Calculating theoretical probability












-4












$begingroup$


I have the following question in one of my tutorial.



Background: A manufacturing company developed 40000 new drugs and they need to be tested. Question The QA checks on the previous batches of drugs found that — it is four times more likely that a drug is able to produce a better result than not. If we take a sample of ten drugs, we need to find the theoretical probability that at most 3 drugs are not able to do a satisfactory job.



I think we need to use the cumulative probability distribution $F(3) = P(Xlt 3).$ However not sure how to calculate it. Any guidance is helpful.






probability probability-distributions







share|cite|improve this question











$endgroup$












  • $begingroup$
    $P(Xleq3) = sum_{0leq ileq3} P(X = i)$ Is that all you're asking? What have you tried?
    $endgroup$
    – learning
    Dec 18 '18 at 14:38








  • 1




    $begingroup$
    "At most" means you want $P(Xleq 3).$
    $endgroup$
    – David K
    Dec 18 '18 at 14:39










  • $begingroup$
    yes. P(X≤3)= P(X=0) + P(X=1) + P(X=2) + P(X=3). But what could be P(X=0), P(X=1) values. should i treat this as independent event or dependent event?
    $endgroup$
    – user1618711
    Dec 18 '18 at 15:05
















-4












$begingroup$


I have the following question in one of my tutorial.



Background: A manufacturing company developed 40000 new drugs and they need to be tested. Question The QA checks on the previous batches of drugs found that — it is four times more likely that a drug is able to produce a better result than not. If we take a sample of ten drugs, we need to find the theoretical probability that at most 3 drugs are not able to do a satisfactory job.



I think we need to use the cumulative probability distribution $F(3) = P(Xlt 3).$ However not sure how to calculate it. Any guidance is helpful.






probability probability-distributions







share|cite|improve this question











$endgroup$












  • $begingroup$
    $P(Xleq3) = sum_{0leq ileq3} P(X = i)$ Is that all you're asking? What have you tried?
    $endgroup$
    – learning
    Dec 18 '18 at 14:38








  • 1




    $begingroup$
    "At most" means you want $P(Xleq 3).$
    $endgroup$
    – David K
    Dec 18 '18 at 14:39










  • $begingroup$
    yes. P(X≤3)= P(X=0) + P(X=1) + P(X=2) + P(X=3). But what could be P(X=0), P(X=1) values. should i treat this as independent event or dependent event?
    $endgroup$
    – user1618711
    Dec 18 '18 at 15:05














-4












-4








-4





$begingroup$


I have the following question in one of my tutorial.



Background: A manufacturing company developed 40000 new drugs and they need to be tested. Question The QA checks on the previous batches of drugs found that — it is four times more likely that a drug is able to produce a better result than not. If we take a sample of ten drugs, we need to find the theoretical probability that at most 3 drugs are not able to do a satisfactory job.



I think we need to use the cumulative probability distribution $F(3) = P(Xlt 3).$ However not sure how to calculate it. Any guidance is helpful.






probability probability-distributions







share|cite|improve this question











$endgroup$




I have the following question in one of my tutorial.



Background: A manufacturing company developed 40000 new drugs and they need to be tested. Question The QA checks on the previous batches of drugs found that — it is four times more likely that a drug is able to produce a better result than not. If we take a sample of ten drugs, we need to find the theoretical probability that at most 3 drugs are not able to do a satisfactory job.



I think we need to use the cumulative probability distribution $F(3) = P(Xlt 3).$ However not sure how to calculate it. Any guidance is helpful.






probability probability-distributions




probability probability-distributions






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 18 '18 at 14:48









amWhy

192k28225439




192k28225439










asked Dec 18 '18 at 14:33









user1618711user1618711

31




31












  • $begingroup$
    $P(Xleq3) = sum_{0leq ileq3} P(X = i)$ Is that all you're asking? What have you tried?
    $endgroup$
    – learning
    Dec 18 '18 at 14:38








  • 1




    $begingroup$
    "At most" means you want $P(Xleq 3).$
    $endgroup$
    – David K
    Dec 18 '18 at 14:39










  • $begingroup$
    yes. P(X≤3)= P(X=0) + P(X=1) + P(X=2) + P(X=3). But what could be P(X=0), P(X=1) values. should i treat this as independent event or dependent event?
    $endgroup$
    – user1618711
    Dec 18 '18 at 15:05


















  • $begingroup$
    $P(Xleq3) = sum_{0leq ileq3} P(X = i)$ Is that all you're asking? What have you tried?
    $endgroup$
    – learning
    Dec 18 '18 at 14:38








  • 1




    $begingroup$
    "At most" means you want $P(Xleq 3).$
    $endgroup$
    – David K
    Dec 18 '18 at 14:39










