Formalization/Verification of a beginner combinatorics problem

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I have the following task:



To build a computer chip, 4 non-distinct components are needed. Of 12 existing components, 2 are defective. Given that 4 components are chosen, what are the probabilities of the following combinations?



  1. No defective components

  2. Exactly one defective component

  3. Exactly two defective components

Progress so far:



The general approach I use is to consider the number of non-defective components over the total number of components for repeated decrements of each set.



  1. $frac1012 cdot frac911 cdot frac810 cdot frac79 approx 0.4242 $

For the next question, I sum the probabilities of each case, depending on if the defective component is chosen last, second last, etc.



  1. $frac1012 cdot frac911 cdot frac810 cdot frac29 + frac1012 cdot frac911 cdot frac210 cdot frac89 + frac1012 cdot frac211 cdot frac910 cdot frac89 + frac212 cdot frac1011 cdot frac910 cdot frac89 $

I could employ a similar intuitive approach for 3. , however I feel like this is messy and not particularly rigorous. How do I answer these questions formally?






probability combinatorics proof-verification







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    up vote
    2
    down vote

    favorite












    I have the following task:



    To build a computer chip, 4 non-distinct components are needed. Of 12 existing components, 2 are defective. Given that 4 components are chosen, what are the probabilities of the following combinations?



    1. No defective components

    2. Exactly one defective component

    3. Exactly two defective components

    Progress so far:



    The general approach I use is to consider the number of non-defective components over the total number of components for repeated decrements of each set.



    1. $frac1012 cdot frac911 cdot frac810 cdot frac79 approx 0.4242 $

    For the next question, I sum the probabilities of each case, depending on if the defective component is chosen last, second last, etc.



    1. $frac1012 cdot frac911 cdot frac810 cdot frac29 + frac1012 cdot frac911 cdot frac210 cdot frac89 + frac1012 cdot frac211 cdot frac910 cdot frac89 + frac212 cdot frac1011 cdot frac910 cdot frac89 $

    I could employ a similar intuitive approach for 3. , however I feel like this is messy and not particularly rigorous. How do I answer these questions formally?






    probability combinatorics proof-verification







    share|cite|improve this question























      up vote
      2
      down vote

      favorite









      up vote
      2
      down vote

      favorite











      I have the following task:



      To build a computer chip, 4 non-distinct components are needed. Of 12 existing components, 2 are defective. Given that 4 components are chosen, what are the probabilities of the following combinations?



      1. No defective components

      2. Exactly one defective component

      3. Exactly two defective components

      Progress so far:



      The general approach I use is to consider the number of non-defective components over the total number of components for repeated decrements of each set.



      1. $frac1012 cdot frac911 cdot frac810 cdot frac79 approx 0.4242 $

      For the next question, I sum the probabilities of each case, depending on if the defective component is chosen last, second last, etc.



      1. $frac1012 cdot frac911 cdot frac810 cdot frac29 + frac1012 cdot frac911 cdot frac210 cdot frac89 + frac1012 cdot frac211 cdot frac910 cdot frac89 + frac212 cdot frac1011 cdot frac910 cdot frac89 $

      I could employ a similar intuitive approach for 3. , however I feel like this is messy and not particularly rigorous. How do I answer these questions formally?






      probability combinatorics proof-verification







      share|cite|improve this question













      I have the following task:



      To build a computer chip, 4 non-distinct components are needed. Of 12 existing components, 2 are defective. Given that 4 components are chosen, what are the probabilities of the following combinations?



      1. No defective components

      2. Exactly one defective component

      3. Exactly two defective components

      Progress so far:



      The general approach I use is to consider the number of non-defective components over the total number of components for repeated decrements of each set.



      1. $frac1012 cdot frac911 cdot frac810 cdot frac79 approx 0.4242 $

      For the next question, I sum the probabilities of each case, depending on if the defective component is chosen last, second last, etc.



