Question

Let $A$ be the event that the absolute difference between two randomly choosen real numbers in the sample space $[0,60]$ is less than or equal to a If $P(A)=\frac{11}{36}$, then a is equal to _____

Hint:

When dealing with probability problems involving geometric areas, express the probability as a ratio of areas, and solve for the unknown by using the geometric properties of the figure.

Answer 1

Correct Answer: 10

$|x-y|<a\Rightarrow -a<x-y<a$
$\Rightarrow x-y<a\text{and}x-y>-a$

$P(A)=\frac{\mathrm{ar}(OACDEG)}{(OBDF)}$
$=\frac{\mathrm{ar}(OBDF)-\mathrm{ar}(ABC)-\mathrm{ar}(EFG)}{\mathrm{ar}(OBDF)}$
$\Rightarrow \frac{11}{36}=\frac{(60{)}^2-\frac{1}{2}(60-a{)}^2-\frac{1}{2}(60-a{)}^2}{3600}$
$\Rightarrow 1100=3600-(60-a{)}^2$
$\Rightarrow (60-a{)}^2=2500\Rightarrow 60-a=50$
$\Rightarrow a=10$
So , the correct answer is 10.

Answer 2

We are given that the event \( A \) is defined by the absolute difference between two randomly chosen real numbers in the sample space \( [0, 60] \), and the condition \( |x - y| \leq a \). 
This implies: \[ -x \leq y \leq x + a \quad \text{and} \quad x - a \leq y \leq x. \] Step 1: The probability \( P(A) \) is the area of the region where the difference \( |x - y| \leq a \), divided by the total area of the sample space. The total area of the sample space is \( 60 \times 60 = 3600 \). 
Step 2: The area corresponding to the condition \( |x - y| \leq a \) is represented as the area of the region \( \text{ABCDE} \) on the diagram. By subtracting the areas of the other regions, we can compute the desired probability: \[ P(A) = \frac{\text{Area of region ABCDE}}{\text{Total Area of square}} = \frac{11}{36}. \] Step 3: Using the formula for the areas: \[ P(A) = \frac{ \text{Area of ABCDE} }{ \text{Area of square} } = \frac{11}{36}. \] Using the geometry of the figure: \[ P(A) = \frac{(60)^2 - (60 - a)^2}{3600} = \frac{11}{36}. \] Solving this, we get: \[ \frac{1100}{3600} = \frac{11}{36}. \] Step 4: Solving for \( a \), we get: \[ (60 - a)^2 = 2500 \quad \Rightarrow \quad 60 - a = 50 \quad \Rightarrow \quad a = 10. \] Thus, the value of \( a \) is 10.