+486 We flip a fair coin 10 times. What is the probability that we get heads in at least 6 of the 10 flips?
+486 Please show all your work for these answers! I will not be using them unless they have work.
Since the problem says "at least", flipping 6, 7, 8, 9, or 10 heads would work. Therefore we have 5 cases.
Case 1 - 6 Heads: We need 6 heads and the other 4 flips have to be tails. Then, we can rearrange the order of the flips (HHHHHHTTTT is different from HHHTTTTHHH). Keeping this in mind, we get: \((\frac{1}{2})^{6}\cdot(\frac{1}{2})^{4}\cdot\binom{10}{6}=\frac{210}{1024}\).
Case 2 - 7 Heads: We can use the same approach as case 1. \((\frac{1}{2})^{7}\cdot(\frac{1}{2})^{3}\cdot\binom{10}{7}=\frac{120}{1024}\).
Case 3 - 8 Heads: \((\frac{1}{2})^{8}\cdot(\frac{1}{2})^{2}\cdot\binom{10}{8}=\frac{45}{1024}\).
Case 4 - 9 Heads: \((\frac{1}{2})^{9}\cdot(\frac{1}{2})^{1}\cdot\binom{10}{9}=\frac{10}{1024}\).
Case 5 - 10 Heads: There is obviously only 1 way for this to happen. So, the probability of this happening is \(\frac{1}{1024}\).
Summing the probabilities together, we get: \(\frac{210+120+45+10+1}{1024}=\frac{386}{1024}=\frac{193}{512}\).