      1. $frac1012 cdot frac911 cdot frac810 cdot frac29 + frac1012 cdot frac911 cdot frac210 cdot frac89 + frac1012 cdot frac211 cdot frac910 cdot frac89 + frac212 cdot frac1011 cdot frac910 cdot frac89 $

      I could employ a similar intuitive approach for 3. , however I feel like this is messy and not particularly rigorous. How do I answer these questions formally?






      probability combinatorics proof-verification




      probability combinatorics proof-verification






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      asked 4 hours ago









      Oscar

      31510




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          3 Answers
          3






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          up vote
          2
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          accepted










          Method 1: The number of ways of selecting a subset of size $k$ from a set with $n$ elements (the number of unordered selections of $k$ elements from a set of $n$ elements) is
          $$binomnk = fracn!k!(n - k)!$$



          There are $$binom124$$ ways to select four of the twelve components.



          The number of ways of selecting exactly $k$ of the $2$ defective components and $4 - k$ of the $12 - 2 = 10$ good components is
          $$binom2kbinom104 - k$$



          Therefore, the probability of selecting exactly $k$ defective components is
          $$Pr(textexactly~k~textdefective) = fracdbinom2kdbinom104 - kdbinom124$$



          Hence,
          beginalign*
          Pr(textno defective components) & = fracdbinom20dbinom104dbinom124\
          Pr(textexactly one defective component) & = fracdbinom21dbinom103dbinom124\
          Pr(textexactly two defective components) & = fracdbinom22dbinom102dbinom124
          endalign*



          Method 2: You have correctly calculated the first two answers. There are only three possibilities. The number of defective components selected is either zero, one, or two. Therefore,
          $$Pr(textexactly two defective components) = 1 - Pr(textno defective components) - Pr(textexactly one defective component)$$






          share|cite|improve this answer





























            up vote
            3
            down vote













            The probability on $kin0,1,2$ defectives among the $4$ chosen is:$$fracbinom4kbinom82-kbinom122$$
            Hypergeometric distribution is used.



            Actually this approach turns things around: there are $4$ chosen items and $8$ non-chosen items. Selecting $2$ items (the defective ones) what is the probability that $k$ of them will come from the chosen ones?






            share|cite|improve this answer
















            • 1




              An interesting approach.
              – N. F. Taussig
              3 hours ago

















            up vote
            1
            down vote













            What you have so far is good. To simplify case 2, you can notice that the probabilities are all the same: for each one you get $12times11times10times 9$ on the bottom and $10times9times8times 2$ on the top in some order. So the total probability is just the probability for a specific "pattern" (e.g. defective, good, good, good) times the number of patterns. The same approach works for 3: here there are $6=binom 42$ patterns, each with probability $frac1012timesfrac911timesfrac210timesfrac19$. This works because the first "good" (wherever you put it) always has $10$ on top, the second "good" always has $9$, and so on.






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              3 Answers
              3






              active

              oldest

              votes








              3 Answers
              3






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes








              up vote
              2
              down vote



              accepted










              Method 1: The number of ways of selecting a subset of size $k$ from a set with $n$ elements (the number of unordered selections of $k$ elements from a set of $n$ elements) is
              $$binomnk = fracn!k!(n - k)!$$



              There are $$binom124$$ ways to select four of the twelve components.



              The number of ways of selecting exactly $k$ of the $2$ defective components and $4 - k$ of the $12 - 2 = 10$ good components is
              $$binom2kbinom104 - k$$



              Therefore, the probability of selecting exactly $k$ defective components is
              $$Pr(textexactly~k~textdefective) = fracdbinom2kdbinom104 - kdbinom124$$



              Hence,
              beginalign*
              Pr(textno defective components) & = fracdbinom20dbinom104dbinom124\
              Pr(textexactly one defective component) & = fracdbinom21dbinom103dbinom124\
              Pr(textexactly two defective components) & = fracdbinom22dbinom102dbinom124
              endalign*



              Method 2: You have correctly calculated the first two answers. There are only three possibilities. The number of defective components selected is either zero, one, or two. Therefore,
              $$Pr(textexactly two defective components) = 1 - Pr(textno defective components) - Pr(textexactly one defective component)$$






              share|cite|improve this answer


























                up vote
                2
                down vote



                accepted










                Method 1: The number of ways of selecting a subset of size $k$ from a set with $n$ elements (the number of unordered selections of $k$ elements from a set of $n$ elements) is
                $$binomnk = fracn!k!(n - k)!$$



                There are $$binom124$$ ways to select four of the twelve components.



