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 = 10Total number of non-negative solutions =10+3−1C3−1                                                              =12C2=66Now let z = 2n∴ x+y=10−2n; n≥0Number of non-negative solutions of above equation is11 + 9 + 7 + 5 + 3 + 1 = 36.∴  Required probability =3666=611