  • $begingroup$
    yes. P(X≤3)= P(X=0) + P(X=1) + P(X=2) + P(X=3). But what could be P(X=0), P(X=1) values. should i treat this as independent event or dependent event?
    $endgroup$
    – user1618711
    Dec 18 '18 at 15:05
















$begingroup$
$P(Xleq3) = sum_{0leq ileq3} P(X = i)$ Is that all you're asking? What have you tried?
$endgroup$
– learning
Dec 18 '18 at 14:38






$begingroup$
$P(Xleq3) = sum_{0leq ileq3} P(X = i)$ Is that all you're asking? What have you tried?
$endgroup$
– learning
Dec 18 '18 at 14:38






1




1




$begingroup$
"At most" means you want $P(Xleq 3).$
$endgroup$
– David K
Dec 18 '18 at 14:39




$begingroup$
"At most" means you want $P(Xleq 3).$
$endgroup$
– David K
Dec 18 '18 at 14:39












$begingroup$
yes. P(X≤3)= P(X=0) + P(X=1) + P(X=2) + P(X=3). But what could be P(X=0), P(X=1) values. should i treat this as independent event or dependent event?
$endgroup$
– user1618711
Dec 18 '18 at 15:05




$begingroup$
yes. P(X≤3)= P(X=0) + P(X=1) + P(X=2) + P(X=3). But what could be P(X=0), P(X=1) values. should i treat this as independent event or dependent event?
$endgroup$
– user1618711
Dec 18 '18 at 15:05










2 Answers
2






active

oldest

votes


















1












$begingroup$

If you let $G$ be the event that a drug produces better results and you let $B$ be the event that a drug produces not as good of results, then the first step in your question is to solve $4P(B)=P(G)$ with the condition that $P(B) + P(G)=1$. From this, we get $P(B)=frac{1}{5}$ and $P(G)=frac{4}{5}$.



Accordingly, if you let $X$ be the event that out of 10 trials, you pick $x$ amount of drugs that do not produce as good of results as the other drug, then $X$ is a random variable with a Binomial Distribution, where $n=10$ and $p=1/5$.



So, $P(X=x)={10 choose x}frac{1}{5}^{x}frac{4}{5}^{10-x}$, for $0 leq x leq 10$.



All that is left is to compute $P(X leq 3)=P(X=0)+P(X=1)+P(X=2)+P(X=3)=sum_{i=0}^{3}{10 choose i}frac{1}{5}^{i}frac{4}{5}^{10-i}$






share|cite|improve this answer











$endgroup$





















    0












    $begingroup$

    The fact that it is a sample means that the event of each being good can be assumed to be independent from each other. This is enough information to calculate the odds of any particular pattern of good vs bad. For example GBGGBGGGBG will turn up with probability $0.8 * 0.2 * 0.8 * 0.8 * 0.2 * 0.8 * 0.8 * 0.8 * 0.2 * 0.8 = 0.2^3 * 0.8^7$



    Now you just have to sum up the probabilities of seeing all patterns that satisfy your rule.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      however, there are ${10 choose 3}$ ways that you can make a chain of a total of ten $G$ and $B$, where there are exactly seven $G$s and three $B$s.
      $endgroup$
      – Live Free or π Hard
      Dec 19 '18 at 2:32








    • 1




      $begingroup$
      @LiveFreeorπHard I do like leaving some work to do for clear homework problems...
      $endgroup$
      – btilly
      Dec 19 '18 at 3:17










    • $begingroup$
      Lol, good point. 🙂 I was just concerned that the poster wouldn’t have seen that combinations, even though you mentioned it in your last sentence.
      $endgroup$
      – Live Free or π Hard
      Dec 19 '18 at 15:56











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    2 Answers
    2






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    2 Answers
    2






    active

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    active

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    active

    oldest

    votes









    1












    $begingroup$

    If you let $G$ be the event that a drug produces better results and you let $B$ be the event that a drug produces not as good of results, then the first step in your question is to solve $4P(B)=P(G)$ with the condition that $P(B) + P(G)=1$. From this, we get $P(B)=frac{1}{5}$ and $P(G)=frac{4}{5}$.



    Accordingly, if you let $X$ be the event that out of 10 trials, you pick $x$ amount of drugs that do not produce as good of results as the other drug, then $X$ is a random variable with a Binomial Distribution, where $n=10$ and $p=1/5$.



    So, $P(X=x)={10 choose x}frac{1}{5}^{x}frac{4}{5}^{10-x}$, for $0 leq x leq 10$.



