Probability that Z is Even for Nonnegative Sum Equation

To find the probability that $z$ is an even number given that non-negative integers $x, y,$ and $z$ satisfy the equation $x + y + z = 10$, we first determine the total number of non-negative integer solutions to the equation and then the number of favorable solutions where $z$ is even.

Step 1: Total Number of Non-negative Integer Solutions

The equation is: $x + y + z = 10 \quad \text{where } x, y, z \ge 0 \text{ and } x, y, z \in \mathbb{Z}$

The number of non-negative integer solutions to $x_1 + x_2 + \dots + x_r = n$ is given by the formula: $N_{\text{total}} = \binom{n + r - 1}{r - 1}$

Here, $n = 10$ and $r = 3$: $N_{\text{total}} = \binom{10 + 3 - 1}{3 - 1} = \binom{12}{2} = \frac{12 \times 11}{2} = 66$

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Step 2: Number of Favorable Solutions ($z$ is Even)

Since $z$ is a non-negative integer and $x + y + z = 10$, $z$ can take any even value in the set $\{0, 2, 4, 6, 8, 10\}$.

Let $z = 2k$, where $k \in \{0, 1, 2, 3, 4, 5\}$. Substituting $z = 2k$ into the equation gives: $x + y = 10 - 2k$

For a given $k$, the number of non-negative integer solutions to $x + y = 10 - 2k$ is: $(10 - 2k) + 1 = 11 - 2k$

Summing the number of solutions for each possible value of $k$:

• For $k = 0 \ (z = 0)$: $11 - 2(0) = 11$ solutions
• For $k = 1 \ (z = 2)$: $11 - 2(1) = 9$ solutions
• For $k = 2 \ (z = 4)$: $11 - 2(2) = 7$ solutions
• For $k = 3 \ (z = 6)$: $11 - 2(3) = 5$ solutions
• For $k = 4 \ (z = 8)$: $11 - 2(4) = 3$ solutions
• For $k = 5 \ (z = 10)$: $11 - 2(5) = 1$ solution

Thus, the total number of favorable solutions is: $N_{\text{favorable}} = 11 + 9 + 7 + 5 + 3 + 1 = 36$

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Step 3: Calculating the Probability

The required probability $P(z \text{ is even})$ is: $P(z \text{ is even}) = \frac{N_{\text{favorable}}}{N_{\text{total}}} = \frac{36}{66} = \frac{6}{11}$

Thus, the correct option is [B].