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. $\frac{5}{11}$
• B. $\frac{1}{2}$
• C. $\frac{6}{11}$
• D. $\frac{36}{55}$

Answer 1

Answer: (C)

$\frac{6}{11}$

Assume $z$ to be $0,2,4,6,8,10$ and find solution of $x+y=10-z$, we get

Case 1: $z=0\Rightarrow x+y=10$ and $x,y\in \{0,1,2,...,10\}$

This can be done in ${}^{10+2-1}{C}_{2-1}={}^{11}{C}_1$ ways.

Case 2: $z=2\Rightarrow x+y=8$ and $x,y\in \{0,1,2,....,8\}$

This can be done in ${}^{8+2-1}{C}_{2-1}={}^9{C}_1$ ways.

Similarly, we can obtain for all the cases as ${}^7{C}_1,{}^5{C}_1,{}^3{C}_1$ and ${}^1{C}_1$ ways.

The total number of solutions for the given expression is ${}^{10+3-1}{C}_{3-1}={}^{12}{C}_2$ ways.

Hence, probability $=\frac{{}^{11}{C}_1{+}^9{C}_1{+}^7{C}_1{+}^5{C}_1{+}^3{C}_1{+}^1{C}_1}{{}^{12}{C}_2}=\frac{36}{66}=\frac{6}{11}$

Hence, option C is correct.