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Thread: Joint marginal conditional probabilities HELP please ;)

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    Exclamation Joint marginal conditional probabilities HELP please ;)




    An organization was surveyed with regard to the number of children
    each member had. The gender of the parent was also considered. The
    results are summarized here:
    1).Males with one child made up 0.15 of the membership.
    2). Females with 2 children also made up 0.15 of the membership.
    3). For those with 3 or more children, the probability that the
    parent was female was 0.5 (I.e., of the members who had
    3 or more children, half of those members were female.)
    4).Males with no children made up 0.05 of the membership.
    5). Males with 3 or more children made up 0.1 of the membership.
    6). For those members who were male, the probability that they
    had no children was 0.125
    7). when all members were considered, the probability of having
    no children was 0.15
    Question:
    a)What is the probability of parent being male, given that
    the member (parent) had 2 or more children?
    b) What is the probability of parent being female, given that
    the member (parent) had fewer than 2 children?


    I draw a table like this:
    0 1 2 3
    male 0,05 0.15 - 0.1
    female 0.1 - 0.15 -
    0.15

    But for finding answers for questions I need to find all elements in my table....I used all given features except 3) and 6). After finding all elements in table which are "-" I will just use conditional probability formula. But before help me please to find elements in the table...;( thank u....

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    Re: Joint marginal conditional probabilities HELP please ;)

    Let say you have the following table

    \begin{tabular}{|c|c|c|c|c|} 
\hline
& $0$ & $1$ & $2$ & $>3$ \\ \hline
male & $a$ & $b$ & $c$ & $d$ \\ \hline
female & $e$ & $f$ & $g$ & $h$ \\ \hline
\end{tabular}

    Here are the results from summary:

    1. b = 0.15

    2. g = 0.15

    3. \frac {h} {d+h} = 0.5 \Rightarrow d = h

    4. a = 0.05

    5. d = 0.1

    6. \frac {a} {a + b + c + d} = 0.125

    7. a + e = 0.15

    At last you may use the fact that the sum of all entries equal to 1 to solve for f

  3. #3
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    Re: Joint marginal conditional probabilities HELP please ;)


    Quote Originally Posted by BGM View Post
    Let say you have the following table

    \begin{tabular}{|c|c|c|c|c|} 
\hline
& $0$ & $1$ & $2$ & $>3$ \\ \hline
male & $a$ & $b$ & $c$ & $d$ \\ \hline
female & $e$ & $f$ & $g$ & $h$ \\ \hline
\end{tabular}

    Here are the results from summary:

    1. b = 0.15

    2. g = 0.15

    3. \frac {h} {d+h} = 0.5 \Rightarrow d = h

    4. a = 0.05

    5. d = 0.1

    6. \frac {a} {a + b + c + d} = 0.125

    7. a + e = 0.15

    At last you may use the fact that the sum of all entries equal to 1 to solve for f
    wow it looks easy! Thank u so much

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