                The number of ways of selecting exactly $k$ of the $2$ defective components and $4 - k$ of the $12 - 2 = 10$ good components is
                $$binom2kbinom104 - k$$



                Therefore, the probability of selecting exactly $k$ defective components is
                $$Pr(textexactly~k~textdefective) = fracdbinom2kdbinom104 - kdbinom124$$



                Hence,
                beginalign*
                Pr(textno defective components) & = fracdbinom20dbinom104dbinom124\
                Pr(textexactly one defective component) & = fracdbinom21dbinom103dbinom124\
                Pr(textexactly two defective components) & = fracdbinom22dbinom102dbinom124
                endalign*



                Method 2: You have correctly calculated the first two answers. There are only three possibilities. The number of defective components selected is either zero, one, or two. Therefore,
                $$Pr(textexactly two defective components) = 1 - Pr(textno defective components) - Pr(textexactly one defective component)$$






                share|cite|improve this answer
























                  up vote
                  2
                  down vote



                  accepted







                  up vote
                  2
                  down vote



                  accepted






                  Method 1: The number of ways of selecting a subset of size $k$ from a set with $n$ elements (the number of unordered selections of $k$ elements from a set of $n$ elements) is
                  $$binomnk = fracn!k!(n - k)!$$



                  There are $$binom124$$ ways to select four of the twelve components.



                  The number of ways of selecting exactly $k$ of the $2$ defective components and $4 - k$ of the $12 - 2 = 10$ good components is
                  $$binom2kbinom104 - k$$



                  Therefore, the probability of selecting exactly $k$ defective components is
                  $$Pr(textexactly~k~textdefective) = fracdbinom2kdbinom104 - kdbinom124$$



                  Hence,
                  beginalign*
                  Pr(textno defective components) & = fracdbinom20dbinom104dbinom124\
                  Pr(textexactly one defective component) & = fracdbinom21dbinom103dbinom124\
                  Pr(textexactly two defective components) & = fracdbinom22dbinom102dbinom124
                  endalign*



                  Method 2: You have correctly calculated the first two answers. There are only three possibilities. The number of defective components selected is either zero, one, or two. Therefore,
                  $$Pr(textexactly two defective components) = 1 - Pr(textno defective components) - Pr(textexactly one defective component)$$






                  share|cite|improve this answer














                  Method 1: The number of ways of selecting a subset of size $k$ from a set with $n$ elements (the number of unordered selections of $k$ elements from a set of $n$ elements) is
                  $$binomnk = fracn!k!(n - k)!$$



                  There are $$binom124$$ ways to select four of the twelve components.



                  The number of ways of selecting exactly $k$ of the $2$ defective components and $4 - k$ of the $12 - 2 = 10$ good components is
                  $$binom2kbinom104 - k$$



                  Therefore, the probability of selecting exactly $k$ defective components is
                  $$Pr(textexactly~k~textdefective) = fracdbinom2kdbinom104 - kdbinom124$$



                  Hence,
                  beginalign*
                  Pr(textno defective components) & = fracdbinom20dbinom104dbinom124\
                  Pr(textexactly one defective component) & = fracdbinom21dbinom103dbinom124\
                  Pr(textexactly two defective components) & = fracdbinom22dbinom102dbinom124
                  endalign*



                  Method 2: You have correctly calculated the first two answers. There are only three possibilities. The number of defective components selected is either zero, one, or two. Therefore,
                  $$Pr(textexactly two defective components) = 1 - Pr(textno defective components) - Pr(textexactly one defective component)$$







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited 3 hours ago

























                  answered 4 hours ago









                  N. F. Taussig

                  40.8k93253




                  40.8k93253




















                      up vote
                      3
                      down vote













                      The probability on $kin0,1,2$ defectives among the $4$ chosen is:$$fracbinom4kbinom82-kbinom122$$
                      Hypergeometric distribution is used.



                      Actually this approach turns things around: there are $4$ chosen items and $8$ non-chosen items. Selecting $2$ items (the defective ones) what is the probability that $k$ of them will come from the chosen ones?






                      share|cite|improve this answer
















                      • 1




                        An interesting approach.
                        – N. F. Taussig
                        3 hours ago














                      up vote
                      3
                      down vote













                      The probability on $kin0,1,2$ defectives among the $4$ chosen is:$$fracbinom4kbinom82-kbinom122$$
                      Hypergeometric distribution is used.



