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 _________

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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 $x-y>-a$

$P(A)=\frac{area(OACDEG)}{area(OBDF)}$

$=\frac{area(OBDF)-area(ABC)-area(EFG)}{area(OBDF)}$

$\Rightarrow \frac{11}{36}=\frac{(60)^{2}-\frac{1}{2}(60-a)^{2}-\frac{1}{2}(60-a)^{2}}{3600}$

$\Rightarrow 1100=3600-(60-a)^{2}$

$\Rightarrow \quad(60-a)^{2}=2500 \Rightarrow 60-a=50$

$\Rightarrow \quad a=10$