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
x + y + z = 10
Total non‑negative solution = 10+3–1C3–1 = 12C2 = = 66
If z is even, then x + y = 10 or 8 or 6 or 4 or 2 or 0
Number of solutions = 11C1 + 9C1 + 7C1 + 5C1 + 3C1 + 1C1 = 11 + 9 + 7 + 5 + 3 + 1 = 36
Probability =.
We are given that three non-negative integers x, y, z satisfy x + y + z = 10. We need to find the probability that z is even.
Step 1: Total number of solutions
The number of non-negative integer solutions to x + y + z = 10 is given by the combination formula with repetition (stars and bars method):
Step 2: Favorable cases where z is even
Let z = 2k, where k is a non-negative integer. Then the equation becomes x + y + 2k = 10, or x + y = 10 - 2k.
For each k, the number of non-negative integer solutions for (x, y) is given by:
k can range from 0 to 5 (since z = 2k ≤ 10). So the total favorable number is the sum over k=0 to 5:
Step 3: Compute the probability
Probability = Favorable / Total = 36 / 66 = 6 / 11.
So the correct answer is .
Related Topics:
This problem involves combinatorics (counting number of solutions) and probability. Key concepts include:
Formulae:
Number of non-negative integer solutions to x1 + x2 + ... + xr = n is C(n + r - 1, r - 1).
Probability = (Number of favorable outcomes) / (Total number of outcomes).