We need $0 < st < 1$, which means $st$ is positive and less than 1.

Check each statement:

I. $s < -1$ and $t > 0$: If $s < -1$ (negative) and $t > 0$ (positive), then $st < 0$ (negative). This violates $st > 0$. ✗

II. $s < -1$ and $t < -1$: Both negative means $st > 0$ ✓. If $s = -2$ and $t = -0.4$, then $st = 0.8$, which satisfies $0 < st < 1$ ✓

Actually, wait. Let's reconsider: if $s < -1$ and $t < -1$, then both are less than $-1$, so both magnitudes are greater than 1. The product would be greater than 1, violating $st < 1$. Let me recalculate: if $s = -1.5$ and $t = -0.5$, then $st = 0.75$, which works. But $t = -0.5$ does not satisfy $t < -1$. If both $s < -1$ and $t < -1$, then $|s| > 1$ and $|t| > 1$, so $|st| > 1$, meaning $st > 1$ since both are negative (making product positive). This violates $st < 1$. ✗

III. $s > -1$ and $t < -1$: This gives us $-1 < s$ and $t < -1$. If $s$ is positive (satisfies $s > -1$) and $t$ is negative with $|t| > 1$, then $st < 0$. ✗ If $s$ is negative with $|s| < 1$ (like $s = -0.5$) and $t < -1$ (like $t = -1.5$), then $st = (-0.5)(-1.5) = 0.75 > 0$ and $< 1$ ✓

Actually, I need to reconsider option I more carefully...

Le Corbusier's Five Points of a New Architecture (1927):

1. Pilotis — columns lifting the building off the ground ✓
2. Free plan — open floor plan enabled by structural columns ✓
3. Free facade — non-load-bearing exterior walls ✓
4. Horizontal windows (ribbon windows) — long panoramic windows ✓
5. Roof garden (flat roof with terrace garden) ✓

Exposed concrete (béton brut) is associated with Brutalism (a later movement). Le Corbusier used it but it's not one of the original Five Points.

This is a binomial distribution with $n=5$, $p=\dfrac{1}{3}$ (correct), $q=\dfrac{2}{3}$ (wrong).

$P(X \geq 4) = P(X=4) + P(X=5)$

$P(X=4) = \binom{5}{4}\left(\dfrac{1}{3}\right)^4\left(\dfrac{2}{3}\right)^1 = 5 \cdot \dfrac{1}{81} \cdot \dfrac{2}{3} = \dfrac{10}{243}$

$P(X=5) = \binom{5}{5}\left(\dfrac{1}{3}\right)^5 = \dfrac{1}{243}$

$P(X \geq 4) = \dfrac{10}{243} + \dfrac{1}{243} = \dfrac{11}{243} = \dfrac{11}{3^5}$