In probability, "A or B" means that A happens, or B happens, or both happen. It does not mean "exactly one" of them happens.

Recall the formula for two events $A$ and $B$: $P(A\text{ or }B) = P(A) + P(B) - P(A\text{ and }B)$. Figure out what $P(A\text{ and }B)$ is for this problem.

In a Desmos expression line, type 1/2 + 1/2 - 1/4 (this represents $P(\text{even}) + P(\text{heads}) - P(\text{even and heads})$). The value that Desmos shows for this expression is the probability that the die shows an even number or the coin shows heads (or both).

Let event $E$ be "the die shows an even number" and event $H$ be "the coin shows heads."

The question asks for the probability that $E$ or $H$ (or both) happens. In probability, "or" means at least one of the events occurs, not exactly one.

A fair six-sided die has 3 even numbers: 2, 4, and 6.

• So $P(E) = 3/6 = 1/2$.

A fair coin has 1 head out of 2 sides.

• So $P(H) = 1/2$.

The die roll and the coin flip are independent, so the probability that both $E$ and $H$ happen is the product of their probabilities:

For any two events $A$ and $B$,

Here, use $A = E$ and $B = H$:

Add and subtract:

• $\frac{1}{2} + \frac{1}{2} = 1$
• $1 - \frac{1}{4} = \frac{3}{4}$

So, the probability that the die shows an even number or the coin shows heads (or both) is $\boxed{\frac{3}{4}}$, which corresponds to choice D.