Largest possible value of a probability












0












$begingroup$


Let's suppose we have $3$ trials each with same probability of success. Let $X$ denote the total number of successes in these trials and put $E(X) = 1.8$. Find the largest and smallest value of $P(X=3)$.



Try



Let $X_i$ be $1$ is trial $i$ is success and $0$ otherwise. We have $X=X_1+X_2+X_3$. Also, $P(X_1=1)=P(X_2=1)=P(X_3=1)=p$, is given. Also,



$$ 1.8 = E(X) = P(X=1) + 2 P(X=2) + 3 P(X=3) $$



notice $X=1$ only when one $X_i=1$. Thus, $P(X=3) = 3 P(X_i=1) = 3p$ and $P(X=2)$ when any two of $X_i=1$ that is $P(X=2) = 3p^2$. Thus,



$$ 1.8 = 3p + 6p^2 + 3P(X=3) $$



$$ P(X=3) = 1.8-p-2p^2 $$



take derivative with respect to $p$ gives



$$ (P(X=3))' = -1 - 4p $$



which means the maximum have to be at $p<0$ which makes no sense. what am I doing wrong?






probability







share|cite|improve this question











$endgroup$








  • 4




    $begingroup$
    Possible duplicate of $E[X]=1.8$, where $X$ is the total number of successes of 3 trials. What is the largest/smallest $P{X=3}$ can be?
    $endgroup$
    – LuxGiammi
    Nov 28 '18 at 19:32
















0












$begingroup$


Let's suppose we have $3$ trials each with same probability of success. Let $X$ denote the total number of successes in these trials and put $E(X) = 1.8$. Find the largest and smallest value of $P(X=3)$.



Try



Let $X_i$ be $1$ is trial $i$ is success and $0$ otherwise. We have $X=X_1+X_2+X_3$. Also, $P(X_1=1)=P(X_2=1)=P(X_3=1)=p$, is given. Also,



$$ 1.8 = E(X) = P(X=1) + 2 P(X=2) + 3 P(X=3) $$



notice $X=1$ only when one $X_i=1$. Thus, $P(X=3) = 3 P(X_i=1) = 3p$ and $P(X=2)$ when any two of $X_i=1$ that is $P(X=2) = 3p^2$. Thus,



$$ 1.8 = 3p + 6p^2 + 3P(X=3) $$



$$ P(X=3) = 1.8-p-2p^2 $$



take derivative with respect to $p$ gives



$$ (P(X=3))' = -1 - 4p $$



which means the maximum have to be at $p<0$ which makes no sense. what am I doing wrong?






probability







share|cite|improve this question











$endgroup$








  • 4




    $begingroup$
    Possible duplicate of $E[X]=1.8$, where $X$ is the total number of successes of 3 trials. What is the largest/smallest $P{X=3}$ can be?
    $endgroup$
    – LuxGiammi
    Nov 28 '18 at 19:32














0












0








0





$begingroup$


Let's suppose we have $3$ trials each with same probability of success. Let $X$ denote the total number of successes in these trials and put $E(X) = 1.8$. Find the largest and smallest value of $P(X=3)$.



Try



Let $X_i$ be $1$ is trial $i$ is success and $0$ otherwise. We have $X=X_1+X_2+X_3$. Also, $P(X_1=1)=P(X_2=1)=P(X_3=1)=p$, is given. Also,



$$ 1.8 = E(X) = P(X=1) + 2 P(X=2) + 3 P(X=3) $$



notice $X=1$ only when one $X_i=1$. Thus, $P(X=3) = 3 P(X_i=1) = 3p$ and $P(X=2)$ when any two of $X_i=1$ that is $P(X=2) = 3p^2$. Thus,



$$ 1.8 = 3p + 6p^2 + 3P(X=3) $$



$$ P(X=3) = 1.8-p-2p^2 $$



take derivative with respect to $p$ gives



$$ (P(X=3))' = -1 - 4p $$



which means the maximum have to be at $p<0$ which makes no sense. what am I doing wrong?






probability







share|cite|improve this question











$endgroup$




Let's suppose we have $3$ trials each with same probability of success. Let $X$ denote the total number of successes in these trials and put $E(X) = 1.8$. Find the largest and smallest value of $P(X=3)$.



