JEE Mathematics 2024 Question with Solution

Given: Two integers $x$ and $y$ are chosen with replacement from the set $\{0,1,2,\dots,10\}$.

Find: The probability that $|x-y|>5$.

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 $6, 7, 8, 9, 10$ $\left(5\text{ values}\right)$
• If $x = 1$, $y$ can be $7, 8, 9, 10$ $\left(4\text{ values}\right)$
• If $x = 2$, $y$ can be $8, 9, 10$ $\left(3\text{ values}\right)$
• If $x = 3$, $y$ can be $9, 10$ $\left(2\text{ values}\right)$
• If $x = 4$, $y$ can be $10$ $\left(1\text{ value}\right)$
• If $x = 5$, there are no possible values of $y$
• If $x = 6$, $y = 0$ $\left(1\text{ value}\right)$
• If $x = 7$, $y = 0, 1$ $\left(2\text{ values}\right)$
• If $x = 8$, $y = 0, 1, 2$ $\left(3\text{ values}\right)$
• If $x = 9$, $y = 0, 1, 2, 3$ $\left(4\text{ values}\right)$
• If $x = 10$, $y = 0, 1, 2, 3, 4$ $\left(5\text{ values}\right)$

Adding these values, the total number of favorable outcomes is:

$$5 + 4 + 3 + 2 + 1 + 1 + 2 + 3 + 4 + 5 = 30$$

Therefore, the required probability is:

$$\frac{30}{121}$$

The correct option is A.

Given: We need the number of ordered pairs $\left(x,y\right)$ from $\{0,1,2,\dots,10\}$ such that $|x-y|>5$.

Find: The required probability.

Use symmetry:

• Count pairs with $x-y>5$
• Double that count to include $x-y<-5$

For $x-y>5$, the counts are:

$$5+4+3+2+1=15$$

By symmetry, the number of pairs with $x-y<-5$ is also $15$.

Hence favorable outcomes:

$$15+15=30$$

Total outcomes:

$$11\times 11=121$$

So the probability is:

$$\frac{30}{121}$$

Therefore, the correct option is A.