                      Actually this approach turns things around: there are $4$ chosen items and $8$ non-chosen items. Selecting $2$ items (the defective ones) what is the probability that $k$ of them will come from the chosen ones?






                      share|cite|improve this answer
















                      • 1




                        An interesting approach.
                        – N. F. Taussig
                        3 hours ago












                      up vote
                      3
                      down vote










                      up vote
                      3
                      down vote









                      The probability on $kin0,1,2$ defectives among the $4$ chosen is:$$fracbinom4kbinom82-kbinom122$$
                      Hypergeometric distribution is used.



                      Actually this approach turns things around: there are $4$ chosen items and $8$ non-chosen items. Selecting $2$ items (the defective ones) what is the probability that $k$ of them will come from the chosen ones?






                      share|cite|improve this answer












                      The probability on $kin0,1,2$ defectives among the $4$ chosen is:$$fracbinom4kbinom82-kbinom122$$
                      Hypergeometric distribution is used.



                      Actually this approach turns things around: there are $4$ chosen items and $8$ non-chosen items. Selecting $2$ items (the defective ones) what is the probability that $k$ of them will come from the chosen ones?







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered 4 hours ago









                      drhab

                      90.5k542124




                      90.5k542124







                      • 1




                        An interesting approach.
                        – N. F. Taussig
                        3 hours ago












                      • 1




                        An interesting approach.
                        – N. F. Taussig
                        3 hours ago







                      1




                      1




                      An interesting approach.
                      – N. F. Taussig
                      3 hours ago




                      An interesting approach.
                      – N. F. Taussig
                      3 hours ago










                      up vote
                      1
                      down vote













                      What you have so far is good. To simplify case 2, you can notice that the probabilities are all the same: for each one you get $12times11times10times 9$ on the bottom and $10times9times8times 2$ on the top in some order. So the total probability is just the probability for a specific "pattern" (e.g. defective, good, good, good) times the number of patterns. The same approach works for 3: here there are $6=binom 42$ patterns, each with probability $frac1012timesfrac911timesfrac210timesfrac19$. This works because the first "good" (wherever you put it) always has $10$ on top, the second "good" always has $9$, and so on.






                      share|cite|improve this answer
























                        up vote
                        1
                        down vote













                        What you have so far is good. To simplify case 2, you can notice that the probabilities are all the same: for each one you get $12times11times10times 9$ on the bottom and $10times9times8times 2$ on the top in some order. So the total probability is just the probability for a specific "pattern" (e.g. defective, good, good, good) times the number of patterns. The same approach works for 3: here there are $6=binom 42$ patterns, each with probability $frac1012timesfrac911timesfrac210timesfrac19$. This works because the first "good" (wherever you put it) always has $10$ on top, the second "good" always has $9$, and so on.






                        share|cite|improve this answer






















                          up vote
                          1
                          down vote










                          up vote
                          1
                          down vote









                          What you have so far is good. To simplify case 2, you can notice that the probabilities are all the same: for each one you get $12times11times10times 9$ on the bottom and $10times9times8times 2$ on the top in some order. So the total probability is just the probability for a specific "pattern" (e.g. defective, good, good, good) times the number of patterns. The same approach works for 3: here there are $6=binom 42$ patterns, each with probability $frac1012timesfrac911timesfrac210timesfrac19$. This works because the first "good" (wherever you put it) always has $10$ on top, the second "good" always has $9$, and so on.






                          share|cite|improve this answer












                          What you have so far is good. To simplify case 2, you can notice that the probabilities are all the same: for each one you get $12times11times10times 9$ on the bottom and $10times9times8times 2$ on the top in some order. So the total probability is just the probability for a specific "pattern" (e.g. defective, good, good, good) times the number of patterns. The same approach works for 3: here there are $6=binom 42$ patterns, each with probability $frac1012timesfrac911timesfrac210timesfrac19$. This works because the first "good" (wherever you put it) always has $10$ on top, the second "good" always has $9$, and so on.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered 4 hours ago









                          Especially Lime

                          20.1k22554




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