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Math Help - anyone can help me in probability!!!

  1. #1
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    Red face anyone can help me in probability!!!

    Hi every body i have small problem in probabiltity
    the problem is


    A university have 5 English sections ,if we distribute another 4 students to these sections


    1_ Find the probability that 2 of them moved to the same section and the other in different sections?



    my answer is :

    (5.1.4.3)/(5.5.5.5)= 60/625

    is it correct or not ????
    i hope to find resopnd to my problem to solved if its wrong...
    Thank you..
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  2. #2
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    Hello, sunrise2007!

    A university have 5 English sections .
    If we distribute another 4 students to these sections,

    1) Find the probability that 2 of them moved to the same section
    and the other in different sections?
    Your denominator is correct.
    . . There are: . 5^4 \:= \:{\color{red}625} possible assignments.


    Two students will be assigned the same section.
    . . There are: . {4\choose2} = {\color{blue}6} ways to choose the two students.
    . . They can be assigned to any of the 5 sections.
    . . The other two students have: . 4 \times 3 \:=\:{\color{blue}12} possible assignments.
    Hence, there are: . 6 \times 5 \times 12 \:=\:{\color{red}360} ways.


    Therefore, the probability is: . \frac{360}{625} \;=\;\boxed{\frac{72}{125}}

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  3. #3
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    I'm sure you're correct, Soroban, but I am coming up with a different answer. Please tell me the error in my logic.

    Choose 2 students out of the 4 to go into the one class. C(4,2)=6

    This can be done in 5 ways for the 5 classes. 6*5=30

    But, the other two students must be placed in the other 4 classes in 4^2=16 ways.

    16*5*6=480

    480/625=96/125
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  4. #4
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    Quote Originally Posted by galactus View Post
    But, the other two students must be placed in the other 4 classes in 4^2=16 ways.
    The 4^2 allows the other two students to be in the same section.
    We want each of them to be in a different section.
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  5. #5
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    Quote Originally Posted by Soroban View Post
    Hello, sunrise2007!

    Your denominator is correct.
    . . There are: . 5^4 \:= \:{\color{red}625} possible assignments.


    Two students will be assigned the same section.
    . . There are: . {4\choose2} = {\color{blue}6} ways to choose the two students.
    . . They can be assigned to any of the 5 sections.
    . . The other two students have: . 4 \times 3 \:=\:{\color{blue}12} possible assignments.
    Hence, there are: . 6 \times 5 \times 12 \:=\:{\color{red}360} ways.


    Therefore, the probability is: . \frac{360}{625} \;=\;\boxed{\frac{72}{125}}
    your answer Mr.Soroban very logic
    my answer was correct but it is mising the permutition of two student and how are the student that choosen from the group ,
    is describtion correct ???
    Thank you so much for every one....

    sorry ,i have another problem
    the problem is
    If P (A1) = 5 P (A2) , AC S.
    P(A/A1)= 0.2 ,P(A/A2) = 0.3
    what is the value of P (A1 /A) ?

    i am trying to solve it but i can not find the relation between the rules...

    From the rules

    P(A/A2)=P(A ∩A2)/P(A2)
    0.3= P(A ∩A2)/P(A2)

    P(A/A1)=P(A ∩A1)/P(A1)
    0.2= P(A ∩A1)/P(A1)
    = P(A ∩A1)/ 5 P(A2)
    Last edited by sunrise2007; November 26th 2007 at 07:52 AM.
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