Probability of exactly one defective unit












2












$begingroup$


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











$endgroup$








  • 1




    $begingroup$
    You can choose the defective in three different ways. Also the number of non-defective units is $95$ not $94$.
    $endgroup$
    – Thomas Shelby
    Dec 9 '18 at 10:49






  • 1




    $begingroup$
    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?
    $endgroup$
    – CruZ
    Dec 9 '18 at 10:53






  • 2




    $begingroup$
    Yes. You will get the same answer mentioned below.
    $endgroup$
    – Thomas Shelby
    Dec 9 '18 at 11:03


















2












$begingroup$


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











$endgroup$








  • 1




    $begingroup$
    You can choose the defective in three different ways. Also the number of non-defective units is $95$ not $94$.
    $endgroup$
    – Thomas Shelby
    Dec 9 '18 at 10:49






  • 1




    $begingroup$
    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?
    $endgroup$
    – CruZ
    Dec 9 '18 at 10:53






  • 2




    $begingroup$
    Yes. You will get the same answer mentioned below.
    $endgroup$
    – Thomas Shelby
    Dec 9 '18 at 11:03
















2












2








2





$begingroup$


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











$endgroup$




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 Dec 9 '18 at 13:33









amWhy

1




1










asked Dec 9 '18 at 10:45









CruZCruZ

406




406








  • 1




    $begingroup$
    You can choose the defective in three different ways. Also the number of non-defective units is $95$ not $94$.
    $endgroup$
    – Thomas Shelby
    Dec 9 '18 at 10:49






  • 1




    $begingroup$
    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?
    $endgroup$
    – CruZ
    Dec 9 '18 at 10:53






  • 2




    $begingroup$
    Yes. You will get the same answer mentioned below.
    $endgroup$
    – Thomas Shelby
    Dec 9 '18 at 11:03
















  • 1




    $begingroup$
    You can choose the defective in three different ways. Also the number of non-defective units is $95$ not $94$.
    $endgroup$
    – Thomas Shelby
    Dec 9 '18 at 10:49






  • 1




    $begingroup$
    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?
    $endgroup$
    – CruZ
    Dec 9 '18 at 10:53






  • 2




    $begingroup$
    Yes. You will get the same answer mentioned below.
    $endgroup$
    – Thomas Shelby
    Dec 9 '18 at 11:03










1




1




$begingroup$
You can choose the defective in three different ways. Also the number of non-defective units is $95$ not $94$.
$endgroup$
– Thomas Shelby
Dec 9 '18 at 10:49




$begingroup$
You can choose the defective in three different ways. Also the number of non-defective units is $95$ not $94$.
$endgroup$
– Thomas Shelby
Dec 9 '18 at 10:49




1




1




$begingroup$
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?
$endgroup$
– CruZ
Dec 9 '18 at 10:53




$begingroup$
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?
$endgroup$
– CruZ
Dec 9 '18 at 10:53




2




2




$begingroup$
Yes. You will get the same answer mentioned below.
$endgroup$
– Thomas Shelby
Dec 9 '18 at 11:03






$begingroup$
Yes. You will get the same answer mentioned below.
$endgroup$
– Thomas Shelby
Dec 9 '18 at 11:03












2 Answers
2






active

oldest

votes


















7












$begingroup$

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









$endgroup$













  • $begingroup$
    I don't understand why it's 500 instead of 100? But I didn't think of the binomial distribution, thank you!
    $endgroup$
    – CruZ
    Dec 9 '18 at 10:59






  • 1




    $begingroup$
    sorry, the 500 was a typo...
    $endgroup$
    – BelowAverageIntelligence
    Dec 9 '18 at 11:00






  • 1




    $begingroup$
    @CruZ This isn't the binomial distribution, it's actually called the hypergeometric distribution.
    $endgroup$
    – Quintec
    Dec 9 '18 at 14:59










  • $begingroup$
    Ah thank you for clearing that up, my mistake! Cheers!
    $endgroup$
    – CruZ
    Dec 9 '18 at 18:07



















7












$begingroup$

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









$endgroup$













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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    7












    $begingroup$

    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









    $endgroup$













    • $begingroup$
      I don't understand why it's 500 instead of 100? But I didn't think of the binomial distribution, thank you!
      $endgroup$
      – CruZ
      Dec 9 '18 at 10:59






    • 1




      $begingroup$
      sorry, the 500 was a typo...
      $endgroup$
      – BelowAverageIntelligence
      Dec 9 '18 at 11:00






