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Math Help - probability problem about cards

  1. #1
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    probability problem about cards

    Hi, this is my question:
    Sean has a set of cards numbered 1,2,3 and 4. He picks a card at random and does not replace it. He then picks another card at random from the remaining cards.

    A) What is the probability that neither of the two cards that Sean picked was 4?

    im not sure but would i do, 3/4 x 2/3= 6/12 = 3/6 = 1/2?
    Answer being 1/2?

    Thanks!
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  2. #2
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    Quote Originally Posted by andyboy179 View Post
    Hi, this is my question:
    Sean has a set of cards numbered 1,2,3 and 4. He picks a card at random and does not replace it. He then picks another card at random from the remaining cards.

    A) What is the probability that neither of the two cards that Sean picked was 4?

    im not sure but would i do, 3/4 x 2/3= 6/12 = 3/6 = 1/2?
    Answer being 1/2?

    Thanks!
    Yes, that's the quick way..

    \binom{4}{2}=6 is the number of ways to choose 2 cards from 4.

    If the pair does not contain 4, there are \binom{3}{2}=3 pairs of two cards not containing 4.

    Listing the options... 1,2....1,3....2,3 do not contain 4

    1,4....2,4,....3,4 contain 4

    half of the two-card pairings do not contain 4.
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  3. #3
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    Thanks alot!!

    the next question says:

    B) What is the probability that the numbers on the two cards that Sean picked add up to 4?

    im not sure how to do this, would it be:
    3/4 x 2/3 = 1/2 again??
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  4. #4
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    Quote Originally Posted by andyboy179 View Post
    Thanks alot!!

    the next question says:

    B) What is the probability that the numbers on the two cards that Sean picked add up to 4?

    im not sure how to do this, would it be:
    3/4 x 2/3 = 1/2 again??
    In order to add to 4, the cards must be 1 and 3.
    That's just 1 out of the 6 selections of 2 cards.

    If you count arrangements, there are two arrangements.....1,3 and 3,1

    but there are are 4(3)=12 arrangements.

    To do it by multiplying probabilities,
    there is a 1/4 probability the first is 1 and a 1/3 probability the second is 3.
    Similarly there is a 1/4 probability the first is 3 and a 1/3 probability the second is a 1.

    That's \frac{1}{4}\ \frac{1}{3}+\frac{1}{4}\ \frac{1}{3}

    Or, there's a 2 in 4 chance the first is a 1 or 3 and a 1 in 3 chance the second is the other

    P=\frac{2}{4}\ \frac{1}{3}
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  5. #5
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    Quote Originally Posted by Archie Meade View Post
    In order to add to 4, the cards must be 1 and 3.
    That's just 1 out of the 6 selections of 2 cards.

    If you count arrangements, there are two arrangements.....1,3 and 3,1

    but there are are 4(3)=12 arrangements.

    To do it by multiplying probabilities,
    there is a 1/4 probability the first is 1 and a 1/3 probability the second is 3.
    Similarly there is a 1/4 probability the first is 3 and a 1/3 probability the second is a 1.

    That's \frac{1}{4}\ \frac{1}{3}+\frac{1}{4}\ \frac{1}{3}

    Or, there's a 2 in 4 chance the first is a 1 or 3 and a 1 in 3 chance the second is the other

    P=\frac{2}{4}\ \frac{1}{3}

    oh, so i could just write ' the probability the two cards sean picks up add up to 4 is 2/4'?
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  6. #6
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    Quote Originally Posted by andyboy179 View Post
    oh, so i could just write ' the probability the two cards sean picks up add up to 4 is 2/4'?
    No, 2/4 is the probability that the first number is one of the numbers that will give a sum of 4.

    You need both numbers.

    Suppose the first number was 3.
    Then the probability the 2nd one is 1 is 1/3.

    Multiply the probabilities of the first being 3 and the 2nd being 1

    this is \frac{1}{4}\ \frac{1}{3}=\frac{1}{12}

    there is also a 1/12 probability of getting a 1 followed by a 3.

    Summing the probabilities gives 2/12=1/6.

    This is the probability of getting the sequence 1,3 or 3,1
    which are the only ways to get a sum of 4 from the two cards.

