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Math Help - probability distribution for 6 fair coins

  1. #1
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    probability distribution for 6 fair coins

    In a certain experiment, 6 fair coins are flipped. Let the random variable x be the number of heads observed. Construct the probability distribution for the experiment.

    The only way i can think to do this may take FOREVER.

    Here is my method

    1-1-1-1-1-1, 1-2-1-1-1-1, 1-1-2-1-1-1, 1-1-1-2-1-1, 1-1-1-1-2-1, 1-1-1-1-1-2, 1-2-2-1-1-1, 1-1-2-2-1-1, ... 6-6-6-6-6-6

    Etc!!!

    Is there a more efficient method to create this probability distribuition than listing all (who knows how many) possiblities out?
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  2. #2
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    Oops sorry i was thinking for some reason that i was dealing with 6 dice.

    Listing all possible things for 6 coins consisting of only heads and tails shouldnt be too bad.

    My mistake.

    Here is my new question: How many possible outcomes are there?
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  3. #3
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    Quote Originally Posted by solace View Post
    Listing all possible things for 6 coins consisting of only heads and tails shouldnt be too bad.
    Here is my new question: How many possible outcomes are there?
    2^6=64
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  4. #4
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    Hello, solace!

    In a certain experiment, 6 fair coins are flipped.
    Let the random variable x be the number of heads observed.
    Construct the probability distribution for the experiment.
    They don't expect you to list all the possible outcomes and count them up.
    . . They probably think that anyone can do that ... given enough time.
    Try some Thinking.

    You're expected to know that are: 2^6 = 64 possible outcomes.

    How many ways are there to get 0 heads?
    . . There is one way: TTTTTT

    How many ways are there to get 1 head (and five tails)?
    . . There are six ways . . . why?
    Because the one head can appear in any of six positions.

    How many ways are there to get 2 heads (and four tails)?
    . . There are C(6,2) = 15 ways. .(Do you see why?)

    How many ways are there to get 3 heads (and three tails)?
    . . There are C(6,3) = 20 ways.


    By now, these values may (or may not) seem familiar to you.
    They are the cofficients for the expansion of (a + b)^6.
    Pascal's Triangle gives us: .1, 6, 15, 20, 15, 6, 1


    Therefore, if x = number of heads, we have:
    . . . P(0)
    = 1/64
    . . . P(1)
    = 6/64
    . . . P(2)
    = 15/64
    . . . P(3)
    = 20/64
    . . . P(4)
    = 15/64
    . . . P(5)
    = 6/64
    . . . P(6)
    = 1/64

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