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Answer is A.
ReplyDeleteBut I need integral calculus to solve it.
By simmetry of circle, the expected value is two times the value on interval [0,pi].
min(x)=0 & max(x)=2
p(x)=1/pi uniformly in [0,pi]. Where p(x) is distribution of probability of variable x.
x/2=1*sin(alpha/2), alpha=angle between sides of triangle formed by ladybugs and center of circle.
x=2*sin(alpha/2) & dx = cos(alpha/2)
E[x]=2*int(0_2){x*p(x))=2*int(0_pi){(1/pi)*sin(alpha)d(alpha)} = 4/pi.
Note: int(a_b){f(x)} is integral between limits a & b of function f.
MIGUE.
Superb!
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