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Math Help - Two students are randomly picked from a class of 10 boys and 5 girls..

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    Two students are randomly picked from a class of 10 boys and 5 girls..

    Two students are randomly picked from a class of 10 boys and 5 girls. What is the probability that the second student picked is a girl, given that the two has at least 1 girl?

    the answer is 1/6 but i dont know how to get that..

    Explanation would be appreciated thanks!
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  2. #2
    MHF Contributor matheagle's Avatar
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    I must be doing something wrong since I got 7/12
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  3. #3
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    Quote Originally Posted by qkqh View Post
    Two students are randomly picked from a class of 10 boys and 5 girls. What is the probability that the second student picked is a girl, given that the two has at least 1 girl?

    the answer is 1/6 but i dont know how to get that..

    Explanation would be appreciated thanks!
    I get 7/12 too..

    P(B|A)=\dfrac{P(B\cap A)}{P(A)}

    A = there's at least one girl chosen
    B = the second one is a girl

    P(A) = 1 - P(\overline{A}) = 1 - \dfrac{\binom{10}{2}}{\binom{15}{2}} = \dfrac{4}{7}

    P(B\cap A) = P(B) = \dfrac{10}{15} \cdot \dfrac{5}{14} + \dfrac{5}{15} \cdot \dfrac{4}{14} = \dfrac{1}{3}

    Note: You could also get P(B) = 1/3 by reasoning that the probability that the second one is a girl is the same as the probability that the first one is a girl.

    Which leads to 7/12...
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  4. #4
    MHF Contributor matheagle's Avatar
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    I used order (BG means boy first, then girl)

    P(BG \cup GG| BG\cup GG\cup GG)

    ={P(BG)+P(GG)\over P(BG)+P(GG)+P(GG)}

    ={ ({10\over 15}) ({5\over 14}) +({5\over 15}) ({4\over 14}) \over ({10\over 15}) ({5\over 14}) +({5\over 15}) ({10\over 14}) + ({5\over 15}) ({4\over 14})}

    multiply everywhere by (15)(14) leads to 70/120 or 7/12.
    Last edited by matheagle; July 16th 2010 at 09:59 PM.
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  5. #5
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    Hello, qkqh!

    I solved it with my usual baby-talk approach.

    And I too have no idead how they got \frac{1}{6}


    Two students are randomly picked from a class of 10 boys and 5 girls.
    What is the probability that the second student picked is a girl,
    given that the two has at least one girl?

    We choose 2 students from a class of 15.
    . . There are: . _{15}C_2 \:=\:{15\choose2} \:=\:210 outcomes.


    \begin{array}{ccccc}<br />
(1) &\text{Girl-Girl} & 5\cdot4 &=& 20 \\<br />
(2) & \text{Girl-Boy} & 5\cdot10 &=& 50 \\<br />
(3) & \text{Boy-Girl} & 10\cdot5 &=& 50 \\<br />
(4) & \text{Boy-Boy} & 10\cdot 9 &=& 90 \\[-3mm]<br />
& & & &-- \\[-2mm] & & & & 210 \end{array}


    Since the two students include at least one girl,
    . . the sample space is: . (1) + (2) + (3) \:=\:20 + 50 + 50 \:=\:120

    The second student is a girl: . (1) + (3) \:=\:20 + 50 \:=\:70


    Therefore: . P(\text{2nd is Girl}\,|\,\text{at least one girl}) \;=\;\dfrac{70}{120} \;=\;\boxed{\frac{7}{12}}


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


    "They" could get \frac{1}{6} with: . \frac{35}{210} or \frac{20}{120}

    . . neither of which makes any sense.
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