JEE Mathematics 2023 Question with Solution

Given: Two real numbers $x$ and $y$ are chosen from the sample space $[0,60]$, and the event is $|x-y|\le a$.

Find: The value of $a$ given that $P(A)=\frac{11}{36}$.

From the condition,

Using areas,

The total area is

The favourable shaded region equals the area of the square minus the two corner triangles, which gives

Now substitute the given probability:

So,

Hence,

Therefore,

which gives

Therefore, the value of $a$ is $10$.

Given: $|x-y|\le a$ for two numbers chosen from $[0,60]$.

Find: $a$ when $P(A)=\frac{11}{36}$.

Interpret the random choice geometrically as a point $(x,y)$ in the square of side $60$. The condition $|x-y|\le a$ means the point lies within distance $a$ from the line $x=y$, measured parallel to the axes through the lines $x-y=\pm a$.

Thus the excluded part consists of two congruent right triangles, each with leg length $60-a$. Their total area is

So the favourable area is

Now,

Multiplying through by $3600$,

Therefore,

Taking the positive root relevant to geometry,

Hence,

So the final answer is $10$.