\%$ of the cases. In what percent of cases are they likely to agree in stating the same fact (in $\%$)?