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Elementry Probability problem

  1. Sep 9, 2009 #1
    An elementary school is offering 3 language classes: one in Spanish, one in French, and one in German. These classes are open to any of the 98 students in the school. There are 38 in the Spanish class, 30 in the French class, and 18 in the German class. There are 13 students that in both Spanish and French, 6 are in both Spanish and German, and 8 are in both French and German. In addition, there are 4 students taking all 3 classes.

    If two students are chosen randomly, what is the probability that neither of them is taking a language class?




    3. The attempt at a solution

    I believe that I have to figure out exactly how many students are not taking any language courses and then use permutation to get the probability. So something like this:

    xPx/98P2

    I have been unable to figure out how to get the right number of students that are not taking a language course. Please help.
     
  2. jcsd
  3. Sep 9, 2009 #2
    This seemed like a difficult problem for me at first, and thought I would have to go into great detail about probability trees and the 'tree-flipping' method to solve for our unknowns here, but looking at it for a while it proved to be quite simple.

    You want to figure out who's not taking any language courses, so it would help to find out how many people ARE taking language courses, and simply finding the complement.

    At first glance it would seem obvious to say there are 38, 30 and 18 students in language courses for Spanish, French and German respectively, but of course there is some overlap so we cannot use these figures.

    We want the number of kids who are ONLY taking Spanish, the number of kids who are ONLY taking French and the number of kids who are ONLY taking German.

    Lets try one at a time.

    Starting with Spanish students, 38, lets subtract those who are taking both Spanish & French (13). Lets also subtract those who are taking Spanish & German (6). Lets finally subtract those who are taking all three (4). That leaves us with 15 students taking ONLY Spanish. Thus, the probability that a student is ONLY in Spanish is 15/98.

    Note that we only subtracted figures that are common with Spanish & one or two more language courses.

    Can you repeat this process for French & German?

    When you do that, you should find that the sum of the three probabilities of students in only ONE of the three classes amounts to the probability of a student being in ANY three of the language classes.

    Knowing that, we can of course find the compliment, and your answer.

    Just remember the question asked for the probability of TWO students not being in a language class. Remember the mulitplication rule.

    My final answer was 63.2%
     
  4. Sep 9, 2009 #3
    I can't seem to understand something.

    If there are total 98 students and 38+30+18=86 go on language class, there are 12 students left which don't go to any of the language classes. :confused:

    By thinking this way you can very easy find the probability that 2 picked students will not go on language classes.

    Am I missing basic logic?
     
  5. Sep 9, 2009 #4
    Yes you are. Remember there is overlap in the students that are in language classes. So the exact number of students in at least 1 language class will infact be much much less than 86. In fact I calculated that there are only 20 students that are in language classes.
     
  6. Sep 9, 2009 #5
    CaptainEvi

    I am trying to comprehend how you got the answer by finding the probability of students being in ONLY 1 class...the way I went about it was finding how many students were in each class ONLY which were Spanish-15, French-5, and German-0. From here add em' all up to get 20 students in ONLY 1 class but don't forget you still have to add in all the students taking multiple classes, 20(total of ONLY 1 class) + 13 (Spanish and French) + 6 (Spanish and German) + 8 (French and German) + 4 (All 3 classes) = 51 students taking a language course.

    You want the probability of selecting 2 students not taking a language course so you subtract 51 from 98 to get 47 students not in a language course.

    For the first student it would be 47/98 and since you can't pick the same student twice you subtract 1 from the students not in a language class and 1 from the total. The probability of choosing the second student that is not in a language course is 46/97.

    So once you get both the probabilities don't you multiply them together to get the final answer of 22.74%?

    Correct me if I'm wrong but I just need to be able to comprehend how you got your answer because mathematics fascinates me and I want to be able to understand a problem like this next time I approach one.
     
  7. Sep 9, 2009 #6
    I just got home and looked over the post. Everyone seems to be going in the same direction I was going in, but still no right answer.

    CaptainEvil - your final answer of 63.2% was not correct.
    Jonathan G- I had already came to the same answer you had, but it's not right either.

    So, I'm pretty lost on this problem. I'm going to keep working on it, but I think the key is figuring out how many students are actually taking a language class versus how many are not.
     
  8. Sep 9, 2009 #7
    I finally got it right. Here is how it works:

    1-(38+30+18-13-6-8+4)/98 = 1 - (63/98) = 35/98 (for one student). For 2 students:

    35*34/98*97 = 1190/9506 = 12.5%

    thanks for the help
     
  9. Sep 9, 2009 #8
    For future reference. This problem requires two facts:

    General Counting Principle for Three Sets

    [tex]|A \cup B \cup C| = |A| + |B| + |C| - |A \cap B| - |A \cap C| - |B \cap C| + |A \cap B \cap C|[/tex]

    where |A| = the number of members of A.

    and

    The Hypergeometric Distribution

    Given N objects of which M are marked. The probability that m marked objects are drawn when n are selected without replacement is

    [tex]P(\text{marked = m out of n})=\frac{(^M_m)(^{N-M}_{n-m})}{(^N_n)}[/tex]

    --Elucidus
     
  10. Sep 9, 2009 #9
    There you go..that is the formula I was trying to remember but I just couldn't come to it so I tried to work it out a different way, which ended up being incorrect and the average joe's mistake.
     
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