Refer to the table below. Of the 36 possible outcomes, determine the number for which the sum (for both dice) is composite.

\begin{tabular}{|c|c|c|c|c|c|c|}
\hline & \multicolumn{6}{|c|}{ Die 2 } \\
\hline Die 1 & [tex]$1$[/tex] & [tex]$2$[/tex] & 3 & 4 & [tex]$5$[/tex] & 6 \\
\hline [tex]$1$[/tex] & [tex]$(1,1)$[/tex] & [tex]$(1,2)$[/tex] & [tex]$(1,3)$[/tex] & [tex]$(1,4)$[/tex] & [tex]$(1,5)$[/tex] & [tex]$(1,6)$[/tex] \\
\hline [tex]$2$[/tex] & [tex]$(2,1)$[/tex] & [tex]$(2,2)$[/tex] & [tex]$(2,3)$[/tex] & [tex]$(2,4)$[/tex] & [tex]$(2,5)$[/tex] & [tex]$(2,6)$[/tex] \\
\hline [tex]$3$[/tex] & [tex]$(3,1)$[/tex] & [tex]$(3,2)$[/tex] & [tex]$(3,3)$[/tex] & [tex]$(3,4)$[/tex] & [tex]$(3,5)$[/tex] & [tex]$(3,6)$[/tex] \\
\hline [tex]$4$[/tex] & [tex]$(4,1)$[/tex] & [tex]$(4,2)$[/tex] & [tex]$(4,3)$[/tex] & [tex]$(4,4)$[/tex] & [tex]$(4,5)$[/tex] & [tex]$(4,6)$[/tex] \\
\hline [tex]$5$[/tex] & [tex]$(5,1)$[/tex] & [tex]$(5,2)$[/tex] & [tex]$(5,3)$[/tex] & [tex]$(5,4)$[/tex] & [tex]$(5,5)$[/tex] & [tex]$(5,6)$[/tex] \\
\hline [tex]$6$[/tex] & [tex]$(6,1)$[/tex] & [tex]$(6,2)$[/tex] & [tex]$(6,3)$[/tex] & [tex]$(6,4)$[/tex] & [tex]$(6,5)$[/tex] & [tex]$(6,6)$[/tex] \\
\hline
\end{tabular}

There are [tex]$\square$[/tex] outcomes where the sum is composite.



Answer :

To determine the number of outcomes for which the sum of the rolls of two dice is composite, we need to first understand the sums that can be obtained and which of these sums are composite.

First, let's list all possible outcomes for rolling two dice, which range from 2 (1+1) to 12 (6+6):

- Sum of 2: [tex]\( (1,1) \)[/tex]
- Sum of 3: [tex]\( (1,2), (2,1) \)[/tex]
- Sum of 4: [tex]\( (1,3), (2,2), (3,1) \)[/tex]
- Sum of 5: [tex]\( (1,4), (2,3), (3,2), (4,1) \)[/tex]
- Sum of 6: [tex]\( (1,5), (2,4), (3,3), (4,2), (5,1) \)[/tex]
- Sum of 7: [tex]\( (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) \)[/tex]
- Sum of 8: [tex]\( (2,6), (3,5), (4,4), (5,3), (6,2) \)[/tex]
- Sum of 9: [tex]\( (3,6), (4,5), (5,4), (6,3) \)[/tex]
- Sum of 10: [tex]\( (4,6), (5,5), (6,4) \)[/tex]
- Sum of 11: [tex]\( (5,6), (6,5) \)[/tex]
- Sum of 12: [tex]\( (6,6) \)[/tex]

Next, let's identify which sums are composite numbers. A composite number has more than two distinct positive divisors. The composite sums between 2 and 12 are:

- 4 (divisors: 1, 2, 4)
- 6 (divisors: 1, 2, 3, 6)
- 8 (divisors: 1, 2, 4, 8)
- 9 (divisors: 1, 3, 9)
- 10 (divisors: 1, 2, 5, 10)
- 12 (divisors: 1, 2, 3, 4, 6, 12)

Now, let's count how many combinations result in each composite sum:

- Sum of 4: [tex]\( (1,3), (2,2), (3,1) \)[/tex] — 3 outcomes
- Sum of 6: [tex]\( (1,5), (2,4), (3,3), (4,2), (5,1) \)[/tex] — 5 outcomes
- Sum of 8: [tex]\( (2,6), (3,5), (4,4), (5,3), (6,2) \)[/tex] — 5 outcomes
- Sum of 9: [tex]\( (3,6), (4,5), (5,4), (6,3) \)[/tex] — 4 outcomes
- Sum of 10: [tex]\( (4,6), (5,5), (6,4) \)[/tex] — 3 outcomes
- Sum of 12: [tex]\( (6,6) \)[/tex] — 1 outcome

Adding these together:

[tex]\(3 + 5 + 5 + 4 + 3 + 1 = 21\)[/tex]

Thus, there are [tex]\(21\)[/tex] outcomes where the sum is composite. This aligns with the number of outcomes where the sum of the dice rolls results in a composite number, as given. Therefore, the answer is:

There are [tex]\(\boxed{21}\)[/tex] outcomes where the sum is composite.