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QJEEM23Ma5e75: JEE Main 2023 Mathematics - Class 12 JEEM23Ma5e75 Solution: Let A be the event that the absolute difference between two…

Let A be the event that the absolute difference between two randomly choosen real numbers in the sample space [0,60][0,60] is less than or equal to a . If P(A)=1136\mathrm{P}(\mathrm{A})=\frac{11}{36} , then a\mathrm{a} is equal to _______.

Step-by-Step Verified Solution

amp;∣x−y∣lt;a⇒−alt;x−ylt;aamp;⇒x−ylt;a and x−ygt;−a\begin{aligned}& |\mathrm{x}-\mathrm{y}|<\mathrm{a} \Rightarrow-\mathrm{a}<\mathrm{x}-\mathrm{y}<\mathrm{a} \\\\& \Rightarrow \mathrm{x}-\mathrm{y}<\mathrm{a} \text { and } \mathrm{x}-\mathrm{y}>-\mathrm{a}\end{aligned}

JEE Main 2023 (Online) 31st January Evening Shift Mathematics - Probability Question 64 English Explanation

amp;P(A)=ar⁡(OACDEG)(OBDF)amp;=ar⁡(OBDF)−ar⁡(ABC)−ar⁡(EFG)ar⁡(OBDF)amp;⇒1136=(60)2−12(60−a)2−12(60−a)23600amp;⇒1100=3600−(60−a)2amp;⇒(60−a)2=2500⇒60−a=50amp;⇒a=10\begin{aligned}& \mathrm{P}(\mathrm{A})=\frac{\operatorname{ar}(\mathrm{OACDEG})}{(\mathrm{OBDF})} \\\\& =\frac{\operatorname{ar}(\mathrm{OBDF})-\operatorname{ar}(\mathrm{ABC})-\operatorname{ar}(\mathrm{EFG})}{\operatorname{ar}(\mathrm{OBDF})} \\\\& \Rightarrow \frac{11}{36}=\frac{(60)^2-\frac{1}{2}(60-\mathrm{a})^2-\frac{1}{2}(60-\mathrm{a})^2}{3600} \\\\& \Rightarrow 1100=3600-(60-\mathrm{a})^2 \\\\& \Rightarrow (60-\mathrm{a})^2=2500 \Rightarrow 60-\mathrm{a}=50 \\\\& \Rightarrow \mathrm{a}=10\end{aligned}

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