Three randomly chosen nonnegative integers x, y and z are found to satisfy the equation x + y + z =

Q: Why can't we just say the probability is 1/2 because $z$ is even or odd?

That is a classic trap! Parity is not uniformly distributed here because the sum is fixed. We are not picking numbers from a set; we are counting partitions. We must count the favorable cases explicitly to see the true distribution.

Q: Why use Stars and Bars instead of just listing them?

Listing is prone to error and becomes impossible as the numbers grow. Stars and Bars is a systematic, algebraic tool that guarantees you won't miss any solutions. It is the professional way to handle combinatorics.

Q: Is there a faster way than summing all these cases?

Yes, you could use generating functions, but for a small sum like 10, the case-by-case method is actually very elegant and less prone to algebraic mistakes. It also helps you visualize the symmetry of the problem.

Question

Three randomly chosen nonnegative integers $x, y$ and $z$ are found to satisfy the equation $x + y + z = 10$. Then the probability that $z$ is even, is

Options:

(A) $\frac{36}{55}$

(B) $\frac{6}{11}$

(C) $\frac{1}{2}$

(D) $\frac{5}{11}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing before a challenge that seems simple on the surface: distribute 10 identical items into three distinct bins, $x$, $y$, and $z$. This is the essence of combinatorics—the art of counting without actually listing every single possibility.

We are looking for the number of non-negative integer solutions to the equation $x + y + z = 10$. This is a classic **Stars and Bars** problem.

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

Here, $n = 10$ and $r = 3$. Plugging these values in, we get:
$$\binom{10+3-1}{3-1} = \binom{12}{2}$$

Calculating this, we find:
$$\binom{12}{2} = \frac{12 	imes 11}{2} = 66$$
So, there are 66 total ways to distribute these items.

### The Parity Constraint

Now, the problem adds a twist: $z$ must be an even non-negative integer. This changes everything. We are no longer looking at the entire universe of 66 solutions; we are looking for a specific subset where $z \in \{0, 2, 4, 6, 8, 10\}$.

Since $x$ and $y$ are non-negative, their sum $x + y$ must be at least 0. From our original equation, $x + y = 10 - z$.

This implies $10 - z \ge 0$, or $z \le 10$. This confirms our range for $z$ is indeed $0$ to $10$.

### The Case-by-Case Analysis

Let's break this down systematically. For each even value of $z$, we need to find the number of solutions for $x + y = 10 - z$.

The number of non-negative solutions to $x + y = N$ is simply $N + 1$. Let's apply this:

*   If $z = 0$, $x + y = 10$, so there are $10 + 1 = 11$ solutions.
*   If $z = 2$, $x + y = 8$, so there are $8 + 1 = 9$ solutions.
*   If $z = 4$, $x + y = 6$, so there are $6 + 1 = 7$ solutions.
*   If $z = 6$, $x + y = 4$, so there are $4 + 1 = 5$ solutions.
*   If $z = 8$, $x + y = 2$, so there are $2 + 1 = 3$ solutions.
*   If $z = 10$, $x + y = 0$, so there is $0 + 1 = 1$ solution.

### The Final Synthesis

Now, we simply sum these favorable outcomes:
$$11 + 9 + 7 + 5 + 3 + 1 = 36$$

We have 36 favorable outcomes out of a total of 66. The probability is the ratio of favorable outcomes to total outcomes:
$$P = \frac{36}{66}$$

Dividing both the numerator and the denominator by 6, we arrive at our elegant final answer:
**$$P = \frac{6}{11}$$**

This problem beautifully demonstrates how a complex constraint can be tamed by breaking it into smaller, manageable cases. Keep practicing this systematic approach, and you will find that even the most daunting combinatorics problems become clear.

Animated Solution for Mathematics - Probability: Three randomly chosen nonnegative integers $x, y$ and $z$ are found to satisfy the equation $x + y + z = 10$ . Then the probability that $z$ is even, is

Select Answer:

Visualized Solution

Problem Setup

Total Outcomes Formula

Calculating Total Solutions

Favorable Condition

Possible Values of $z$

Case 1: $z = 0$

All Favorable Cases

Total Favorable Outcomes

Final Probability Calculation

The Sigma Insight: Classical Definition of Probability

Analyzing the Setup

The Parity Constraint

The Case-by-Case Analysis

The Final Synthesis

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