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Conditional probability - Random number of dice

  1. Aug 25, 2005 #1
    Can someone help me with this question ?

    A random number N of dice is thrown. Let Ai be the event that N = i, and assume that P(Ai) = 1/(2)^i, i >= 1. The sum of the scores is S. Find the probability that:

    (a) S = 4, given N is even;
    (b) the largest number shown by any die is r, where S is unknown.
     
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  3. Aug 26, 2005 #2

    EnumaElish

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    (a) [tex]P(S=4 \text{ and N is even})/P(\text{N is even})=[/tex]
    [tex]P\left(\left\{N=2 \text{ and }(S=1+3 \text{ or }S=2+2 \text{ or }S=3+1)\right\}\text{ OR }\{N=4\text{ and }S=1+1+1+1\}\right)\left/ \sum_{k=1}^\infty 2^{-2k}\right.[/tex]

    If the world is just, then the denominator should add up to 1/2. Because P(odd) = P(even) and P(odd) + P(even) = 1, so P(odd) = P(even) = 1/2.

    So calculating P(S=4 and N is even)/P(N is even) is a matter of calculating the numerator and then multiplying it with 2.
     
  4. Aug 26, 2005 #3

    NateTG

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    For part b:

    The chance that [itex]r[/itex] is the largest value shown is going to be
    [tex]\sum_{i=1}^{\infty} P(A_i) \times p(r,i)[/tex]
    where
    [tex]p(r,i)[/itex]
    is the probability that [itex]r[/itex] is the largest value shown by [itex]i[/itex] dice.

    Now, the chance of all of the dice being less than or equal to [itex]r[/itex], assuming that [itex]r\in{1,2,3,4,5,6}[/itex] is going to be
    [tex]\left(\frac{r}{6}\right)^{i}[/itex]
    and of those
    [tex]\left(\frac{r-1}{6}\right)^{i}[/itex]
    don't contain any [itex]r[/itex]
    So we have
    [tex]\sum_{i=1}^{\infty} \frac{r^i-(r-1)^i}{12^i}=\sum_{i=1}^{\infty}\left(\frac{r}{12}\right)^i - \sum_{i=1}^{\infty}\left(\frac{r-1}{12}\right)^i=\frac{r}{12-r}-\frac{r-1}{13-r}=\frac{13r-r^2-12r+r^2+12-r}{156-25r+r^2}=\frac{12}{r^2-25r+156}[/tex]
    So
    [tex]r=6 \rightarrow \frac{12}{156+36-150}= \frac{12}{42} = \frac{2}{7}[/tex]
    [tex]r=5 \rightarrow \frac{12}{156+25-125}= \frac{12}{56} = \frac{3}{14}[/tex]
    [tex]r=4 \rightarrow \frac{12}{156+16-100}= \frac{12}{72} = \frac{1}{6}[/tex]
    [tex]r=3 \rightarrow \frac{12}{156+9-75}=\frac{12}{90}= \frac{2}{15}[/tex]
    [tex]r=2 \rightarrow \frac{12}{156+4-50}=\frac{12}{110}=\frac{6}{55}[/tex]
    [tex]r=1 \rightarrow \frac{12}{156+1-25}=\frac{12}{132}=\frac{1}{11}[/tex]
     
  5. Aug 28, 2005 #4
    The question of the part (a) is P(S = 4 | N is even) and not P(S = 4 and N is even | N is even).
     
  6. Aug 28, 2005 #5

    Galileo

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    That's what EnumaElish did. By the definition of conditional probability:

    [tex]P(A|C)=\frac{P(A \cap C)}{P(C)}[/tex]

    Which is, by the way, trivially equal to [itex]P(A\cap C|C)[/itex].
     
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