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Math Help - Expected Value

  1. #1
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    Expected Value

    Hey Everyone

    I'm a little stuck on a problem and I know its most likely pretty easy.

    I need to find the expected value of the number of heads after flipping a fair coin three times.

    I know the probability for heads or tails is 1/2. I know the probability for each outcome of the three flips is 1/8. Just not really sure where to go from there.
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  2. #2
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    Dear BlakeRobertsonMD,

    definition:

    E = p1*v1 + p2*v2 + ... + pn*vn, where
    n means the amount of events,
    p2 means the probality of the second event,
    v2 means the value of the second event.

    Now n = 4, because "the number of heads" can be 0, 1, 2 or 3.
    So v1=0, v2=1, v3=2, v4=3.
    p1 = p(nothing heads) = 1/8.
    p4 = p(all heads) = 1/8.
    p2 = p(exactly one head) = p(the first head others tails) + p(the second head others tails) + p(the 3th head others tails) = 1/8 + 1/8 + 1/8 = 3/8.

    I leave for you p3 and E.
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  3. #3
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    E(X)=np=3\left(\frac{1}{2}\right)=1.5
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by Sean12345 View Post
    E(X)=np=3\left(\frac{1}{2}\right)=1.5
    A bald statement like this is not very informative, explain why.

    CB
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  5. #5
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    Quote Originally Posted by CaptainBlack View Post
    A bald statement like this is not very informative, explain why.

    CB
    All I know is that for a discrete variable such as this the expected value is the product of the number of trials and the probability of X happening on each trial. Since in this case the probability of X(Heads being shown) happening on each flip of the coin is the same (1/2) then the expected value is np where n is the number of trials (3) and p is the probability of a heads being shown (1/2) .
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  6. #6
    Grand Panjandrum
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    Quote Originally Posted by Sean12345 View Post
    All I know is that for a discrete variable such as this the expected value is the product of the number of trials and the probability of X happening on each trial. Since in this case the probability of X(Heads being shown) happening on each flip of the coin is the same (1/2) then the expected value is np where n is the number of trials (3) and p is the probability of a heads being shown (1/2) .
    Which in fact explains nothing, in a question like this we should relate the definition of the expected value to the data. So we sould either use the defintion that:

    E(x)=\sum ip(i)

    or that the number of heads is the sum of three RV's each of which has expectation 1/2 (which using the definition of expected value is: \sum_{i=0}^1 i(1/2)=1/2 ) and as expectation is additive the expected value of the sum is three times the individual expected values.

    The bald statement in your post has no pedagogical (teaching) value so needs amplification.

    CB
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