    • 1




      $begingroup$
      @CruZ This isn't the binomial distribution, it's actually called the hypergeometric distribution.
      $endgroup$
      – Quintec
      Dec 9 '18 at 14:59










    • $begingroup$
      Ah thank you for clearing that up, my mistake! Cheers!
      $endgroup$
      – CruZ
      Dec 9 '18 at 18:07
















    7












    $begingroup$

    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









    $endgroup$













    • $begingroup$
      I don't understand why it's 500 instead of 100? But I didn't think of the binomial distribution, thank you!
      $endgroup$
      – CruZ
      Dec 9 '18 at 10:59






    • 1




      $begingroup$
      sorry, the 500 was a typo...
      $endgroup$
      – BelowAverageIntelligence
      Dec 9 '18 at 11:00






    • 1




      $begingroup$
      @CruZ This isn't the binomial distribution, it's actually called the hypergeometric distribution.
      $endgroup$
      – Quintec
      Dec 9 '18 at 14:59










    • $begingroup$
      Ah thank you for clearing that up, my mistake! Cheers!
      $endgroup$
      – CruZ
      Dec 9 '18 at 18:07














    7












    7








    7





    $begingroup$

    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









    $endgroup$



    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 Dec 9 '18 at 10:56









    BelowAverageIntelligenceBelowAverageIntelligence

    5021213




    5021213












    • $begingroup$
      I don't understand why it's 500 instead of 100? But I didn't think of the binomial distribution, thank you!
      $endgroup$
      – CruZ
      Dec 9 '18 at 10:59






    • 1




      $begingroup$
      sorry, the 500 was a typo...
      $endgroup$
      – BelowAverageIntelligence
      Dec 9 '18 at 11:00






    • 1




      $begingroup$
      @CruZ This isn't the binomial distribution, it's actually called the hypergeometric distribution.
      $endgroup$
      – Quintec
      Dec 9 '18 at 14:59










    • $begingroup$
      Ah thank you for clearing that up, my mistake! Cheers!
      $endgroup$
      – CruZ
      Dec 9 '18 at 18:07


















    • $begingroup$
      I don't understand why it's 500 instead of 100? But I didn't think of the binomial distribution, thank you!
      $endgroup$
      – CruZ
      Dec 9 '18 at 10:59






    • 1




      $begingroup$
      sorry, the 500 was a typo...
      $endgroup$
      – BelowAverageIntelligence
      Dec 9 '18 at 11:00






    • 1




      $begingroup$
      @CruZ This isn't the binomial distribution, it's actually called the hypergeometric distribution.
      $endgroup$
      – Quintec
      Dec 9 '18 at 14:59










    • $begingroup$
      Ah thank you for clearing that up, my mistake! Cheers!
      $endgroup$
      – CruZ
      Dec 9 '18 at 18:07
















    $begingroup$
    I don't understand why it's 500 instead of 100? But I didn't think of the binomial distribution, thank you!
    $endgroup$
    – CruZ
    Dec 9 '18 at 10:59




    $begingroup$
    I don't understand why it's 500 instead of 100? But I didn't think of the binomial distribution, thank you!
    $endgroup$
    – CruZ
    Dec 9 '18 at 10:59




    1




    1




    $begingroup$
    sorry, the 500 was a typo...
    $endgroup$
    – BelowAverageIntelligence
    Dec 9 '18 at 11:00




    $begingroup$
    sorry, the 500 was a typo...
    $endgroup$
    – BelowAverageIntelligence
    Dec 9 '18 at 11:00




    1




    1




    $begingroup$
    @CruZ This isn't the binomial distribution, it's actually called the hypergeometric distribution.
    $endgroup$
    – Quintec
    Dec 9 '18 at 14:59




    $begingroup$
    @CruZ This isn't the binomial distribution, it's actually called the hypergeometric distribution.
    $endgroup$
    – Quintec
    Dec 9 '18 at 14:59












    $begingroup$
    Ah thank you for clearing that up, my mistake! Cheers!
    $endgroup$
    – CruZ
    Dec 9 '18 at 18:07




    $begingroup$
    Ah thank you for clearing that up, my mistake! Cheers!
    $endgroup$
    – CruZ
    Dec 9 '18 at 18:07











    7












    $begingroup$

    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









    $endgroup$


















      7












      $begingroup$

      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









      $endgroup$
















        7












        7








        7





        $begingroup$

        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









        $endgroup$



        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 Dec 9 '18 at 10:55









        trancelocationtrancelocation

        9,9451722




        9,9451722






























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