What is the probability of getting exactly 6 heads when flipping a fair coin 9 times?

To find the probability of getting exactly 6 heads when flipping a fair coin 9 times, we can use the binomial probability formula. The binomial probability formula is:

P(X = k) = C(n, k) * p^k * (1-p)^(n-k)

Where:

  • P(X = k) is the probability of getting exactly k successes in n trials.
  • C(n, k) is the binomial coefficient, calculated as C(n, k) = n! / (k!(n-k)!), which counts the number of ways to choose k successes from n trials.
  • p is the probability of success on a single trial (for a fair coin, this is 0.5 for heads).
  • n is the total number of trials (in this case, 9 flips).
  • k is the number of successful outcomes we are interested in (in this case, 6 heads).

For our problem, we have:

  • n = 9 (the total number of flips)
  • k = 6 (the number of heads we want)
  • p = 0.5 (the probability of getting heads on any given flip)

Now, we’ll calculate each part:

  • C(9, 6) = 9! / (6!(9-6)!) = 9! / (6!3!) = (9 × 8 × 7) / (3 × 2 × 1) = 84

Next, we calculate the probability:

  • P(X = 6) = C(9, 6) * (0.5)^6 * (0.5)^(9-6)
  • P(X = 6) = 84 * (0.5)^6 * (0.5)^3
  • P(X = 6) = 84 * (0.5)^9
  • P(X = 6) = 84 * (1/512)
  • P(X = 6) = 84 / 512 = 21 / 128

So, the probability of getting exactly 6 heads when flipping a fair coin 9 times is 21/128, which is approximately 0.1641 or 16.41%.

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