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Let $\mathrm{A}$ be the event that the absolute difference between two randomly choosen real numbers in the sample space $\mathrm{ [0,60]}$ is less than or equal to $\mathrm{a}$ . If $\mathrm{P(A)=\frac{11}{36},}$ then a is equal to
Option: 1
10

Option: 2
-

Option: 3
-

Option: 4
-

Answers (1)

best_answer

$\mathrm{\mid x-y\mid< a\rightarrow -a< x-y\, \, \&\: \: x-y< a}$

$\mathrm{x,y \in\left [ 0,60 \right ]}$

$\mathrm{ P(A)=\frac{\text { Shaded area }}{\text { Total area }}=\frac{(60)^2-\left[\frac{1}{2}(60-a)^2+\frac{1}{2} \times(60-a)^2\right]}{(60)^2} }$
$\mathrm{ P(A)=\frac{(60)^2-(60-a)^2}{(60)^2} }$
$\mathrm{ \frac{11}{36}=\frac{120 a-a^2}{3600} }$
$\mathrm{1100=120 a-a^2 }$
$\mathrm{a^2-120 a+1100=0 }$
$\mathrm{ a^2-110 a-10 a+1100=0 }$
$\mathrm{ a(a-110)-10(a-110)=0= }$
$\mathrm{ (a-10)(a-110)=0 }$
$\mathrm{ \text { Ans. } a=10 }$
$\mathrm{ (\because \text { for } a=110, P(A)=1)}$

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