Probability of exactly one defective unit











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2
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Assume 5 out of 100 units are defective. We pick 3 out of the 100 units at random.



What is the probability that exactly one unit is defective?




My answer would be



$P(text{Defect}=1) = P(text{Defect})times P(text{Not defect})times P(text{Not defect}) = 5/100 times 95/99 times 94/98$



However, I am not sure whether or not this is correct or not. Can someone verify?






probability







share|cite|improve this question




















  • 1




    You can choose the defective in three different ways. Also the number of non-defective units is $95$ not $94$.
    – Thomas Shelby
    17 hours ago






  • 1




    Sorry, my mistake, I meant 95/99 and 94/98. But then are you saying that I have to calculate like this: P(D=1) = {DDN}U{DND}U{NDD}, where D is af defective unit and N is a non-defective unit?
    – CruZ
    17 hours ago






  • 2




    Yes. You will get the same answer mentioned below.
    – Thomas Shelby
    17 hours ago

















up vote
2
down vote

favorite












Assume 5 out of 100 units are defective. We pick 3 out of the 100 units at random.



What is the probability that exactly one unit is defective?




My answer would be



$P(text{Defect}=1) = P(text{Defect})times P(text{Not defect})times P(text{Not defect}) = 5/100 times 95/99 times 94/98$



However, I am not sure whether or not this is correct or not. Can someone verify?






probability







share|cite|improve this question




















  • 1




    You can choose the defective in three different ways. Also the number of non-defective units is $95$ not $94$.
    – Thomas Shelby
    17 hours ago






  • 1




    Sorry, my mistake, I meant 95/99 and 94/98. But then are you saying that I have to calculate like this: P(D=1) = {DDN}U{DND}U{NDD}, where D is af defective unit and N is a non-defective unit?
    – CruZ
    17 hours ago






  • 2




    Yes. You will get the same answer mentioned below.
    – Thomas Shelby
    17 hours ago















up vote
2
down vote

favorite









up vote
2
down vote

favorite











Assume 5 out of 100 units are defective. We pick 3 out of the 100 units at random.



What is the probability that exactly one unit is defective?




My answer would be



$P(text{Defect}=1) = P(text{Defect})times P(text{Not defect})times P(text{Not defect}) = 5/100 times 95/99 times 94/98$



However, I am not sure whether or not this is correct or not. Can someone verify?






probability







share|cite|improve this question















Assume 5 out of 100 units are defective. We pick 3 out of the 100 units at random.



What is the probability that exactly one unit is defective?




My answer would be



$P(text{Defect}=1) = P(text{Defect})times P(text{Not defect})times P(text{Not defect}) = 5/100 times 95/99 times 94/98$



However, I am not sure whether or not this is correct or not. Can someone verify?






probability




probability






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 15 hours ago









amWhy

191k27223439




191k27223439










asked 17 hours ago









CruZ

396




396








  • 1




    You can choose the defective in three different ways. Also the number of non-defective units is $95$ not $94$.
    – Thomas Shelby
    17 hours ago






  • 1




    Sorry, my mistake, I meant 95/99 and 94/98. But then are you saying that I have to calculate like this: P(D=1) = {DDN}U{DND}U{NDD}, where D is af defective unit and N is a non-defective unit?
    – CruZ
    17 hours ago






  • 2




    Yes. You will get the same answer mentioned below.
    – Thomas Shelby
    17 hours ago
















  • 1




    You can choose the defective in three different ways. Also the number of non-defective units is $95$ not $94$.
    – Thomas Shelby
    17 hours ago






  • 1




    Sorry, my mistake, I meant 95/99 and 94/98. But then are you saying that I have to calculate like this: P(D=1) = {DDN}U{DND}U{NDD}, where D is af defective unit and N is a non-defective unit?
    – CruZ
    17 hours ago






  • 2




    Yes. You will get the same answer mentioned below.
    – Thomas Shelby
    17 hours ago










1




1




You can choose the defective in three different ways. Also the number of non-defective units is $95$ not $94$.
– Thomas Shelby
17 hours ago




