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Math Help - Probabilities

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    Probabilities

    Roll 2 dice and sum the spots. Then flip a coin that number of times. Find the exact probability of rolling a total of two if the experiment produced exactly as many heads as tails.

    I recognize that it's a conditional probability but I don't know how to find the probability of getting exactly as many heads as tails.
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    Quote Originally Posted by fulltwist8 View Post
    Roll 2 dice and sum the spots. Then flip a coin that number of times. Find the exact probability of rolling a total of two if the experiment produced exactly as many heads as tails.

    I recognize that it's a conditional probability but I don't know how to find the probability of getting exactly as many heads as tails.
    Probability of heads is the same as tails (50%).
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    but the probability of getting exactly the same number of heads and tails is not .5

    That's what I'm having trouble finding
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    Quote Originally Posted by fulltwist8 View Post
    but the probability of getting exactly the same number of heads and tails is not .5
    Exactly!
    In order to have an equal number of head & tails, E, you must have rolled an even sum: 2 4 6 8 10 or 12.
    Here is an example. The probability of E and a sum of 4 is
    P(E4) = P(E|4)P(4) = \left( {\frac{\binom {4}{2}}{{2^4 }}} \right)\left( {\frac{3}{{36}}} \right).

    So you must work out the following:
    P(E) = P(E2) + P(E4) + P(E6) + P(E8) + P(E10) + P(E12).
    That is your denominator in the answer to this question which is:
    P(2|E) = \frac{{P(E2)}}{{P(E)}} .
    Last edited by Plato; February 28th 2008 at 10:05 AM.
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    Hello, fulltwist8!

    This will take a while . . .


    Roll two dice and sum the spots.
    Then flip a coin that number of times.
    Find the probability of rolling a total of 2
    if there were exactly as many heads as tails.

    We want:. P[\text{(sum of 2) }|\text{ (H = T)}] \;=\; \frac{P[\text{(sum of 2)} \wedge \text{(H = T)}]} {P[\text{H = T}] }


    The numerator is: . P[\text{(sum of 2) } \wedge \text{ (H = T)}]

    . . P[\text{sum of 2}] \:=\:\frac{1}{36}
    . . With two tosses, P[\text{H = T}] \:=\:\frac{2}{4} \:=\:\frac{1}{2}
    . . . . Hence: . P[\text{(sum of 2) }\wedge\text{ (H = T)}] \;=\;\frac{1}{36}\cdot\frac{1}{2} \:=\:\frac{1}{72}


    The denominator is: . P[\text{H = T}]
    We must consider all twelve oucomes of the dice.

    \begin{array}{ccccccccc}<br />
P[\text{sum of 2}] &=& \frac{1}{36} & & P[\text{H = T}] & = & \frac{1}{2} \\<br />
P[\text{sum of 3}] &=& \frac{2}{36} & & P[\text{H = T}] &=& 0 \\ <br />
P[\text{sum of 4}] &=&\frac{3}{36} & & P[\text{H = T}] &=&{4\choose2}\left(\frac{1}{2}\right)^4 & = & \frac{3}{8} \\<br />
P[\text{sum of 5}] &=&\frac{4}{36} & & P[\text{H = T}] &=& 0 \end{array}
    \begin{array}{ccccccccc}P[\text{sum of 6}] &=&\frac{5}{36} & & P[\text{H = T}] &=&{6\choose3}\left(\frac{1}{2}\right)^6 &=& \frac{5}{16} \\<br />
P[\text{sum of 7}] &=&\frac{6}{36} & & P[\text{H = T}] &=& 0 \\<br />
P[\text{sum of 8}] &=&\frac{5}{36} & & P[\text{H = T}] &=& {8\choose4}\left(\frac{1}{2}\right)^8 &=& \frac{35}{128} \\<br />
P[\text{sum of 9}] &=&\frac{4}{36} & & P[\text{H = T}] &=& 0 \end{array}
    \begin{array}{ccccccccc}P[\text{sum of 10}] &=& \frac{3}{36} & & P[\text{H = T}] &=&{10\choose5}\left(\frac{1}{2}\right)^{10} &=& \frac{63}{256} \\<br />
P[\text{sum of 11}] &=&\frac{2}{36} & & P[\text{H = T}] &=&0 \\<br />
P[\text{sum of 12}] &=&\frac{1}{36} & & P[\text{H = T}] &=&{12\choose6}\left(\frac{1}{2}\right)^{12} &=& {\color{blue}\frac{231}{512}}\end{array}

    Hence: . P[\text{H = T}] \;=\;\frac{1}{36}\!\cdot\!\frac{1}{2} + \frac{3}{36}\!\cdot\!\frac{3}{8} + \frac{5}{36}\!\cdot\!\frac{5}{16} + \frac{5}{36}\!\cdot\!\frac{35}{128} + \frac{3}{36}\!\cdot\!\frac{63}{256} + \frac{1}{36}\!\cdot\!\frac{231}{512} \;=\;\frac{2941}{18432}


    Therefore:. P[\text{(sum of 2) }|\text{ (H = T)}] \;=\; \frac{P[\text{(sum of 2)} \wedge \text{(H = T)}]} {P[\text{H = T}] } \;=\;\frac{\frac{1}{72}}{\frac{2941}{18432}} \;=\;\boxed{\frac{256}{2941}}


    I hope my arithmetic is correct . . . this time
    Thanks for the heads-up, Plato!
    .
    Last edited by Soroban; February 28th 2008 at 01:02 PM.
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    Quote Originally Posted by Soroban View Post
    \begin{array}{ccccccccc}P[\text{sum of 10}] &=& \frac{3}{36} & & P[\text{H = T}] &=&{10\choose5}\left(\frac{1}{2}\right)^{10} &=& \frac{63}{256} \\<br />
P[\text{sum of 11}] &=&\frac{2}{36} & & P[\text{H = T}] &=&0 \\<br />
P[\text{sum of 12}] &=&\frac{1}{36} & & P[\text{H = T}] &=&{12\choose6}\left(\frac{1}{2}\right)^{12} &=& \color{red}\frac{231}{1024}\end{array}
    I hope my arithmetic is correct . . ..
    One very minor arithmetic error.
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