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Q.
The probablilty that the length of a randomly chosen chord
of a circle lies between $\frac{2}{3}$ and $\frac{5}{6}$ of its diameter is
Probability - Part 2
Solution:
Let $l$ be the length of the chord $A B$ of the given circle of radius $a$ and $r$ be the distance of the mid point $D$ of the chord from the centre $C$, then $r=a \cos \theta$ and $l=2 a \sin \theta$. According to given condition:
$\frac{2}{3}(2 a)<\,2 a \sin \theta<\frac{5}{6}(2 a)$
$\Rightarrow \frac{2}{3}<\,\sin \theta<\frac{5}{6} $
$\Rightarrow \frac{\sqrt{11}}{6}<\,\cos \theta<\frac{\sqrt{5}}{3}$
$\Rightarrow \frac{\sqrt{11}}{6} a<\, r<\,\frac{\sqrt{5}}{3} a$
$\therefore $ The given condition is satisfied if the mid point of the chord lies within the region
between the concentric circles of radii $\frac{\sqrt{11}}{6} a$ and $\frac{\sqrt{5}}{3} a$.
Hence, the required probability
$=\frac{\pi\left(\frac{\sqrt{5}}{3} a\right)^{2}-\pi\left(\frac{\sqrt{11}}{6} a\right)^{2}}{\pi a^{2}}=\frac{1}{4}$
