<p>Three randomly chosen non-negative integers \(x\), \(y\) and \(z\) are found to satisfy the equation \(x + y + z = 10\). Then the probability that \(z\) is even, is</p>
Step-by-Step Solution
Key Concept: Use stars and bars to count total solutions to x+y+z=10, then count solutions where z is even by substituting z=2k and using the constraint that k must be a non-negative integer with 2k≤10.
<p><strong>Step 1: Count total solutions</strong></p><p>For non-negative integers x+y+z=10, using stars and bars: total solutions = C(10+3-1, 3-1) = C(12,2) = 66</p><p><strong>Step 2: Count solutions where z is even</strong></p><p>Let z = 2k where k ≥ 0. Then x+y+2k = 10, so x+y = 10-2k</p><p>For valid solutions: 10-2k ≥ 0 ⟹ k ≤ 5, so k ∈ {0,1,2,3,4,5}</p><p><strong>Step 3: For each k, count solutions</strong></p><p>For each value of k, the number of non-negative integer solutions to x+y = 10-2k is:</p><p>C(10-2k+1, 1) = 11-2k</p><p>Total favorable cases = Σ(11-2k) for k=0 to 5 = 11+9+7+5+3+1 = 36</p><p><strong>Step 4: Calculate probability</strong></p><p>P(z is even) = 36/66 = 6/11</p><p>∴ Answer: C</p>
Correct Answer: C