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Three sides of a regular polygon with 8 sides are chosen at random. Find the probability that, when these sides are extended, they form a triangle containing the polygon.



Answer :

konrad509
[tex]|\Omega|={8\choose 3}=\dfrac{8!}{3!5!}=\dfrac{6\cdot7\cdot8}{6}=7\cdot8=56[/tex]

There are two possibilities of choosing three sides of the polygon, so when they are extended, they form a triangle containing the polygon (pictures in the attachment). Multiplying it by the number of sides, gives us 16 possibilities in total.

[tex]|A|=16\\\\ P(A)=\dfrac{|A|}{|\Omega|}\\ P(A)=\dfrac{16}{56}=\dfrac{2}{7}[/tex]
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