CAT4MBA

Number of possible equilateral triangles = 4


Let’s find number of triangles with 2 equal sides

Let the equal sides be 1 : other side cant be 4 or 7 (Sum of two sides should more than or equal to the third side) , number of possible triangles = 1

If the equal sides are 2 : 7 cant be the third side, So total number of triangles = 2

If the equal sides are 4 – number of triangles = 3

If the equal sides are 7 – number of triangles = 3

So total number of bilateral triangles are = 9

Finally we will try to find the number of triangles with all different sides
7 can’t be the side of any of the triangle. (The other two large sides 2 + 4 < 7)
And we cant have any triangle with 1, 2, 4

So the total number of triangles = 13.

Thus the probability that one of the triangles is equilateral is 4/13

I think we had missed 2 triangles from the total list of triangles. Beacause the answer is given as 4/11.

No.of equilateral triangles - 4C1 = 4
No.of Scalene triangles - 4C1 * 3C1 = 12
No.of normal triangles - 4C3 = 4

Among the total number of triangles 9 are removed (Example (2,2,7) (1,1,4) ....)

(But i was unable to trace the entire 9.)

So, the remaining triangles are = 20-9 = 11.

Therefore Probability = 4/11.

Thanks.

As per your solution total number of possible triangles = 20 .( correct)
And number of triangles failing the principle (sum of the two sides shd be more than equal to 3rd side ) is 9

Among the total number of triangles 9 are removed (Example (2,2,7) (1,1,4) ....)

But I cldnt find out all the possible nine sets I cld find only 7 and those are
(1, 1, 4), (1, 1, 7), (1, 2, 4), (1, 2, 7), (1, 4, 7), (2, 2, 7), (2, 4, 7)

So for me its still 20- 7 =13
And the probability is 4/13