east how mary tines do you have to roll a fair die to be sure that the probability of rolling at least one 6 is greater than 7 in 10 Sip? number of rolls needed is $\square$. pe a whole number:)

Step 1 ：The problem is asking for the minimum number of times we need to roll a fair die such that the probability of getting at least one 6 is greater than 0.7.

Step 2 ：A fair die has 6 faces, so the probability of rolling a 6 in one roll is \( \frac{1}{6} \), and the probability of not rolling a 6 is \( \frac{5}{6} \).

Step 3 ：If we roll the die n times, the probability of not rolling a 6 at all is \( \left(\frac{5}{6}\right)^n \). Therefore, the probability of rolling at least one 6 is \( 1 - \left(\frac{5}{6}\right)^n \).

Step 4 ：We need to find the smallest integer n such that \( 1 - \left(\frac{5}{6}\right)^n > 0.7 \).

Step 5 ：We can solve this inequality by taking the natural logarithm of both sides, but since the inequality involves an exponent, it's easier to solve it by trial and error.

Step 6 ：By trying different values of n, we find that the smallest integer n that satisfies the inequality is 7.

Step 7 ：Final Answer: The minimum number of times we need to roll a fair die such that the probability of getting at least one 6 is greater than 0.7 is \( \boxed{7} \).