3 of the 6 vertices of a regular hexagon are chosen at random.The
probability that the triangle with these vertices is equilateral is
1. 1/10 2. 3/10 3. 1/5 4. 4/10

• i think the ans is 1/10 as we can choose 3 vertices in 6c3 ways.. now the possibility to mk an equiltrl triangle is 2.. so 2/6c3= 1/10
• 10 years agoHelpfull: Yes(12) No(2)
• 1/10
  
  event = 6c3 and only 2 possible selection to made a eq. tringle.
  2/20 =1/10
  
  
  
• 10 years agoHelpfull: Yes(3) No(0)
• a regular hexagon contains 6 eq. triangles.
  
  so 6/6c3 =6/20=3/10.
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• Ans is 1/10
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• ans is 1/10 as total no. of combinations will be 20 and favorable combinations are 2
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• 6c3=20
  to form 2 equi trianlge from hexagon is 2
  therfore prob=2/10=1/10
• 10 years agoHelpfull: Yes(1) No(0)
• Here sample space is 6c3 (choosing any three vertices)
  number of vertices for equilateral=6c1*1*1(selecting any vertex from 6 vertices after that we have no choice we have to choose fix one)
  so answer is
  6c1/6c3
  =3/10
• 10 years agoHelpfull: Yes(0) No(1)