Let $ABC$ be a triangle. Choose points $A'$, $B'$ and $C'$ independently on side segments $BC$, $CA$ and $AB$ respectively with a uniform distribution. For a point $Z$ in the plane, let $p(Z)$ denote the probability that $Z$ is contained in the triangle enclosed by lines $AA'$, $BB'$ and $CC'$. For which interior point $Z$ in triangle $ABC$ is $p(Z)$ maximised?
1. **Define Points and Variables:**
   Let \( AZ \cap BC = D \), \( BZ \cap CA = E \), and \( CZ \cap AB = F \). Suppose \( \frac{AF}{AB} = a \), \( \frac{BD}{BC} = b \), and \( \frac{CE}{CA} = c \). 

2. **Establish Relationship:**
   We have the relationship \( abc = (1-a)(1-b)(1-c) \). This is derived from the fact that the product of the ratios of the segments created by the cevians (lines from vertices to opposite sides) in a triangle is equal to the product of the remaining segments.

3. **Probability Expression:**
   The probability \( p(Z) \) that point \( Z \) is contained in the triangle enclosed by lines \( AA' \), \( BB' \), and \( CC' \) is given by:
   \[
   p(Z) = abc + (1-a)(1-b)(1-c)
   \]
   Since \( abc = (1-a)(1-b)(1-c) \), we have:
   \[
   p(Z) = 2abc
   \]

4. **Substitute Variables:**
   Let \( x = \frac{a}{1-a} \), \( y = \frac{b}{1-b} \), and \( z = \frac{c}{1-c} \). This implies:
   \[
   a = \frac{x}{1+x}, \quad b = \frac{y}{1+y}, \quad c = \frac{z}{1+z}
   \]
   Given \( xyz = 1 \), we need to maximize:
   \[
   2abc = \frac{2xyz}{(1+x)(1+y)(1+z)}
   \]

5. **Apply AM-GM Inequality:**
   Using the Arithmetic Mean-Geometric Mean (AM-GM) inequality:
   \[
   (1+x)(1+y)(1+z) \geq 8\sqrt[3]{xyz} = 8
   \]
   Therefore:
   \[
   \frac{2xyz}{(1+x)(1+y)(1+z)} \leq \frac{2 \cdot 1}{8} = \frac{1}{4}
   \]

6. **Equality Condition:**
   Equality in AM-GM holds when \( 1+x = 1+y = 1+z \), which implies \( x = y = z = 1 \). Thus:
   \[
   a = b = c = \frac{1}{2}
   \]

7. **Conclusion:**
   The maximum value of \( p(Z) \) is attained when \( a = b = c = \frac{1}{2} \), which corresponds to \( Z \) being the centroid of the triangle \( ABC \).

\(\blacksquare\)

The final answer is \( \boxed{ Z } \) is the centroid of the triangle \( ABC \).