Probability Question 12
Question 12 - 31 January - Shift 2
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 _________
Show Answer
Answer: 10
Solution:
Formula: Area under curve, Probability of occurrence of an event
$|x-y|<a \Rightarrow-a<x-y<a$
$\Rightarrow x-y<a$ and $x-y>-a$
$P(A)=\frac{area(OACDEG)}{area(OBDF)}$
$=\frac{area(OBDF)-area(ABC)-area(EFG)}{area(OBDF)}$
$\Rightarrow \frac{11}{36}=\frac{(60)^{2}-\frac{1}{2}(60-a)^{2}-\frac{1}{2}(60-a)^{2}}{3600}$
$\Rightarrow 1100=3600-(60-a)^{2}$
$\Rightarrow \quad(60-a)^{2}=2500 \Rightarrow 60-a=50$
$\Rightarrow \quad a=10$
