Let $a,b \in [0,1], c \in [-1,1]$ be reals chosen independently and uniformly at random. What is the probability that $p(x) = ax^2+bx+c$ has a root in $[0,1]$?
1. **Case Analysis for \( c \):**
   - If \( c > 0 \), then \( p(0) = c > 0 \). Since \( a, b \geq 0 \) and \( x \in [0, 1] \), \( ax^2 + bx \geq 0 \). Therefore, \( p(x) = ax^2 + bx + c > 0 \) for all \( x \in [0, 1] \). Hence, \( p(x) \) cannot have a root in \([0, 1]\) if \( c > 0 \).
   - If \( c < 0 \), then \( p(0) = c < 0 \). For \( p(x) \) to have a root in \([0, 1]\), by the Intermediate Value Theorem, we need \( p(1) \geq 0 \). Thus, \( a + b + c \geq 0 \) or equivalently \( a + b \geq -c \).

2. **Volume Calculation:**
   - We need to find the probability that \( a + b \geq -c \) given \( a, b \in [0, 1] \) and \( c \in [-1, 1] \).
   - Consider the 3D space where \( a, b \in [0, 1] \) and \( c \in [-1, 1] \). The total volume of this space is \( 1 \times 1 \times 2 = 2 \).

3. **Region of Interest:**
   - The region where \( a + b \geq -c \) and \( a, b \in [0, 1] \) can be visualized as follows:
     - For \( c \in [-1, 0] \), the inequality \( a + b \geq -c \) becomes \( a + b \geq -c \). This is always true since \( a, b \geq 0 \) and \( -c \leq 1 \).
     - For \( c \in [0, 1] \), the inequality \( a + b \geq -c \) becomes \( a + b \geq -c \). This is always true since \( a, b \geq 0 \) and \( -c \leq 0 \).

4. **Volume of the Region:**
   - For \( c \in [-1, 0] \), the volume of the region where \( a + b \geq -c \) is the entire unit square in the \( a \)-\( b \) plane, which has an area of 1.
   - For \( c \in [0, 1] \), the volume of the region where \( a + b \geq -c \) is also the entire unit square in the \( a \)-\( b \) plane, which has an area of 1.

5. **Total Volume:**
   - The total volume of the region where \( a + b \geq -c \) is \( 1 \times 1 + 1 \times 1 = 2 \).

6. **Probability Calculation:**
   - The probability is the ratio of the volume of the region of interest to the total volume of the space.
   - Therefore, the probability is \( \frac{2}{2} = 1 \).

The final answer is \(\boxed{1}\)