You can choose the defective in three different ways. Also the number of non-defective units is $95$ not $94$.
– Thomas Shelby
17 hours ago




1




1




Sorry, my mistake, I meant 95/99 and 94/98. But then are you saying that I have to calculate like this: P(D=1) = {DDN}U{DND}U{NDD}, where D is af defective unit and N is a non-defective unit?
– CruZ
17 hours ago




Sorry, my mistake, I meant 95/99 and 94/98. But then are you saying that I have to calculate like this: P(D=1) = {DDN}U{DND}U{NDD}, where D is af defective unit and N is a non-defective unit?
– CruZ
17 hours ago




2




2




Yes. You will get the same answer mentioned below.
– Thomas Shelby
17 hours ago






Yes. You will get the same answer mentioned below.
– Thomas Shelby
17 hours ago












2 Answers
2






active

oldest

votes

















up vote
7
down vote



accepted










Your answer should be $$frac{binom{5}{1}binom{95}{2}}{binom{100}{3}}$$ Since we want the total number of ways to choose 3 meeting the criteria over the total number of ways to choose 3 out of the 100.






share|cite|improve this answer





















  • I don't understand why it's 500 instead of 100? But I didn't think of the binomial distribution, thank you!
    – CruZ
    17 hours ago






  • 1




    sorry, the 500 was a typo...
    – BelowAverageIntelligence
    17 hours ago






  • 1




    @CruZ This isn't the binomial distribution, it's actually called the hypergeometric distribution.
    – Quintec
    13 hours ago










  • Ah thank you for clearing that up, my mistake! Cheers!
    – CruZ
    10 hours ago


















up vote
7
down vote













Here is a suggestion how to proceed as ordering does not play a role




  • Choose one defective item: $binom{5}{1}$

  • Choose two non-defective ones: $binom{95}{2}$

  • Chose any three: $binom{100}{3}$
    $$P(mbox{"exactly 1 defective"}) = frac{binom{5}{1}cdot binom{95}{2}}{binom{100}{3}}$$






share|cite|improve this answer





















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    2 Answers
    2






    active

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    2 Answers
    2






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    active

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    up vote
    7
    down vote



    accepted










    Your answer should be $$frac{binom{5}{1}binom{95}{2}}{binom{100}{3}}$$ Since we want the total number of ways to choose 3 meeting the criteria over the total number of ways to choose 3 out of the 100.






    share|cite|improve this answer





















    • I don't understand why it's 500 instead of 100? But I didn't think of the binomial distribution, thank you!
      – CruZ
      17 hours ago






    • 1




      sorry, the 500 was a typo...
      – BelowAverageIntelligence
      17 hours ago






    • 1




      @CruZ This isn't the binomial distribution, it's actually called the hypergeometric distribution.
      – Quintec
      13 hours ago










    • Ah thank you for clearing that up, my mistake! Cheers!
      – CruZ
      10 hours ago















    up vote
    7
    down vote



    accepted










    Your answer should be $$frac{binom{5}{1}binom{95}{2}}{binom{100}{3}}$$ Since we want the total number of ways to choose 3 meeting the criteria over the total number of ways to choose 3 out of the 100.






    share|cite|improve this answer





















    • I don't understand why it's 500 instead of 100? But I didn't think of the binomial distribution, thank you!
      – CruZ
      17 hours ago






    • 1




      sorry, the 500 was a typo...
      – BelowAverageIntelligence
      17 hours ago






    • 1




      @CruZ This isn't the binomial distribution, it's actually called the hypergeometric distribution.
      – Quintec
      13 hours ago










    • Ah thank you for clearing that up, my mistake! Cheers!
      – CruZ
      10 hours ago













    up vote
    7
    down vote



    accepted







    up vote
    7
    down vote



    accepted






    Your answer should be $$frac{binom{5}{1}binom{95}{2}}{binom{100}{3}}$$ Since we want the total number of ways to choose 3 meeting the criteria over the total number of ways to choose 3 out of the 100.






