JEE MAIN - Mathematics (2023 - 31st January Evening Shift - No. 21)
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 $\mathrm{P}(\mathrm{A})=\frac{11}{36}$, then $\mathrm{a}$ is equal to _______.
Answer
10
Explanation
$$
\begin{aligned}
& |\mathrm{x}-\mathrm{y}|<\mathrm{a} \Rightarrow-\mathrm{a}<\mathrm{x}-\mathrm{y}<\mathrm{a} \\\\
& \Rightarrow \mathrm{x}-\mathrm{y}<\mathrm{a} \text { and } \mathrm{x}-\mathrm{y}>-\mathrm{a}
\end{aligned}
$$
_31st_January_Evening_Shift_en_21_1.png)
$$ \begin{aligned} & \mathrm{P}(\mathrm{A})=\frac{\operatorname{ar}(\mathrm{OACDEG})}{(\mathrm{OBDF})} \\\\ & =\frac{\operatorname{ar}(\mathrm{OBDF})-\operatorname{ar}(\mathrm{ABC})-\operatorname{ar}(\mathrm{EFG})}{\operatorname{ar}(\mathrm{OBDF})} \\\\ & \Rightarrow \frac{11}{36}=\frac{(60)^2-\frac{1}{2}(60-\mathrm{a})^2-\frac{1}{2}(60-\mathrm{a})^2}{3600} \\\\ & \Rightarrow 1100=3600-(60-\mathrm{a})^2 \\\\ & \Rightarrow (60-\mathrm{a})^2=2500 \Rightarrow 60-\mathrm{a}=50 \\\\ & \Rightarrow \mathrm{a}=10 \end{aligned} $$
$$ \begin{aligned} & \mathrm{P}(\mathrm{A})=\frac{\operatorname{ar}(\mathrm{OACDEG})}{(\mathrm{OBDF})} \\\\ & =\frac{\operatorname{ar}(\mathrm{OBDF})-\operatorname{ar}(\mathrm{ABC})-\operatorname{ar}(\mathrm{EFG})}{\operatorname{ar}(\mathrm{OBDF})} \\\\ & \Rightarrow \frac{11}{36}=\frac{(60)^2-\frac{1}{2}(60-\mathrm{a})^2-\frac{1}{2}(60-\mathrm{a})^2}{3600} \\\\ & \Rightarrow 1100=3600-(60-\mathrm{a})^2 \\\\ & \Rightarrow (60-\mathrm{a})^2=2500 \Rightarrow 60-\mathrm{a}=50 \\\\ & \Rightarrow \mathrm{a}=10 \end{aligned} $$
Comments (0)
