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Math Help - Probability

  1. #1
    mlg
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    Probability

    Hi
    I need help with this problem:
    10 playing cards are numbered 1 to 10. 3 different cards are randomly selected. Calculate the probability that the sum of the numbers selected are at most 9?
    Any help will be greatly appreciated.
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    Re: Probability

    Quote Originally Posted by mlg View Post
    Hi
    I need help with this problem:
    10 playing cards are numbered 1 to 10. 3 different cards are randomly selected. Calculate the probability that the sum of the numbers selected are at most 9?
    Any help will be greatly appreciated.
    You need to make a list of sums that meet the criteria.
    $123,~124,~125,~126,~234~,\cdots$ ect.

    Now $C_3^{10}=\dbinom{10}{3}=~?$
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  3. #3
    mlg
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    Re: Probability

    Thank you for your help.
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    mlg
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    Re: Probability

    Can anyone see anything wrong with the following approach?
    10^3 possible outcomes.
    84 of them sum to less than or equal to 9.
    P = 84/1000 = 21/250
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    Re: Probability

    Quote Originally Posted by mlg View Post
    Can anyone see anything wrong with the following approach?
    10^3 possible outcomes.
    84 of them sum to less than or equal to 9.
    P = 84/1000 = 21/250
    Almost every thing is wrong with it. Did you complete the list I started?
    $123,~124,~125,~126,~134,~135,~234$ That is only seven favorable outcomes. Are there any others?
    How many total outcomes are there. See here.

    What answer do you get?
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    mlg
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    Re: Probability

    Thanks again.
    Regarding favourable outcomes, for example 123, can you also take the number 132 etc? (or is this repetition?).
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    Re: Probability

    Quote Originally Posted by mlg View Post
    Thanks again.
    Regarding favourable outcomes, for example 123, can you also take the number 132 etc? (or is this repetition?).
    NO! The adding the numbers in the set $\{1,~2,~3\}$ gives the sum of six no matter how one orders the numbers.
    Hence this problem comes-down to counting the number of three element subsets that have a sum $~S\le 9$.

    Now $6\le S\le 27$, $1+2+3=6~\&~8+9+10=27$. How many of those subsets have a sum $\le 9~?$

    This is a question about content and not about order.
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    mlg
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    Re: Probability

    Is the answer 42/120 correct?
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    Re: Probability

    Quote Originally Posted by mlg View Post
    Is the answer 42/120 correct?
    NO! Order has nothing to do with this question.

    So $\dfrac{7}{120}$.
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  10. #10
    mlg
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    Re: Probability

    Thanks again.
    I was taking all the different orders, e.g.. 123, 132, 231 etc.
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    Lightbulb Re: Probability

    This should be clearly solved using a Tree Diagram,
    with branches each of which the First card has \frac{1}{10} probability, Second card \frac{1}{9} and Third card has \frac{1}{8} as he is picking out one card after the other without replacement.

    Each of these 4 basic and unique entries sums to 9 and each occurred in 3! = 6 times (ex. 1 2 6, 1 6 2, 2 1 6, 2 6 1, 6 1 2, 6 2 1 etc.) throughout the tree diagrams.

    1 1 7
    1 2 6
    1 3 5
    1 4 4

    The probability sum of the 4 \times 3! outcomes:

    \left (\frac{1}{10}\times \frac{1}{9}\times \frac{1}{8}  \right )\times 4\times 3! = \frac{1}{30} correct!


    But I think it is wrong to consider 7 entries 3 of which are repeated combinations, because they are already considered in the \times 3!:

    1 1 7
    1 2 6
    1 3 5
    1 4 4
    1 5 3
    1 6 2
    1 7 1


    Giving,

    \left (\frac{1}{10}\times \frac{1}{9}\times \frac{1}{8}  \right )\times 7\times 3! = \frac{7}{120} wrong !?!
    Last edited by zikcau25; March 16th 2014 at 01:33 PM.
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    Re: Probability

    Again, I do mistakes on Combinations word.
    I meant Repeated Arrangements instead of Repeated Combinations above. Sorry.
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    Re: Probability

    Quote Originally Posted by zikcau25 View Post
    This should be clearly solved using a Tree Diagram,
    with branches each of which the First card has \frac{1}{10} probability, Second card \frac{1}{9} and Third card has \frac{1}{8} as he is picking out one card after the other without replacement.
    Each of these 4 basic and unique entries sums to 9 and each occurred in 3! = 6 times (ex. 1 2 6, 1 6 2, 2 1 6, 2 6 1, 6 1 2, 6 2 1 etc.) throughout the tree diagrams.
    1 1 7
    1 2 6
    1 3 5
    1 4 4
    @zikcau25;813647], did you bother to read the OP much less the entire theard?
    Obviously not, or else you don't understand decks of cards.
    Here is the OP:
    Quote Originally Posted by mlg View Post
    10 playing cards are numbered 1 to 10. 3 different cards are randomly selected. Calculate the probability that the sum of the numbers selected are at most 9?
    Any help will be greatly appreciated.
    There are only ten cards numbered 1 thru 10.
    Thus there is no way you can choose three of then and have $144$.
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    Re: Probability

    Umm, example:

    1 4 4 (represents 1 arrangement of the 3 numbered cards and that the sum of the 3 numbers on these 3 cards add up to 9 only)

    1: the number written on the first card,
    4: the number written on the second card,
    4: the number written on the third card


    Precisely there with spaces in between, I mean 1, 4, 4 taken separately or 1+4+4= 9 (sum) and not 144 as a whole value alone.
    Thanks.
    Last edited by zikcau25; March 16th 2014 at 02:23 PM.
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  15. #15
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    Re: Probability

    Quote Originally Posted by zikcau25 View Post
    Umm, example:

    1 4 4

    1 on the first card,
    4 on the second card,
    4 on the third card
    Precisely there with spaces in between, I mean 1, 4, 4 taken separately or 1+4+4= 9 (sum) and not 144 as a whole value alone.
    But that is impossible.
    There is only one 4 in the whole deck,


    Cards dealt out only once. If the first is a 4 the no other card can possibly be a 4.
    This is not sampling with replacement. You are confused on that point.
    Thanks from zikcau25
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