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If 12 fair coins are flipped once, what is the probability of a result as extreme as or more extreme than 10 heads? (Note, "as or more extreme than" means in *either* direction -- that is to say, as or more extreme both in terms of getting that many or more heads, and in terms of getting that many or more tails.)

Sagot :

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Using the binomial probability concept ; the probability of obtaining 10 or more heads is 0.0457

Recall :

P(x = x) = nCx * p^x * q^(n-x)

Where :

  • n = number of trials = 12
  • x ≥ 10
  • p = probability of success = 0.5
  • q = 1 - q = 0.5

P(x ≥ 10) = P(x = 10) + P(x = 11) + P(x = 12)

Using a binomial probability calculator :

P(x = 10) =

P(x = 11) =

P(x = 12) =

P(x ≥ 10) = 0.01611 + 0.02930 + 0.000244

P(x ≥ 10) = 0.045654

Therefore, probability of getting as extreme or more extreme than 10 heads is 0.0457.

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