11. What is the probability, if I flip a fair coin with heads and tails ten times in a row, that I get at least 8 heads? .0547 .1131 0.4395 .00977

Step 1 ：Define the problem: What is the probability, if I flip a fair coin with heads and tails ten times in a row, that I get at least 8 heads?

Step 2 ：Define the binomial probability function: \(P(X=k) = C(n, k) \cdot p^k \cdot (1-p)^{n-k}\), where \(C(n, k)\) is the number of combinations of \(n\) items taken \(k\) at a time, \(p\) is the probability of success, and \(k\) is the number of successes.

Step 3 ：Set the probability of getting a head, \(p_{head}\), to 0.5.

Step 4 ：Set the number of flips, \(n_{flips}\), to 10.

Step 5 ：Calculate the probability of getting 8 heads, \(prob_8\), using the binomial probability function: \(prob_8 = 0.0439453125\).

Step 6 ：Calculate the probability of getting 9 heads, \(prob_9\), using the binomial probability function: \(prob_9 = 0.009765625\).

Step 7 ：Calculate the probability of getting 10 heads, \(prob_10\), using the binomial probability function: \(prob_10 = 0.0009765625\).

Step 8 ：Add the probabilities of getting 8, 9, and 10 heads to get the total probability: \(total_{prob} = prob_8 + prob_9 + prob_10 = 0.0546875\).

Step 9 ：Final Answer: The probability of getting at least 8 heads in 10 flips of a fair coin is \(\boxed{0.0547}\).