To get exactly 2 heads in 3 flips, first count how many outcomes have exactly 2 heads: HHT, HTH, and THH. There are 3 such outcomes. Since the coin is fair, each specific sequence of 3 flips has probability ${\left(\frac{1}{2}\right)}^3=\frac{1}{8}$. Therefore, the probability of getting exactly 2 heads is $3{\left(\frac{1}{2}\right)}^3=\frac{3}{8}$. So the equivalent expression is $3{\left(\frac{1}{2}\right)}^3$.

There are $4$ possible values for $x$ and $3$ possible values for $y$, so the grid contains $4\times 3=12$ integer points. A point lies on the line $y=x$ when the $x$- and $y$-coordinates are equal. Among the given points, those are $(1,1)$, $(2,2)$, and $(3,3)$. The point $(4,4)$ is not included because $y$ can be at most $3$. So there are $3$ favorable points out of $12$ total points. The probability is $\frac{3}{12}=\frac{1}{4}$.

Let $r$ be the number of red tickets. Since there are 12 blue tickets, the total number of tickets is $r+12$. The probability of drawing a red ticket is therefore $\frac{r}{r+12}$. Set this equal to the given probability: $\frac{r}{r+12}=\frac{3}{5}$. Cross-multiply to get $5r=3(r+12)$. Then $5r=3r+36$, so $2r=36$, and $r=18$. Therefore, there are 18 red tickets.

To find the probability that a randomly selected student does not ride the bus, first count the students who use other methods: walk or car. That is $18+15=33$. Next find the total number of students surveyed: $18+27+15=60$. Therefore, the probability is $\frac{33}{60}$.

There are $4+6=10$ total marbles in the bag. Since 4 of them are red, the probability of choosing a red marble is $\frac{4}{10}=\frac{2}{5}$. Therefore, the statement that must be true is choice A.