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Math Help - If a coin is tossed four times...(PLEASE HELP ASAP!!!)

  1. #1
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    If a coin is tossed four times...(PLEASE HELP ASAP!!!)

    If a coin is tossed four times, what is the probability that heads will come up twice?

    I cant quite seem to get the right answer. anything will help
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    Behold, the power of SARDINES!
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    Quote Originally Posted by rkaminky View Post
    If a coin is tossed four times, what is the probability that heads will come up twice?

    I cant quite seem to get the right answer. anything will help
    This is a binomial probability.

    The prob of getting exactly two heads is

    \left( \frac{1}{2}\right)^2 \left( \frac{1}{2}\right)^2

    but this \binom{4}{2}=6 ways

    so we get

    6 \cdot \left( \frac{1}{2}\right)^4=\frac{6}{16}=\frac{3}{8}
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    Behold, the power of SARDINES!
    TheEmptySet's Avatar
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    The above post if for exactly two heads. If you want 2 or more you need to add the prob of 3 heads and prob of 4 heads.

    Good luck.
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    Quote Originally Posted by rkaminky View Post
    If a coin is tossed four times, what is the probability that heads will come up twice?

    I cant quite seem to get the right answer. anything will help
    Many approaches are possible. Are you familiar with the binomial distribution?

    Alternatively, consider all the different arrangements of HHTT. There are six. The probability of each is (1/2)^4 (assuming a fair coin). Therefore .....
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    Quote Originally Posted by rkaminky View Post
    If a coin is tossed four times, what is the probability that heads will come up twice?

    I cant quite seem to get the right answer. anything will help
    \Omega = HHHH ; HHHT ; HHTH ; HTHH ; HHTT ; HTTT ; HTTH ; HTHT ; TTTT ; TTTH ; TTHT ; THTT ; TTHH ; THHH ; THHT ; THTH

    Let A be the event that heads come up twice.

    A = HHTT ; HTTH ; HTHT ; TTHH ; THHT ; THTH

    P(A) = \frac{6}{16} = \frac{3}{8}
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    Bar0n janvdl's Avatar
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    Quote Originally Posted by mr fantastic View Post
    Many approaches are possible. Are you familiar with the binomial distribution?
    Let us try Mr F's suggestion of binomial distribution.

    P(T) = {4 \choose 2} \left( \frac{1}{2} \right) ^{2} \left( \frac{1}{2} \right) ^{2} = \frac{6}{16} = \frac{3}{8}
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