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Math Help - Card probability help

  1. #1
    Junior Member
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    Card probability help

    One card is drawn at random from 27 cards numbered from 1 to 27. What is the probability that the number is divisible by 3 or 4?
    A second card is drawn from the remaining cards what is the probability that the sum of the numbers is odd?

    Hi, i'm stuck on the first part I cant figure out how to do it can someone help please?
    Thanks
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  2. #2
    Senior Member
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    Hi.

    Quote Originally Posted by Detanon View Post
    One card is drawn at random from 27 cards numbered from 1 to 27. What is the probability that the number is divisible by 3 or 4?
    A second card is drawn from the remaining cards what is the probability that the sum of the numbers is odd?

    Hi, i'm stuck on the first part I cant figure out how to do it can someone help please?
    Thanks
    There are 9 numbers divisible by 3:
    3
    6
    9
    12
    15
    18
    21
    24
    27
    (Because 27 : 3 = 9)

    and there are 6 numbers divisible by 4:
    4
    8
    12
    16
    20
    24
    (27 : 4 = 6,...)

    The probability of picking a particular card (e. g. the card has the number 5 on it) is 1/27. This is obvious, is it?

    There are 6+9 = 15 of 27 numbers divisible by 3 or 4, so the asked probabilty is 15/27

    Edit: downthesun01 is right, I counted some numbers twice
    Last edited by Rapha; May 31st 2010 at 09:33 PM.
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  3. #3
    Senior Member
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    Thanks
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    Quote Originally Posted by Rapha View Post
    Hi.


    There are 9 numbers divisible by 3:
    3
    6
    9
    12
    15
    18
    21
    24
    27
    (Because 27 : 3 = 9)

    and there are 6 numbers divisible by 4:
    4
    8
    12
    16
    20
    24
    (27 : 4 = 6,...)

    The probability of picking a particular card (e. g. the card has the number 5 on it) is 1/27. This is obvious, is it?

    There are 6+9 = 15 of 27 numbers divisible by 3 or 4, so the asked probabilty is 15/27
    This is incorrect. As you can see, two of the numbers (12 and 24) appear in both lists.

    So the actual probability is \frac{13}{27}

    For the second part, the only way to get an odd sum is to add an even number to an odd number. Therefore, the answer the solution to the second part is the probability if the first drawn card being odd multiplied by the second drawn card being even plus the probability of the first card being even multiplied by the second card drawn being odd.

    pr(sum is odd)=[pr(first card is even)*pr(second card is odd)]+[pr(first card is odd)*pr(second card is even)]=...
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