Question

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

(A) 36/55

(B) 6/11

(C) 1/2

(D) 5/11

Answer 1

Answer is (B) 6/11

It is given that the integers x, y and z satisfy

x + y + z = 10

The total number of non-negative integers satisfying this equation is

Suppose z is even, let z = 2k.

Therefore,

x + y + 2k = 10

⇒ x + y = 10 – 2k

Then, the total number of non-negative solutions is 

11 + 9 + 7 + 5 + 3 + 1 = 36

Therefore, the probability that the integer z should be even is 

36/66 = 6/11