    There are other ways to do it, but the probability will be 1/6

    If you list the possibilities

    12 13 14
    21 23 24
    31 32 34
    41 42 43

    Only 2 of these 12 arrangements give a sum of 4

    12 13 14
    23 24
    34

    Only 1 of these 6 selections gives a sum of 4
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  7. #7
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    oh i understand now, thanks!

    the last part is:

    Sean now picks a third card from the remaining two cards.

    C) Calculate the probability that the numbers on the first two cards add up to the number on the third card.

    im not sure what to do on this one aswell, so could i have some more help please?
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  8. #8
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    Quote Originally Posted by andyboy179 View Post
    oh i understand now, thanks!

    the last part is:

    Sean now picks a third card from the remaining two cards.

    C) Calculate the probability that the numbers on the first two cards add up to the number on the third card.

    im not sure what to do on this one aswell, so could i have some more help please?
    Ok,

    first examine what the possibilities are...

    the first two cards can be

    1 and 2
    1 and 3
    1 and 4
    2 and 3
    2 and 4
    3 and 4

    This rules out four of these pairings since the highest number card is 4

    The first two cards can only be 1 and 2 or 1 and 3.

    If the first two cards are 1 and 2 then the 3rd card must be 3.
    If the first two cards are 1 and 3 then the third card must be 4.

    Now that the problem is clear, you can use arrangements or selections to calculate the probabilities.

    1,2,3 has a \frac{1}{4}\ \frac{1}{3}\ \frac{1}{2} probability

    because there is a 1 in 4 chance of picking 1 first,
    a 1 in 3 chance of picking 2 second
    and a 1 in 2 chance of picking 4 third.

    Also, the 2 could come before 1, but the probability is the same.

    So the probability of a 1 and 2 followed by 3 is 2\frac{1}{4}\ \frac{1}{3}\ \frac{1}{2}=\frac{1}{12}

    Working out the probability of 1,3,4 or 3,1,4 is the same procedure.

    Adding up the probabilities, you get 1/12+1/12=2/12=1/6

    You can list the options also

    1,2,3
    1,2,4
    1,3,2
    1,3,4
    1,4,2
    1,4,3
    2,1,3
    2,1,4
    2,3,1
    2,3,4
    2,4,1
    2,4,3
    3,1,2
    3,1,4
    3,2,1
    3,2,4
    3,4,1
    3,4,2
    4,1,2
    4,1,3
    4,2,1
    4,2,3
    4,3,1
    4,3,2

    There are 24 arrangements and 4 of these are in the required order
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  9. #9
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    Quote Originally Posted by Archie Meade View Post
    Ok,

    first examine what the possibilities are...

    the first two cards can be

    1 and 2
    1 and 3
    1 and 4
    2 and 3
    2 and 4
    3 and 4

    This rules out four of these pairings since the highest number card is 4

    The first two cards can only be 1 and 2 or 1 and 3.

    If the first two cards are 1 and 2 then the 3rd card must be 3.
    If the first two cards are 1 and 3 then the third card must be 4.

    Now that the problem is clear, you can use arrangements or selections to calculate the probabilities.

    1,2,3 has a \frac{1}{4}\ \frac{1}{3}\ \frac{1}{2} probability

    because there is a 1 in 4 chance of picking 1 first,
    a 1 in 3 chance of picking 2 second
    and a 1 in 2 chance of picking 4 third.

    Also, the 2 could come before 1, but the probability is the same.

    So the probability of a 1 and 2 followed by 3 is 2\frac{1}{4}\ \frac{1}{3}\ \frac{1}{2}=\frac{1}{12}

    Working out the probability of 1,3,4 or 3,1,4 is the same procedure.

    Adding up the probabilities, you get 1/12+1/12=2/12=1/6

    You can list the options also

    1,2,3
    1,2,4
    1,3,2
    1,3,4
    1,4,2
    1,4,3
    2,1,3
    2,1,4
    2,3,1
    2,3,4
    2,4,1
    2,4,3
    3,1,2
    3,1,4
    3,2,1
    3,2,4
    3,4,1
    3,4,2
    4,1,2
    4,1,3
    4,2,1
    4,2,3
    4,3,1
    4,3,2

    There are 24 arrangements and 4 of these are in the required order

    when adding 1/4 + 1/3 +1/2 how does it = 1/12??
    wouldn't it be 1/9?
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  10. #10
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    Quote Originally Posted by andyboy179 View Post
    when adding 1/4 + 1/3 +1/2 how does it = 1/12??
    wouldn't it be 1/9?
    We sum the probabilities of each way to get the desired result.

