Let A be the event that the absolute difference between two randomly chosen real numbers i…

Question

Let $A$ be the event that the absolute difference between two randomly chosen real numbers in $[0,60]$ is $\le a$. If $P(A)=\dfrac{11}{36}$, then $a$ is equal to \_\_\_\_.

Answer: 10

Accepted Answer

Using the geometric (area) probability model on $[0,60]^2$:

$$P(A) = \frac{3600-(60-a)^2}{3600} = \frac{11}{36}$$

Actually: $\dfrac{11}{36}=\dfrac{3600-(60-a)^2}{3600}\Rightarrow(60-a)^2=3600-1100=2500\Rightarrow60-a=50\Rightarrow a=10$.

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