Now give the current a choice: two resistors side by side, each on its own branch, make a parallel circuit. More paths make the trip EASIER, so the total resistance drops below the smallest branch — always. That is your built-in answer check: a parallel total bigger than either branch is wrong before you even look at the working.

For two branches the recipe is product over sum: $R_t = \dfrac{R_1 R_2}{R_1 + R_2}$ (read aloud: R-total equals R-one R-two, over R-one plus R-two) — same name tags as last lesson: $R_1$ and $R_2$ are the two branches in ohms, $R_t$ the one resistance that replaces them both. Try it in your head: 4 Ω beside 12 Ω gives $\dfrac{48}{16} = 3\ \Omega$ — smaller than 4, just as promised.