    Each way involves the multiplication of the probabilities at each stage of choosing cards for that way.

    In other words......

    1,2,3 is 1 followed by 2 followed by 3.

    This is a "successful" sequence from the perspective of the number of the 3rd card
    being the sum of the numbers of the first two cards.

    This is.... 1 and 2 and 3 together.

    So, to calculate the probability of this happening, we calculate

    (i) the probability that 1 came first, this is 1/4
    (ii) the probability that 2 came second, this is 1/3 as there are three cards left and 2 is one of them
    (iii) the probability that 3 came third.

    The probability that 1,2,3 happened in that order is the product of those "stage-by-stage" probabilities.

    The probability of 1 and 2 and 3 in that order is \frac{1}{4}\ \frac{1}{3}\ \frac{1}{2}

    That's where we multiply probabilities.

    Now we can also have 2,1,3 so that probability must be worked out.
    Also 1,3,4 and 3,1,4 are successful outcomes.

    Hence we calculate the probabilities for those 4 ways of getting a successful
    outcome and then add them all up

    In other words....1,2,3 or 2,1,3 or 1,3,4 or 3,1,4

    is a successful outcome.

    Or.... add probabilities
    And...multiply probabilities

    is a way to think of it.

    You can check from the sequences listed how the probabilities should work out
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  11. #11
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    i think i understand it now.
    so for 1,2,3 i would do 1/4 x 1/3 x 1/2= 1/24= 1/12?
    then would i do the same for the other 3 so i would be left with 1/12, 1/12, 1/12 and 1/12?
    than would i add them all up like this, 1/12 + 1/12 + 1/12 + 1/12 = 4/48= 2/24= 1/12= 1/6??

    Andswer= the probability that the numbers on the first 2 cards add up to the number on the third card is 1/6???
    Last edited by andyboy179; May 30th 2010 at 02:40 AM. Reason: typo
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  12. #12
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    Quote Originally Posted by andyboy179 View Post
    i think i understand it now.
    so for 1,2,3 i would do 1/4 x 1/3 x 1/2= 1/24= 1/12?
    then would i do the same for the other 3 so i would be left with 1/12, 1/12, 1/12 and 1/12?
    then, would i add them all up like this, 1/12 + 1/12 + 1/12 + 1/12 = 4/48= 2/24= 1/12= 1/6??

    Answer= the probability that the numbers on the first 2 cards add up to the number on the third card is 1/6???
    You're almost there apart from those fractions...

    \frac{1}{4}\ \frac{1}{3}\ \frac{1}{2}=\frac{1}{12}\ \frac{1}{2}=\frac{1}{24}

    That's the probability of 1,2,3

    2,1,3 and 1,3,4 and 3,1,4 have the same probabilities, so the final answer is

    \frac{1}{24}+\frac{1}{24}+\frac{1}{24}+\frac{1}{24  }=\frac{4}{24}=\frac{4}{4(6)}=\frac{1}{6}

    You need to review why you are writing \frac{1}{24}=\frac{1}{12}

    In an earlier post, you mentioned.... shouldn't \frac{1}{4}+\frac{1}{3}+\frac{1}{2}=\frac{1}{9} ?

    If those fractions were being added, you cannot sum the denominators like that...

    You can sum numerators if the denominators are the same

    2+3=5 is 2 units + 3 units = 5 units, but they must be the same units

    2 apples + 3 apples = 5 apples.

    2 quarters + 3 quarters = 5 quarters

    \frac{2}{4}+\frac{3}{4}=\frac{2+3}{4}=\frac{5}{4}

    1/4, 1/3 and 1/2 do not have the same denominator, but we can write them with a common denominator,
    by multiplying them by 1

    \frac{1}{4}+\frac{1}{3}+\frac{1}{2}=\frac{3}{3}\ \frac{1}{4}+\frac{4}{4}\ \frac{1}{3}+\frac{6}{6}\ \frac{1}{2}=\frac{3}{12}+\frac{4}{12}+\frac{6}{12}  =\frac{3+4+6}{12}=\frac{13}{12}
    Last edited by Archie Meade; May 30th 2010 at 07:50 AM. Reason: small typo
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  13. #13
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    i know now, thankyou very much for all you help!!!
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