    share|cite|improve this answer












    Your answer should be $$frac{binom{5}{1}binom{95}{2}}{binom{100}{3}}$$ Since we want the total number of ways to choose 3 meeting the criteria over the total number of ways to choose 3 out of the 100.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered 17 hours ago









    BelowAverageIntelligence

    5121213




    5121213












    • I don't understand why it's 500 instead of 100? But I didn't think of the binomial distribution, thank you!
      – CruZ
      17 hours ago






    • 1




      sorry, the 500 was a typo...
      – BelowAverageIntelligence
      17 hours ago






    • 1




      @CruZ This isn't the binomial distribution, it's actually called the hypergeometric distribution.
      – Quintec
      13 hours ago










    • Ah thank you for clearing that up, my mistake! Cheers!
      – CruZ
      10 hours ago


















    • I don't understand why it's 500 instead of 100? But I didn't think of the binomial distribution, thank you!
      – CruZ
      17 hours ago






    • 1




      sorry, the 500 was a typo...
      – BelowAverageIntelligence
      17 hours ago






    • 1




      @CruZ This isn't the binomial distribution, it's actually called the hypergeometric distribution.
      – Quintec
      13 hours ago










    • Ah thank you for clearing that up, my mistake! Cheers!
      – CruZ
      10 hours ago
















    I don't understand why it's 500 instead of 100? But I didn't think of the binomial distribution, thank you!
    – CruZ
    17 hours ago




    I don't understand why it's 500 instead of 100? But I didn't think of the binomial distribution, thank you!
    – CruZ
    17 hours ago




    1




    1




    sorry, the 500 was a typo...
    – BelowAverageIntelligence
    17 hours ago




    sorry, the 500 was a typo...
    – BelowAverageIntelligence
    17 hours ago




    1




    1




    @CruZ This isn't the binomial distribution, it's actually called the hypergeometric distribution.
    – Quintec
    13 hours ago




    @CruZ This isn't the binomial distribution, it's actually called the hypergeometric distribution.
    – Quintec
    13 hours ago












    Ah thank you for clearing that up, my mistake! Cheers!
    – CruZ
    10 hours ago




    Ah thank you for clearing that up, my mistake! Cheers!
    – CruZ
    10 hours ago










    up vote
    7
    down vote













    Here is a suggestion how to proceed as ordering does not play a role




    • Choose one defective item: $binom{5}{1}$

    • Choose two non-defective ones: $binom{95}{2}$

    • Chose any three: $binom{100}{3}$
      $$P(mbox{"exactly 1 defective"}) = frac{binom{5}{1}cdot binom{95}{2}}{binom{100}{3}}$$






    share|cite|improve this answer

























      up vote
      7
      down vote













      Here is a suggestion how to proceed as ordering does not play a role




      • Choose one defective item: $binom{5}{1}$

      • Choose two non-defective ones: $binom{95}{2}$

      • Chose any three: $binom{100}{3}$
        $$P(mbox{"exactly 1 defective"}) = frac{binom{5}{1}cdot binom{95}{2}}{binom{100}{3}}$$






      share|cite|improve this answer























        up vote
        7
        down vote










        up vote
        7
        down vote









        Here is a suggestion how to proceed as ordering does not play a role




        • Choose one defective item: $binom{5}{1}$

        • Choose two non-defective ones: $binom{95}{2}$

        • Chose any three: $binom{100}{3}$
          $$P(mbox{"exactly 1 defective"}) = frac{binom{5}{1}cdot binom{95}{2}}{binom{100}{3}}$$






        share|cite|improve this answer












        Here is a suggestion how to proceed as ordering does not play a role




        • Choose one defective item: $binom{5}{1}$

        • Choose two non-defective ones: $binom{95}{2}$

        • Chose any three: $binom{100}{3}$
          $$P(mbox{"exactly 1 defective"}) = frac{binom{5}{1}cdot binom{95}{2}}{binom{100}{3}}$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 17 hours ago









        trancelocation

        8,7121520




        8,7121520






























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