Question

Two integers $x$ and $y$ are chosen with replacement from the set $\\{0, 1, 2, 3, \ldots, 10\\}$. Then the probability that $|x - y| > 5$ is:

• A. $\frac{30}{121}$
• B. $\frac{62}{121}$
• C. $\frac{60}{121}$
• D. $\frac{31}{121}$

Answer 1

Answer: (A)

$\frac{30}{121}$

Answer 2

The total number of outcomes when choosing $x$ and $y$ with replacement from the set $\\{0, 1, 2, \dots, 10\\}$ is:
$11 \times 11 = 121$
To satisfy $|x - y| > 5$, we need $x - y > 5$ or $x - y < -5$. We count the favorable pairs by analyzing each possible value of $x$:
If $x = 0$, $y$ can be 6, 7, 8, 9, 10 (5 values)
If $x = 1$, $y$ can be 7, 8, 9, 10 (4 values)
If $x = 2$, $y$ can be 8, 9, 10 (3 values)
If $x = 3$, $y$ can be 9, 10 (2 values)
If $x = 4$, $y$ can be 10 (1 value)
If $x = 5$, there are no possible values of $y$
If $x = 6$, $y = 0$ (1 value)
If $x = 7$, $y = 0, 1$ (2 values)
If $x = 8$, $y = 0, 1, 2$ (3 values)
If $x = 9$, $y = 0, 1, 2, 3$ (4 values)
If $x = 10$, $y = 0, 1, 2, 3, 4$ (5 values)
Adding these values, the total number of favorable outcomes is:
$5 + 4 + 3 + 2 + 1 + 1 + 1 + 2 + 3 + 4 + 5 = 30$
The required probability is:
$\frac{30}{121}$
Final Answer: $\frac{30}{121}$