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.
So x=earned x>200 they paid their parrents 28 for supplies (this assumes that they paied 28 altogether and not 28 each) so subtract 28 from pay new pay=y y>200-28 y>172 divided it among themselves (3) 172/3=57 and 1/3=57.33 they each earned more that $57.33
