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Math Help - One more question.

  1. #1
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    One more question.

    Just looking over some more abstract questions and foundthis..

    Four fair coins are tossed. Find the Probability of obtaining 2 heads and two tails

    I found the answer already. its 4C2(1/2)^2 x (1/2)^2 = 3/8
    But im wondering if there is a different way to get this answer
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by 3deltat View Post
    Just looking over some more abstract questions and foundthis..

    Four fair coins are tossed. Find the Probability of obtaining 2 heads and two tails

    I found the answer already. its 4C2(1/2)^2 x (1/2)^2 = 3/8
    But im wondering if there is a different way to get this answer
    this is the easiest way to do it i think. the other way i can think of is to do all the combinations manually, that is,

    P(getting heads with first coin, tails with second, heads with third, tails with fourth) + P(heads with first coin, heads with second, tails with third, tails with fourth) + .....

    and try to make sure you get all combinations of two heads and two tails, which would be a pain, Bernoulli Trials is definatley the way to go here, which is what you did.

    Maybe Soroban can come up with some other way, he's somewhat of a genius with probability
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  3. #3
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    Quote Originally Posted by 3deltat View Post
    Four fair coins are tossed. Find the Probability of obtaining 2 heads and two tails: its 4C2(1/2)^2 x (1/2)^2 = 3/8
    But im wondering if there is a different way to get this answer
    There are several ways to model this. But all of are equivalent to what you have done.

    Basically, there are 2^4 outcomes. The number of ways to arrange the string “TTHH” is {4 \choose 2} .
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  4. #4
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    Alright thanks.
    I asked this question because i found myself trying to answer this question a different way than what i ended up doing, but I agree that this proly would be the easiest way
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