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If a fair die is rolled 7 times, what is the probability, rounded to the nearest thousandth, of getting at least 6 threes?
Answer Attempt 1 out of 2
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Step 1 ：Calculate the probability of rolling a three on a fair die: $p_{three}=\frac{1}{6}$

Step 2 ：Calculate the probability of getting exactly 6 threes in 7 rolls using the binomial probability formula: $pro{b}_{6\mathrm{\_}threes}=(\genfrac{}{}{0ex}{}{7}{6})\times {\left(\frac{1}{6}\right)}^6\times {\left(\frac{5}{6}\right)}^1$

Step 3 ：Calculate the probability of getting exactly 7 threes in 7 rolls using the binomial probability formula: $pro{b}_{7\mathrm{\_}threes}=(\genfrac{}{}{0ex}{}{7}{7})\times {\left(\frac{1}{6}\right)}^7\times {\left(\frac{5}{6}\right)}^0$

Step 4 ：Sum the probabilities of getting exactly 6 threes and exactly 7 threes to find the total probability: $total\mathrm{\_}prob=pro{b}_{6\mathrm{\_}threes}+pro{b}_{7\mathrm{\_}threes}$

Step 5 ：Round the total probability to the nearest thousandth: $rounded\mathrm{\_}total\mathrm{\_}prob=0.000$

Step 6 ：Final Answer: $\boxed{{\displaystyle 0.000}}$