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by Anonymous on April 26th, 2008

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Suppose you have a loaded coin that lands with probability 2/3 on each flip. Suppose you flip the coin 8 times. What is the probability that in those 8 flips you get an even number of heads? 1

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Answers. 2 helpful answers below.

  • by Quirkie on April 27th, 2008

    Quirkie

    if (2/3) is the probability for heads then:
    P(i heads) = nCi (1/3)^(n-i) (2/3)^i
    So put n=8
    and add the results for P(0),P(2),P(4),P(6),P(8)
    There is formula for that, but it's not worth working out.

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  • by EL1 2 on April 26th, 2008

    EL1 2

    Let P(x) be the probability of getting x heads.

    Then it's P(0) + P(2) + P(4) + P(6) + P(8).

    Since it's a binomial distribution, we know P(x) = n! / (k! * (n-k)!) * p^k * (1-p)^(n-k)

    Plug things in:

    P(0) = 8! / (0! * 8!) * (2/3)^0 * (1/3)^8 = 1 * 1 * (1/3)^8 = 1 / 6561

    I think you can do the rest.

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  • Suppose you have a loaded coin that lands with probability 2/3 on each flip. Suppose you flip the coin 8 times. What is the probability that in those 8 flips you get an even number of heads?

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