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Math Help - Probability, Christmas presents

  1. #1
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    Probability, Christmas presents

    I think i need some help with this (pls forgive me for bad English)

    Presents under the Christmas tree. 8 for Lisa, 7 for John, 6 for mom, and 5 for dad. It is not possible to see who the different presents belongs to.

    I'm pretty sure i got question a) and b) right, but the last part I'm not sure about:

    John and Lisa agreed to open one present each. First John opens one, and then Lisa.

    c) What is the probability that they both find a present that belongs to them?
    d) What is the probability that neither of them find a present that belongs to them?

    on c) I thought something like this: P(J|L)= 7/26*8/25 = 56/650 = 28/325

    on d) : 19/26*17/25 = 323/325

    But I'm not sure I got it right. I think it's more complicated... What bothers me is what if John opens one of Lisas presents? The text does'nt say anything about this, but it says that the last part of the task is rather difficult. Any suggestions?

    Thanks
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  2. #2
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    thinking...

    Not just me that finds this complex? Maybe I have to make some assumptions/limitations before i start. Like that John will not show Lisa what's inside. Just keep it if it is his, and wrap it up perfectly and mix it with the other presents if it is not? Hope someone understands what I'm trying to say here. English is not my native language...
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  3. #3
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    The idea is that there are a total of 26 presents before john picks a present, so he has a 7/26 chance of choosing one of his presents. so, 7/26. Lisa has a total of 25 presents, so 8/25.
    When multiplying fractions you do the following, sorry if this is hard to understand:
    7/26*8/25. Find a common denominator, this being 650. we then multiply 7 and 8, this is 56.
    so, 56/650 is equal to 28/325. However, probability is on a scale of 0 to 1. It works out at 0.08615384, recurring from 6 to 4. So, ultimately, 0.086 should satisfy the teacher. In A level however, 28/325 would work.

    Question d follows the same pattern, but in reverse. so, 26-7=19. 19/26. 25 left. 25-8=17. 17/25. Again, 19/26*17/25. 25*26=650, 19*17=323. These cannot be cancelled down, which means it is a total of 323/650. or in decimal terms, 0.49692037, recurring from the 6 to the 7, or 0.497 rounded to 3 decimal places.

    Hope this helped you.
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  4. #4
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    Thanks for helping me, greatersanta616

    The way you explain is the way I first thought would lead me to the right answer. I drew a probability tree, starting with two different options. One for John picking his own present, and one for John picking a present for Lisa or the parents. This tree helped me to find the probability of both John and Lisa finding a present that belongs to them. Like we both found out:

    P(J|L)= 7/26*8/25 = 56/650 = 28/325

    When I startet with "d" I again thought like you, using the same tree: P= 19/26*17/25 = 323/325

    But like I said, I was not satisfied with this solution. It was just too easy. So I phoned my math teacher today and explained. He gave me the following hint: "Yes, your probability tree works fine for solving "c", but not so fine for solving "d". Something have to be split up"

    It made me realize that it is also relevant to know if John picks either Lisa's present or one of his parent's present to find the exact answer. It is not enough just to know it is not for him. So I split the first part of the tree in three different possibilities (see drawing), where the green and blue lines shows two different possibilities that have to be added together:

    green: P = 8/26 * (7/25 + 11/25) = 8/26 * 18/25 = 144/650

    blue: P = 11/26 * (7/25 + 10/25) = 11/26 * 17/25 = 187/650

    Added together the right answer for "d" should be 331/650. In decimal terms, 0.509, rounded to 3 decimal places.

    mattemann
    Attached Thumbnails Attached Thumbnails Probability, Christmas presents-tree.gif  
    Last edited by mattemann; September 22nd 2010 at 01:23 PM.
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