JEE Mathematics 2024 Question with Solution

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

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

The total number of outcomes is

$$11 \times 11 = 121$$

because there are $11$ choices for $x$ and $11$ choices for $y$.

Now,

$$|x-y|>5$$

means either

$$x-y>5$$

or

$$x-y<-5$$

Count the favourable pairs:

If $x=0$, then $y=6,7,8,9,10$ gives $5$ values.

If $x=1$, then $y=7,8,9,10$ gives $4$ values.

If $x=2$, then $y=8,9,10$ gives $3$ values.

If $x=3$, then $y=9,10$ gives $2$ values.

If $x=4$, then $y=10$ gives $1$ value.

If $x=5$, there is no possible value of $y$.

If $x=6$, then $y=0$ gives $1$ value.

If $x=7$, then $y=0,1$ gives $2$ values.

If $x=8$, then $y=0,1,2$ gives $3$ values.

If $x=9$, then $y=0,1,2,3$ gives $4$ values.

If $x=10$, then $y=0,1,2,3,4$ gives $5$ values.

Hence the total number of favourable outcomes is

$$5+4+3+2+1+0+1+2+3+4+5=30$$

Therefore, the required probability is

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

So, the correct option is A.

Given: Two integers are selected with replacement from $\{0,1,2,\ldots,10\}$.

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

Total outcomes:

$$11 \times 11 = 121$$

Now split the condition into two parts:

$$x-y>5$$

and

$$x-y<-5$$

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

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

For $x-y<-5$, by symmetry, the possible counts are again

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

Hence total favourable outcomes are

$$15+15=30$$

So the probability is

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

Therefore, the correct option is A.