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Math Help - Review for Exam...Urgent

  1. #1
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    Review for Exam...Urgent

    I am reviewing for an exam today, and, I was given a review sheet. However, I do not see how the answer is what is listed. Here is the question.

    Two distinct even numbers are selected at random from the first ten even numbers greater than zero. What is the probability that the sum is 30?

    I don't know, maybe he miss worded it, but as is I'm going to assume the sample space contains 10 numbers....2,4,6,8,10,12,14,16,18,20

    Which, would leave me with combinations, 10,20 14,16 and 12,18. However, more if I include their reverse 20,10 16,14 and 18,12, but I am assuming again. Well, the first way I get 3/10 for an answer.What am I doing wrong? It shows an answer of 1/15 on the review sheet.The teacher could be wrong, I already found a couple wrong and I am sure of it. This problem could also have a sample space of 20 the way he worded it...I still have about 8 hours before my final...
    Last edited by WhoaBlackBetty; December 15th 2008 at 05:01 AM.
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    Also, I would just like to add a big thank you to everyone who has helped me. After this class I am taking a cryptography and logic course and I am sure I will be back in January! Thanks Guys and/or Gals. It definitely helps the learning process! What would I have done without a PC. It truly helps people in this day and age!
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  3. #3
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    Quote Originally Posted by WhoaBlackBetty View Post
    I am reviewing for an exam today, and, I was given a review sheet. However, I do not see how the answer is what is listed. Here is the question.

    Two distinct even numbers are selected at random from the first ten even numbers greater than zero. What is the probability that the sum is 30?

    I don't know, maybe he miss worded it, but as is I'm going to assume the sample space contains 10 numbers....2,4,6,8,10,12,14,16,18,20

    Which, would leave me with combinations, 10,20 14,16 and 12,18. However, more if I include their reverse 20,10 16,14 and 18,12, but I am assuming again. Well, the first way I get 3/10 for an answer.What am I doing wrong? It shows an answer of 1/15 on the review sheet.The teacher could be wrong, I already found a couple wrong and I am sure of it. This problem could also have a sample space of 20 the way he worded it...
    Since the numbers have to be distinct, there are (10)(9) = 90 possible combinations. Out of those 90, six will have a sum equal to 30:

    (10, 20), (20, 10), (14, 16), (16, 14), (12, 18) and (18, 12).

    Therefore the probability is 6/90 = 1/15.
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  4. #4
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    Thanks for that Fantastic! I don't remember reading anything about distinct in the book, but could you explain where you got the 9 from? 10*9? Everything else you said makes sense...
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    Quote Originally Posted by WhoaBlackBetty View Post
    Thanks for that Fantastic! I don't remember reading anything about distinct in the book, but could you explain where you got the 9 from? 10*9? Everything else you said makes sense...
    Once the first even number is chosen (ten possible choices) there are only nine choices left if the second even number has to be different.

    Quote Originally Posted by WhoaBlackBetty View Post
    [snip]
    Two distinct even numbers are selected at random from the first ten even numbers greater than zero. What is the probability that the sum is 30?
    [snip]
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    Thanks! I should of realized that...It makes perfect sense. See you next term!
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