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.
The nth term for the given sequence can be written as:
Step-by-step explanation:
Notice that this arithmetic sequence is created by adding to each term a common difference of 6 units:
7 + 6 = 13
13 + 6 = 19
19 + 6 = 25
Then, using the general expression for the nth term of an arithmetic sequence of first term "" and common difference "<em>d</em>" we can write the nth term as: