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# Consider three straight lines A, B and C in a plane. 4 unique points a

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Re: Consider three straight lines A, B and C in a plane. 4 unique points a [#permalink]
total ways of choosing 3 points out of (4+3+6) (because to form a triangle, we need to select any three points.)
=13C3=13*12*11/3!=26*11=286
We have to minus some triangles from those points which are collinear.
number of triangles could have been formed, had these numbers been non-collinear.
= 4C3+3C3+6C3=4+1+20=25
Therefore, the total number of triangles is = 286-25=261..
Probability of choosing 261 ways out of 286 ways = 261/286
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Re: Consider three straight lines A, B and C in a plane. 4 unique points a [#permalink]
chetan2u wrote:
Kritisood wrote:
Consider three straight lines A, B and C in a plane. 4 unique points are marked on line A, 3 unique points are marked on line B and 6 unique points are marked on line C. If we randomly select three marked points, find the probability that the three points form a triangle. Note that no other points, except the points on the same line are collinear

A. 25/286
B. 72/286
C. 90/286
D. 189/286
E. 261/286

Now making a triangle is SAME as choosing 3 points, so total ways of choosing 3 points out of (4+3+6)=13C3=13*12*11/3!=26*11=286

BUt while choosing these points we have also chosen 3 co-linear points ..
Ways to choose 3 co-linear points = 4C3+3C3+6C3=4+1+20=25

So ways of forming triangles = 286-25=261..
Probability of choosing 261 ways out of 286 ways = 261/286

E

Hi chetan2u i understand your approach completely. I did as follows could you tell me where I faltered?

total points possible = 286
to form a triangle I can choose 1 point from the first line in 4c1 ways, one point from the second line in 3c1 ways, and 1 point from the third line in 6c1 way. Total = 4*3*6=72

I understand i am missing many cases, but why is this approach wrong?
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Re: Consider three straight lines A, B and C in a plane. 4 unique points a [#permalink]
1 - (4c3+3c3+6c3) / 13c3 = 261/289
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Re: Consider three straight lines A, B and C in a plane. 4 unique points a [#permalink]
2
Kudos
Kritisood wrote:
chetan2u wrote:
Kritisood wrote:
Consider three straight lines A, B and C in a plane. 4 unique points are marked on line A, 3 unique points are marked on line B and 6 unique points are marked on line C. If we randomly select three marked points, find the probability that the three points form a triangle. Note that no other points, except the points on the same line are collinear

A. 25/286
B. 72/286
C. 90/286
D. 189/286
E. 261/286

Now making a triangle is SAME as choosing 3 points, so total ways of choosing 3 points out of (4+3+6)=13C3=13*12*11/3!=26*11=286

BUt while choosing these points we have also chosen 3 co-linear points ..
Ways to choose 3 co-linear points = 4C3+3C3+6C3=4+1+20=25

So ways of forming triangles = 286-25=261..
Probability of choosing 261 ways out of 286 ways = 261/286

E

Hi chetan2u i understand your approach completely. I did as follows could you tell me where I faltered?

total points possible = 286
to form a triangle I can choose 1 point from the first line in 4c1 ways, one point from the second line in 3c1 ways, and 1 point from the third line in 6c1 way. Total = 4*3*6=72

I understand i am missing many cases, but why is this approach wrong?

What you have taken is when you take one point from each side. Now three points in one line cannot form a triangle but two points on a line can surely make a triangle.
So there will be many cases where you take 2 points on one line and 3rd point on the other line.
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Re: Consider three straight lines A, B and C in a plane. 4 unique points a [#permalink]
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Re: Consider three straight lines A, B and C in a plane. 4 unique points a [#permalink]
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