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Question: Three randomly chosen non-negative integers x, y and z are found to satisfy the equation x + y + z =...

Three randomly chosen non-negative integers x, y and z are found to satisfy the equation x + y + z = 10, then what is the probability that z is even?
(a) 511\dfrac{5}{11}
(b) 12\dfrac{1}{2}
(c) 611\dfrac{6}{11}
(d) 3655\dfrac{36}{55}

Explanation

Solution

To solve this question we need to know that the number of non-negative solutions of a linear equation in r variables in which all variables has coefficient 1 and sum of all the variables is equal to n is given by n+r−1Cr−1^{n+r-1}{{C}_{r-1}}. Like in equation x + y + z = 8, total number of non-negative solutions will be 8+3−1C3−1=10C2^{8+3-1}{{C}_{3-1}}{{=}^{10}}{{C}_{2}}. After finding total number of non-negative solutions we will make different possible cases of values which are valid for z i.e. even values (0, 2, 4, 6, 8, 10), then after we will apply the same formula n+r−1Cr−1^{n+r-1}{{C}_{r-1}} to find the number of ways of solving in each case.

Complete step-by-step answer :
We will solve this problem using the formula, n+r−1Cr−1^{n+r-1}{{C}_{r-1}}
According to this formula number of non-negative solutions of the equation of the of r variables whose degree is 1 and sum of r variables is equal to n is given by n+r−1Cr−1^{n+r-1}{{C}_{r-1}}.
So using this formula in the equation, x + y + z =10 we get
In x + y + z =10
n = 10 and r = 3
so total number of non-negative solutions are = 10+3−1C3−1=12C2^{10+3-1}{{C}_{3-1}}{{=}^{12}}{{C}_{2}}
now, we will make cases as z can only take even values,
so case 1: z = 0
The number of ways of solving the equation x + y + 0 = 10 is given by n+r−1Cr−1^{n+r-1}{{C}_{r-1}} where n = 10 and r = 2.
=11C1^{11}{{C}_{1}}
Case 2: z = 2
The number of ways of solving the equation x + y = 8 is given by n+r−1Cr−1^{n+r-1}{{C}_{r-1}} where n = 8 and r = 2.
=9C1^{9}{{C}_{1}}
Case 3: z = 4
The number of ways of solving the equation x + y = 6 is given by n+r−1Cr−1^{n+r-1}{{C}_{r-1}} where n = 6 and r = 2.
=7C1^{7}{{C}_{1}}
Similarly, case 4: z = 6 we get number of solutions as 5C1^{5}{{C}_{1}}
case 5: z = 8 we get number of solutions as 3C1^{3}{{C}_{1}}
case 6: z = 10 we get number of solutions as 1C1^{1}{{C}_{1}}
Now to find the probability that z is even we will divide the total number of favourable cases by the total number of cases.
Total number of favourable cases is the sum of all the cases we made for z.
And total number of cases we solved as 10+3−1C3−1=12C2^{10+3-1}{{C}_{3-1}}{{=}^{12}}{{C}_{2}}.
Hence we get the probability as,
probability⁡=11C1+9C1+7C1+5C1+3C1+1C112C2\operatorname{probability}=\dfrac{^{11}{{C}_{1}}{{+}^{9}}{{C}_{1}}{{+}^{7}}{{C}_{1}}{{+}^{5}}{{C}_{1}}{{+}^{3}}{{C}_{1}}{{+}^{1}}{{C}_{1}}}{^{12}{{C}_{2}}}
Now we know that value of nCr^{n}{{C}_{r}} is given by n!r!(n−r)!\dfrac{n!}{r!\left( n-r \right)!}, hence
By solving for each value we get,
probability⁡=11+9+7+5+3+166=611\operatorname{probability}=\dfrac{11+9+7+5+3+1}{66}=\dfrac{6}{11}
Hence we get option (c) as the correct answer.

Note : You need to make sure that while making the cases for z, you only have to take cases till z reaches the value of 10 else if we take greater than 10 then x and y will have to be negative and it is not possible in our problem. And also remember the formula n+r−1Cr−1^{n+r-1}{{C}_{r-1}} to find the number of non-negative solutions for more problems of this kind.