To construct a random triangle with side sides 1 through 6 centimeters, we role a fair die three times. The number shown on the die indicates the size of a potential side in centimeters.
Thanks Kosto. I have a different answer. Could you show me how you obtained yours?
I have just count the cases which satisfy the triangle inequalities. I counted them by EXCEL.
Thanks again. I did the same rather than formulated. However, I also used the condition that no side's measure should exceed the sum of the other two. Based on this assumption, I had 120 choices. Perhaps, I have miscounted, do you agree?
By the way what formula did you use for Excel?
I also found 111/216, using the rule that each side must be less than the sum of other two.
Thanks very much Professor Kamberaj. Yes it is the same answer Professor Mitov submitted a few days ago. I aapreciate your time.
It seems this is a duplicated question. So I repeat here the answer I gave earlier in a similar thread.
There is actually no need to compute one by one each valid instance. The probability you are looking for is
P = 1/2 + 1/(2n²) where n is the number of faces of the die.
For a standard die (n = 6), we have P = 1/2 + 1/72 = 37/72.
Some details might be worth mentioning. let N be the total number of valid instances. We can rewrite N = E + 3I + 6S where
E = Total number of equilateral triangles
3I = Total number of Isosceles
6S = Total number of scalene
The factors 3 and 6 take into account permutations.
Of course, E = n but S and I will depend on the parity of n
If n is even then
I = (3n - 4)(n / 4)
S = n(2n - 5)(n - 2)/24
if n is odd then
I = (3n - 1)(n - 1)/4
S = (2n - 1)(n - 1)(n - 3)/24
The above expressions for S are valid only if n > 3. Otherwise take S = 0.
It turns out that 3I + 6S = n(n² - 1)/2 whether n is even or odd. Hence the final result
E + 3I + 6S = N = (n/2)(n² + 1)
and hence the probability P = N/(n^3) = 1/2 + 1/(2n²)
In summary, for a standard die (n = 6) we have
P = 37/72
P(Equilateral) = 1/36
P(isosceles) = 7/24
P(scalene) = 7/36
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