So we need to know the likelihood for each sum. the first sum is 2, and there is no way for one of the die to equal 6 if the sum is 2, therefore the probability is 0. The same applies when the sum is 3, 4, 5, and 6. Once the sum gets to 7, you must evaluate all possible options. For 7, your options are 1&6, 2&5, 3&4, 4&3, 5&2, 6&1, where the number before the ampersand is the first die, and the number after is the second. there is only one option of the 6 choices where the first die is 6, therefore the probability is 1/6. For 8, the options are 2&6, 3&5, 4&4, 5&3, 6&2. so of the 5 choices, there is only one option, therefore the probability is 1/5. For 9, the choices are 3&6, 4&5, 5&4, 6&3. So of the 4 choices, there is 1 option, therefore the probability is 1/4. For 10, the options are 4&6, 5&5, 6&4. Of the 3 choices, there is 1 option, therefore the probability is 1/3. For 11, the options are 5&6, 6&5. Of the 2 choices, there is 1 option, therefore the probability is 1/2. Finally, for 12, the only option is 6&6. There is only one choice, so the probability is 1.
It is given that the Points B, D, and F are midpoints of the sides of ΔACE. EC = 38 and DF = 16.
The midpoint theorem states that the if a line segments connecting two midpoints then the line is parallel to the third side and it's length is half of the third side.
Since F and D are midpoints of AE and EC respectively.
So by midpoint theorem length of AC is twice of DF.
