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Question

Three points P,Q,R are selected at random from the circumference of a circle. The probability p that the points lie on a semicircle is k4.find k?

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Solution

Let the arc length (¯¯¯¯¯¯¯¯PQ)=x;l(¯¯¯¯¯¯¯¯PR)=y and let the length of circumference be 2s. (as shown in figure (i))Clearly, 0≤x≤2s;0≤y≤2s⇒ Total region =2s×2s=4s2Now following four case arise which are favorable to Case (i): x,y≤s, then clearly the three points lie in semicircle ∴x,y∈[0,s] is favorable region (as shown in figure (ii))Case (ii): x≤s,y≥x+s (as shown in figure (iii))Clearly P,Q,R lie on semi-circle x∈[0,s];y∈[x+s,2s] is favorable region Case (iii) : x≥s,y≥s clearly P,Q,R lie on semi-circle with favorable region x,y∈[s,2s] (as shown in figure (iv)) Case (iv): y≤s;x≥y+s∴ Favorable region is y∈[0,s];x∈[y+s,2s] (as shown in figure (v)) The favorable region is ∴ P(Event) =s2+12s2+s2+12s24s2=34

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