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Math Help - 3 coins

  1. #1
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    3 coins

    If 3 coins are thrown, the probability to have exactly two heads is:

    a) 3/8
    b) 2/6

    Explain.
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  2. #2
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    \begin{pmatrix}3\\2\end{pmatrix}p^{2}q

    Where p=q=.5

    You could try expanding (p+q)^{3} and looking at the binomial expansion.
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  3. #3
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    Is there an easier way to find out the answer, is it that hard?
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  4. #4
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    That isn't difficult.

    It may look imposing if you don't understand the notation.

    The number of ways to choose 2 out of 3 objects is 3.

    You have 3p^{2}q

    Enter in your probabilities of success and failure. If you flip a coin the probabilities are 50-50.

    So, you have 3(\frac{1}{2})^2(\frac{1}{2})=\frac{3}{8}

    Maybe I should've explained it this way to start with:

    Throw the 3 coins. How many possible arrangements can you have?.

    2^{3}=8. Right?.

    Let's list them:

    HTT
    HHH
    TTT
    HHT
    THH
    HTH
    THT
    TTH

    Now, how many of those have exactly 2 H's?.....3. See?.
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  5. #5
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    Hello, Bret!

    Here's yet another approach . . .


    If 3 coins are thrown, the probability to have exactly two heads is:

    . . . a)\;\frac{3}{8}\qquad b)\:\frac{2}{6}

    Explain.

    Suppose we want the coins to turn up H\!HT in that order.

    Prob. that coin #1 is Heads: \frac{1}{2}
    Prob. that coin #2 is Heads: \frac{1}{2}
    Prob. that coin #3 is Tails: \frac{1}{2}

    Hence: . P(HHT) \: =\:\left(\frac{1}{2}\right)^3\:=\:\frac{1}{8}


    Since the problem did not insist on a specific order of Heads and Tails,
    . . there are three ways to get two Heads: . HHT,\;HTH,\;THH

    Therefore: . P(\text{2H, 1T})\;=\;3 \times \frac{1}{8}\;= \; \boxed{\frac{3}{8}}

    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    A handy tip

    If there are 3 coins, how can we list all the outcomes?

    Since each coin can turn up in 2 ways, there are: 2^3 = 8 outcomes.


    Write 8\;H's\text{ and }T's . . . first half H's, the second half Tis.

    . . . \left \begin{array}{cccc}H\\H\\H\\H\end{array}\;\right\}
    . . . \left \begin{array}{cccc}T\\T\\T\\T\end{array}\;\right\}


    Write 8\;H's\text{ and }T's . . . this time "change every two".

    . . . \left \begin{array}{cc}H\\H\end{array}\;\right\}
    . . . \left \begin{array}{cc}T\\T\end{array}\;\right\}
    . . . \left \begin{array}{cc}H\\H\end{array}\;\right\}
    . . . \left \begin{array}{cc}T\\T\end{array}\;\right\}


    Write 8\;H's\text{ and }T's . . . this time "change every one" (alternate).

    . . . \begin{array}{cccc}\left H\;\;\right\} \\ \left T\;\;\right\} \\ \left H\;\;\right\} \\ \left T\;\;\right\}\end{array}
    . . . \begin{array}{cccc} \left H\;\;\right\} \\ \left T\;\;\right\} \\ \left H\;\;\right\} \\ \left T\;\;\right\}\end{array}


    Write the three columns side-by-side:

    . . . \begin{array}{cccccccc}1)&H&H&H\\2)&H&H&T\\3)&H&T&  H\\4)&H&T&T\\5)&T&H&H\\6)&T&H&T\\7)&T&T&H\\8)&T&T&  T\end{array}

    . . . and we have the eight possible outcomes!

    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    Of course, this procedure works for 4 coins (16 outcomes)
    . . and 5 coins (32 outcomes) . . . it just takes longer.

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