    All that is left is to compute $P(X leq 3)=P(X=0)+P(X=1)+P(X=2)+P(X=3)=sum_{i=0}^{3}{10 choose i}frac{1}{5}^{i}frac{4}{5}^{10-i}$






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      If you let $G$ be the event that a drug produces better results and you let $B$ be the event that a drug produces not as good of results, then the first step in your question is to solve $4P(B)=P(G)$ with the condition that $P(B) + P(G)=1$. From this, we get $P(B)=frac{1}{5}$ and $P(G)=frac{4}{5}$.



      Accordingly, if you let $X$ be the event that out of 10 trials, you pick $x$ amount of drugs that do not produce as good of results as the other drug, then $X$ is a random variable with a Binomial Distribution, where $n=10$ and $p=1/5$.



      So, $P(X=x)={10 choose x}frac{1}{5}^{x}frac{4}{5}^{10-x}$, for $0 leq x leq 10$.



      All that is left is to compute $P(X leq 3)=P(X=0)+P(X=1)+P(X=2)+P(X=3)=sum_{i=0}^{3}{10 choose i}frac{1}{5}^{i}frac{4}{5}^{10-i}$






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        If you let $G$ be the event that a drug produces better results and you let $B$ be the event that a drug produces not as good of results, then the first step in your question is to solve $4P(B)=P(G)$ with the condition that $P(B) + P(G)=1$. From this, we get $P(B)=frac{1}{5}$ and $P(G)=frac{4}{5}$.



        Accordingly, if you let $X$ be the event that out of 10 trials, you pick $x$ amount of drugs that do not produce as good of results as the other drug, then $X$ is a random variable with a Binomial Distribution, where $n=10$ and $p=1/5$.



        So, $P(X=x)={10 choose x}frac{1}{5}^{x}frac{4}{5}^{10-x}$, for $0 leq x leq 10$.



        All that is left is to compute $P(X leq 3)=P(X=0)+P(X=1)+P(X=2)+P(X=3)=sum_{i=0}^{3}{10 choose i}frac{1}{5}^{i}frac{4}{5}^{10-i}$






        share|cite|improve this answer











        $endgroup$



        If you let $G$ be the event that a drug produces better results and you let $B$ be the event that a drug produces not as good of results, then the first step in your question is to solve $4P(B)=P(G)$ with the condition that $P(B) + P(G)=1$. From this, we get $P(B)=frac{1}{5}$ and $P(G)=frac{4}{5}$.



        Accordingly, if you let $X$ be the event that out of 10 trials, you pick $x$ amount of drugs that do not produce as good of results as the other drug, then $X$ is a random variable with a Binomial Distribution, where $n=10$ and $p=1/5$.



        So, $P(X=x)={10 choose x}frac{1}{5}^{x}frac{4}{5}^{10-x}$, for $0 leq x leq 10$.



        All that is left is to compute $P(X leq 3)=P(X=0)+P(X=1)+P(X=2)+P(X=3)=sum_{i=0}^{3}{10 choose i}frac{1}{5}^{i}frac{4}{5}^{10-i}$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 19 '18 at 2:35

























        answered Dec 19 '18 at 2:24









        Live Free or π HardLive Free or π Hard

        479213




        479213























            0












            $begingroup$

            The fact that it is a sample means that the event of each being good can be assumed to be independent from each other. This is enough information to calculate the odds of any particular pattern of good vs bad. For example GBGGBGGGBG will turn up with probability $0.8 * 0.2 * 0.8 * 0.8 * 0.2 * 0.8 * 0.8 * 0.8 * 0.2 * 0.8 = 0.2^3 * 0.8^7$



            Now you just have to sum up the probabilities of seeing all patterns that satisfy your rule.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              however, there are ${10 choose 3}$ ways that you can make a chain of a total of ten $G$ and $B$, where there are exactly seven $G$s and three $B$s.
              $endgroup$
              – Live Free or π Hard
              Dec 19 '18 at 2:32








            • 1




              $begingroup$
              @LiveFreeorπHard I do like leaving some work to do for clear homework problems...
              $endgroup$
              – btilly
              Dec 19 '18 at 3:17










            • $begingroup$
              Lol, good point. 🙂 I was just concerned that the poster wouldn’t have seen that combinations, even though you mentioned it in your last sentence.
              $endgroup$
              – Live Free or π Hard
              Dec 19 '18 at 15:56
















            0












            $begingroup$

            The fact that it is a sample means that the event of each being good can be assumed to be independent from each other. This is enough information to calculate the odds of any particular pattern of good vs bad. For example GBGGBGGGBG will turn up with probability $0.8 * 0.2 * 0.8 * 0.8 * 0.2 * 0.8 * 0.8 * 0.8 * 0.2 * 0.8 = 0.2^3 * 0.8^7$



