### Instruction: We flip a fair coin 10 times.  What is the probability that we get heads in at least 6 of the 10 flips?### Response: To solve this problem, I need to use the binomial distribution formula, which tells me the probability of getting exactly k successes (heads) in n trials (flips) with a fixed probability of success p (0.5 for a fair coin).
The formula is P(k) = nCk * p^k * (1-p)^(n-k), where nCk is the number of ways to choose k objects from n objects, also known as the binomial coefficient.
I want to find the probability of getting at least 6 heads, which means I need to add up the probabilities of getting 6, 7, 8, 9, or 10 heads.
I can use a calculator or a table to find the binomial coefficients, or I can use the formula nCk = n! / (k! * (n-k)!), where n! is the factorial of n, which is the product of all positive integers up to n.
Using either method, I find that 10C6 = 210, 10C7 = 120, 10C8 = 45, 10C9 = 10, and 10C10 = 1.
Plugging these values into the binomial distribution formula, I get:
P(6) = 210 * 0.5^6 * 0.5^4 = 0.2051
P(7) = 120 * 0.5^7 * 0.5^3 = 0.1172
P(8) = 45 * 0.5^8 * 0.5^2 = 0.0439
P(9) = 10 * 0.5^9 * 0.5^1 = 0.0098
P(10) = 1 * 0.5^10 * 0.5^0 = 0.00098
Adding these probabilities together, I get: