Tiles in 2 Boxes Twenty tiles are numbered 1 through 20 and are placed into box $A$. Twenty other tiles numbered 11 through 30 are placed into box $B$. One tile is randomly drawn from each box. What is the probability that the tile from box $A$ is less than 15 and the tile from box $B$ is either even or greater than 25? Express your answer as a common fraction.
Since the two events are independent, we consider each separately. The probability of the tile from A being less than 15 is equal to $\frac{14}{20} = \frac{7}{10}$. The probability of a tile from B being even or greater than 25 is $\frac{10+2}{20} = \frac{3}{5}$. So we multiply the probabilities for the independent events, giving us probability $\frac{7}{10} \cdot \frac{3}{5} = \boxed{\frac{21}{50}}$.
 A: The probability that the tile from box A is less than $15$ is $\frac{14}{20} = \frac{7}{10}$, because there are $14$ tiles which are less than $15$ in box A, out of the $20$ total tiles it has. 
 For box B, let $E_1$ be the event of the tile being greater than $25$ and $E_2$ be the event of the tile being even. We want to calculate the proability of at least one of these happening, which can be represented by $P(E_1 \text{ or } E_2)$. By the inclusion-exclusion principle, $P(E_1 \text{ or } E_2) = P(E_1) + P(E_2) - P(E_1 \text{ and } E_2)$. If you don't know what that is, just think of the sets of situations where events $E_1$ and $E_2$ happen. Picture these sets in a Venn diagram and notice that we want the events that happen in the union of these sets. And this union can be thought of as the sum of the quantity of situations in which $E_1$ happens plus the same for $E_2$ minus the situations where they both happen (these were counted twice so we need to subtract them one time).
$P(E_1) = \frac{5}{20}$ because there are $5$ tiles greater than $25$, $P(E_2) = \frac{10}{20}$ because there are $10$ even numbers from $11$ to $30$ and $P(E_1 \text{ and } E_2) = \frac{3}{20}$ because there are only $3$ even tiles greater than $25$ in box B.
So $P(E_1 \text{ or } E_2) = \frac{5}{20} + \frac{10}{20} - \frac{3}{20} = \frac{12}{20} = \frac{3}{5}$. The problem asks for the probability of two independent events (taking tiles from one box doesn't influence the choice of tiles over the other one), therefore we just need to multiply their corresponding probabilities: $$ p = \frac{7}{10} \cdot \frac{3}{5} = \frac{21}{50}$$