            Now you just have to sum up the probabilities of seeing all patterns that satisfy your rule.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              however, there are ${10 choose 3}$ ways that you can make a chain of a total of ten $G$ and $B$, where there are exactly seven $G$s and three $B$s.
              $endgroup$
              – Live Free or π Hard
              Dec 19 '18 at 2:32








            • 1




              $begingroup$
              @LiveFreeorπHard I do like leaving some work to do for clear homework problems...
              $endgroup$
              – btilly
              Dec 19 '18 at 3:17










            • $begingroup$
              Lol, good point. 🙂 I was just concerned that the poster wouldn’t have seen that combinations, even though you mentioned it in your last sentence.
              $endgroup$
              – Live Free or π Hard
              Dec 19 '18 at 15:56














            0












            0








            0





            $begingroup$

            The fact that it is a sample means that the event of each being good can be assumed to be independent from each other. This is enough information to calculate the odds of any particular pattern of good vs bad. For example GBGGBGGGBG will turn up with probability $0.8 * 0.2 * 0.8 * 0.8 * 0.2 * 0.8 * 0.8 * 0.8 * 0.2 * 0.8 = 0.2^3 * 0.8^7$



            Now you just have to sum up the probabilities of seeing all patterns that satisfy your rule.






            share|cite|improve this answer









            $endgroup$



            The fact that it is a sample means that the event of each being good can be assumed to be independent from each other. This is enough information to calculate the odds of any particular pattern of good vs bad. For example GBGGBGGGBG will turn up with probability $0.8 * 0.2 * 0.8 * 0.8 * 0.2 * 0.8 * 0.8 * 0.8 * 0.2 * 0.8 = 0.2^3 * 0.8^7$



            Now you just have to sum up the probabilities of seeing all patterns that satisfy your rule.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 18 '18 at 22:11









            btillybtilly

            94068




            94068












            • $begingroup$
              however, there are ${10 choose 3}$ ways that you can make a chain of a total of ten $G$ and $B$, where there are exactly seven $G$s and three $B$s.
              $endgroup$
              – Live Free or π Hard
              Dec 19 '18 at 2:32








            • 1




              $begingroup$
              @LiveFreeorπHard I do like leaving some work to do for clear homework problems...
              $endgroup$
              – btilly
              Dec 19 '18 at 3:17










            • $begingroup$
              Lol, good point. 🙂 I was just concerned that the poster wouldn’t have seen that combinations, even though you mentioned it in your last sentence.
              $endgroup$
              – Live Free or π Hard
              Dec 19 '18 at 15:56


















            • $begingroup$
              however, there are ${10 choose 3}$ ways that you can make a chain of a total of ten $G$ and $B$, where there are exactly seven $G$s and three $B$s.
              $endgroup$
              – Live Free or π Hard
              Dec 19 '18 at 2:32








            • 1




              $begingroup$
              @LiveFreeorπHard I do like leaving some work to do for clear homework problems...
              $endgroup$
              – btilly
              Dec 19 '18 at 3:17










            • $begingroup$
              Lol, good point. 🙂 I was just concerned that the poster wouldn’t have seen that combinations, even though you mentioned it in your last sentence.
              $endgroup$
              – Live Free or π Hard
              Dec 19 '18 at 15:56
















            $begingroup$
            however, there are ${10 choose 3}$ ways that you can make a chain of a total of ten $G$ and $B$, where there are exactly seven $G$s and three $B$s.
            $endgroup$
            – Live Free or π Hard
            Dec 19 '18 at 2:32






            $begingroup$
            however, there are ${10 choose 3}$ ways that you can make a chain of a total of ten $G$ and $B$, where there are exactly seven $G$s and three $B$s.
            $endgroup$
            – Live Free or π Hard
            Dec 19 '18 at 2:32






            1




            1




            $begingroup$
            @LiveFreeorπHard I do like leaving some work to do for clear homework problems...
            $endgroup$
            – btilly
            Dec 19 '18 at 3:17




            $begingroup$
            @LiveFreeorπHard I do like leaving some work to do for clear homework problems...
            $endgroup$
            – btilly
            Dec 19 '18 at 3:17












            $begingroup$
            Lol, good point. 🙂 I was just concerned that the poster wouldn’t have seen that combinations, even though you mentioned it in your last sentence.
            $endgroup$
            – Live Free or π Hard
            Dec 19 '18 at 15:56




            $begingroup$
            Lol, good point. 🙂 I was just concerned that the poster wouldn’t have seen that combinations, even though you mentioned it in your last sentence.
            $endgroup$
            – Live Free or π Hard
            Dec 19 '18 at 15:56


















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            .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid ...
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