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n points are equally spaced on a circle. where n is an even
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26 May 2020, 10:47
26 May 2020, 10:47
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n points are equally spaced on a circle. where n is an even number greater than 3. If 3 of me n points are to be chosen at random. what is the probability that a triangle having the 3 points chosen as vertices will be a right triangle?
A. \(\frac{n-1}{6}\)
B. \(\frac{n+2}{6}\)
C. \(\frac{2}{3n+2}\)
D. \(\frac{3}{n-1}\)
E. \(\frac{6}{n+4}\)
A. \(\frac{n-1}{6}\)
B. \(\frac{n+2}{6}\)
C. \(\frac{2}{3n+2}\)
D. \(\frac{3}{n-1}\)
E. \(\frac{6}{n+4}\)
Re: n points are equally spaced on a circle. where n is an even
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26 May 2020, 10:47
26 May 2020, 10:47
Expert Reply
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Re: n points are equally spaced on a circle. where n is an even
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09 Jun 2020, 00:17
09 Jun 2020, 00:17
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Probability = favorable cases/Total number of cases
= choose 3 points such that they form right triangle/choose 3 points on the circle
Choose 3 points such that they form right triangle
To get a right angled triangle, one of the sides must be diameter (2 points as ends of diameter)and the other point on the perimeter of the circle
If there are n points on the circle, we can form n/2 diagonals. We can choose one diagonal in n/2 ways.
After picking up the diagonal (2 points), we can choose the third point from the available n-2 points in (n-2)C1 ways.
Therefore, total number of ways to choose 3 points to form right angled triangle = n/2 X (n-2) ways
Total number of cases to choose 3 points
nC3
Therefore probability = (n/2 X (n-2))/nC3
= 3/(n-1)
= choose 3 points such that they form right triangle/choose 3 points on the circle
Choose 3 points such that they form right triangle
To get a right angled triangle, one of the sides must be diameter (2 points as ends of diameter)and the other point on the perimeter of the circle
If there are n points on the circle, we can form n/2 diagonals. We can choose one diagonal in n/2 ways.
After picking up the diagonal (2 points), we can choose the third point from the available n-2 points in (n-2)C1 ways.
Therefore, total number of ways to choose 3 points to form right angled triangle = n/2 X (n-2) ways
Total number of cases to choose 3 points
nC3
Therefore probability = (n/2 X (n-2))/nC3
= 3/(n-1)
