SOLUTION: If the sides of a triangle are decided by the throw of a single dice thrice, what is the probability that the triangle is of maximum area given that it is an isosceles triang

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Question 1041294: If the sides of a triangle are decided by the throw of
a single dice thrice, what is the probability that the
triangle is of maximum area given that it is an isosceles
triangle?

Answer by Edwin McCravy(20077) About Me  (Show Source):
You can put this solution on YOUR website!
The triangle has maximum area when all three dice are 6's.
That's the case of the three rolls being (6,6,6).  So 
there is only one way the triangle can have maximum area.
So the numerator of the probability will be 1.

Now we calculate the denominator of the probability.

Since we are given that the three dice rolls must result in an 
isosceles triangle, we must count the number of rolls that would 
result in such.  

In doing so, we must realize that rolls such as (2,2,6) do not 
form an isosceles triangle, even though two rolls are the same.  
To form a triangle, the sum of the shortest two of the sides must 
be greater than the longest of the three sides.  That is not true 
in the case of rolls (2,2,6) since 2+2 = 4 < 6.

We calculate the number of rolls with two or three rolls the same
which will produce an isosceles triangle.

Case 1:  the three rolls are equal resulting in an equilateral 
triangle.

There are obviously 6 of these: (1,1,1),(2,2,2),...,(6,6,6)

Case 2: Two of the sides are equal, and the third side not equal,
resulting in a non-equilateral isosceles triangle.

First we find what three rolls result in a non-equilateral isosceles
triangle, then we'll order them as to which of the three roll 
numbers 1st, 2nd, and 3rd they were obtained in.

Sub-case 1: the two equal rolls are 1's.
There is no way to choose an unequal roll that will
result in a triangle. [the case (1,1,1) is counted in case 1]
That's 0 ways.

Sub-case 2: the two equal rolls are 2's.
The unequal roll can only be 1 or 3 for it to
result in a non-equilateral isosceles triangle.
That's 2 ways.
 
Sub-case 3: the two equal rolls are 3's.
The unequal roll can only be 1,2,4, or 5 for it to
result in a non-equilateral isosceles triangle.
That's 4 ways.

Sub-case 4: the two equal rolls are 4's.
The unequal roll can only be 1,2,3,5, or 6 for it to
result in a non-equilateral isosceles triangle.
That's 5 ways.

Sub-case 5: the two equal rolls are 5's.
The unequal roll can only be 1,2,3,4, or 6 for it to
result in a non-equilateral isosceles triangle.
That's 5 ways.

Sub-case 6: the two equal rolls are 5's.
The unequal roll can only be 1,2,3,4 or 5 for it to
result in a non-equilateral isosceles triangle.
That's 5 ways.

So for case 2 there are 0+2+4+5+5+5 = 21 ways to choose
the 3 rolls, unordered.

As we mentioned earlier, there are three "roll numbers", 1st roll, 
2nd roll and 3rd roll.

We must consider these orders for case 2 where there is one 
unequal roll.

We can choose the roll number for the unequal roll in 3 ways,
and the equal pair of rolls will automatically be in the remaining
two roll numbers.
  
So there are 3 ways to assign roll numbers to each of the 21
cases of 2 equal rolls and 1 unequal roll that will result is
a non-equilateral isosceles triangle.

That's 21*3 = 63 ways to choose the three rolls, where
exactly two are equal.

So for cases 1 and 2 there are 6+63 = 69 rolls that will result
in isosceles triangles.

So the desired conditional probability is 1/69.

Edwin