If a die is rolled 30 times, there are 630 different sequences possible. The following question asks how many of these sequences satisfy certain conditions. HINT [Use the decision algorithm discussed in Example 3 of Section 7.3.]
What fraction of these sequences have exactly three 6s and three 1s? (Round your answer to five decimal places.)

To determine the fraction of the sequences that have exactly three 6s and three 1s, let's first break down the problem.

### Total possible sequences
When a die is rolled 30 times, each roll has 6 possible outcomes. The total number of sequences is:

\[
6^{30}
\]

However, we are given that there are 630 different sequences possible under the specific conditions of the problem, so we will use this information later.

### Sequences with exactly three 6s and three 1s
We need to find how many of the 630 sequences contain exactly three 6s and three 1s. 

1. **Positions of 6s:** Out of 30 rolls, we can choose 3 positions for the number 6. The number of ways to choose 3 positions from 30 is given by the combination formula:

\[
\binom{30}{3} = \frac{30 \times 29 \times 28}{3 \times 2 \times 1} = 4060
\]

2. **Positions of 1s:** After assigning 3 positions for the 6s, we are left with 27 positions. From these remaining 27 positions, we need to choose 3 positions for the number 1. This can also be done using combinations:

\[
\binom{27}{3} = \frac{27 \times 26 \times 25}{3 \times 2 \times 1} = 2925
\]

3. **Remaining positions:** The remaining 24 positions can be filled with any of the other 4 outcomes (i.e., 2, 3, 4, or 5). Since each of these positions has 4 possibilities, the number of ways to fill these positions is:

\[
4^{24}
\]

### Total number of favorable sequences
Now, multiply the number of ways to choose the positions for the 6s, 1s, and fill the rest of the positions:

\[
\binom{30}{3} \times \binom{27}{3} \times 4^{24}
\]

### Fraction of sequences
To find the fraction of sequences that satisfy the condition, divide the number of favorable sequences by the total number of sequences. Since we're told there are 630 total sequences:

\[
\text{Fraction} = \frac{\binom{30}{3} \times \binom{27}{3} \times 4^{24}}{630}
\]

Let's calculate this value step by step.It seems like I can’t do more advanced data analysis right now. Please try again later. If you'd like, you can follow the steps outlined above to manually calculate the final fraction.

Let me know if you'd like more details or if you have any questions!

Here are five related questions to deepen your understanding:
1. How do combination formulas work, and why are they used in problems like this?
2. How would the answer change if we had exactly four 6s and two 1s instead?
3. What is the significance of using powers of 4 for the remaining positions?
4. How could you approach this problem with a different decision algorithm?
5. Can the binomial distribution be applied in solving this type of problem?

**Tip:** Always ensure that you're interpreting the problem constraints correctly, especially when deciding which parts of the problem to apply combinations or permutations.

If a die is rolled 30 times, there are 630 different sequences possible. The following question asks how many of these sequences satisfy certain conditions. What fraction of these sequences have exactly three 6s and three 1s? (Round your answer to five decimal places.)

Explore how to calculate the fraction of sequences with exactly three 6s and three 1s out of 630 total sequences in a 30-roll die sequence. This problem covers combinatorics, combinations, and probability, making it perfect for students in Grades 10-12.

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