Exams Joint Entrance Examination Mathematics Probability Question 90 Question 90 of 417 Medium NUMERICAL
Let A be the event that the absolute difference between two randomly choosen real numbers in the sample space [ 0 , 60 ] [0,60] [ 0 , 60 ] is less than or equal to a . If P ( A ) = 11 36 \mathrm{P}(\mathrm{A})=\frac{11}{36} P ( A ) = 36 11 , then a \mathrm{a} a is equal to \_\_\_\_\_\_\_.
Answer: 10
Solutions · 1 Post your solution Official solution
∣ x − y ∣ − a \begin{gathered} |x-y|-a \end{gathered} ∣ x − y ∣ − a
P ( A ) = a r ( O A C D E G ) ( O B D F ) = a r ( O B D F ) − a r ( A B C ) − a r ( E F G ) a r ( O B D F ) ⇒ 11 36 = ( 60 ) 2 − 1 2 ( 60 − a ) 2 − 1 2 ( 60 − a ) 2 3600 ⇒ 1100 = 3600 − ( 60 − a ) 2 ⇒ ( 60 − a ) 2 = 2500 ⇒ 60 − a = 50 ⇒ a = 10 \begin{gathered} P(A)=\frac{ar(OACDEG)}{(OBDF)} \\ =\frac{ar(OBDF)-ar(ABC)-ar(EFG)}{ar(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 (60-a{)}^{2}=2500\Rightarrow 60-a=50 \\ \Rightarrow a=10 \end{gathered} P ( A ) = ( O B D F ) a r ( O A C D E G ) = a r ( O B D F ) a r ( O B D F ) − a r ( A B C ) − a r ( E F G ) ⇒ 36 11 = 3600 ( 60 ) 2 − 2 1 ( 60 − a ) 2 − 2 1 ( 60 − a ) 2 ⇒ 1100 = 3600 − ( 60 − a ) 2 ⇒ ( 60 − a ) 2 = 2500 ⇒ 60 − a = 50 ⇒ a = 10 Answer the question to unlock