Try



Let $X_i$ be $1$ is trial $i$ is success and $0$ otherwise. We have $X=X_1+X_2+X_3$. Also, $P(X_1=1)=P(X_2=1)=P(X_3=1)=p$, is given. Also,



$$ 1.8 = E(X) = P(X=1) + 2 P(X=2) + 3 P(X=3) $$



notice $X=1$ only when one $X_i=1$. Thus, $P(X=3) = 3 P(X_i=1) = 3p$ and $P(X=2)$ when any two of $X_i=1$ that is $P(X=2) = 3p^2$. Thus,



$$ 1.8 = 3p + 6p^2 + 3P(X=3) $$



$$ P(X=3) = 1.8-p-2p^2 $$



take derivative with respect to $p$ gives



$$ (P(X=3))' = -1 - 4p $$



which means the maximum have to be at $p<0$ which makes no sense. what am I doing wrong?






probability




probability






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 28 '18 at 19:51









LuxGiammi

17410




17410










asked Nov 28 '18 at 18:37









NeymarNeymar

354114




354114








  • 4




    $begingroup$
    Possible duplicate of $E[X]=1.8$, where $X$ is the total number of successes of 3 trials. What is the largest/smallest $P{X=3}$ can be?
    $endgroup$
    – LuxGiammi
    Nov 28 '18 at 19:32














  • 4




    $begingroup$
    Possible duplicate of $E[X]=1.8$, where $X$ is the total number of successes of 3 trials. What is the largest/smallest $P{X=3}$ can be?
    $endgroup$
    – LuxGiammi
    Nov 28 '18 at 19:32








4




4




$begingroup$
Possible duplicate of $E[X]=1.8$, where $X$ is the total number of successes of 3 trials. What is the largest/smallest $P{X=3}$ can be?
$endgroup$
– LuxGiammi
Nov 28 '18 at 19:32




$begingroup$
Possible duplicate of $E[X]=1.8$, where $X$ is the total number of successes of 3 trials. What is the largest/smallest $P{X=3}$ can be?
$endgroup$
– LuxGiammi
Nov 28 '18 at 19:32










1 Answer
1






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2












$begingroup$

If your trials are independent



$P(X=3)= 0.6^3$



But, supposing you are biopsying some diseased tissue. There is a probability $p$ chance that the tissue has disease, and there is some probability $q$, that any sample will show disease if the tissue is diseased.



All of the trials have equal likelihood of showing disease, but they are not independent.



If $q$ is big... i.e. let $q = 1$



Either all three samples show disease or none do.



$P(X=3) = 0.6$



At the other extreme we can set up scenarios where at least one sample will always positive, but it is impossible for a 3 to be positive.



$P(X=3) = 0$






share|cite|improve this answer











$endgroup$













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    $begingroup$

    If your trials are independent



    $P(X=3)= 0.6^3$



    But, supposing you are biopsying some diseased tissue. There is a probability $p$ chance that the tissue has disease, and there is some probability $q$, that any sample will show disease if the tissue is diseased.



    All of the trials have equal likelihood of showing disease, but they are not independent.



    If $q$ is big... i.e. let $q = 1$



    Either all three samples show disease or none do.



    $P(X=3) = 0.6$



    At the other extreme we can set up scenarios where at least one sample will always positive, but it is impossible for a 3 to be positive.



    $P(X=3) = 0$






    share|cite|improve this answer











    $endgroup$


















      2












      $begingroup$

      If your trials are independent



      $P(X=3)= 0.6^3$



      But, supposing you are biopsying some diseased tissue. There is a probability $p$ chance that the tissue has disease, and there is some probability $q$, that any sample will show disease if the tissue is diseased.



      All of the trials have equal likelihood of showing disease, but they are not independent.



      If $q$ is big... i.e. let $q = 1$



      Either all three samples show disease or none do.



      $P(X=3) = 0.6$



      At the other extreme we can set up scenarios where at least one sample will always positive, but it is impossible for a 3 to be positive.



      $P(X=3) = 0$






      share|cite|improve this answer











      $endgroup$
















        2












        2








        2





        $begingroup$

        If your trials are independent



        $P(X=3)= 0.6^3$



        But, supposing you are biopsying some diseased tissue. There is a probability $p$ chance that the tissue has disease, and there is some probability $q$, that any sample will show disease if the tissue is diseased.



        All of the trials have equal likelihood of showing disease, but they are not independent.



        If $q$ is big... i.e. let $q = 1$



        Either all three samples show disease or none do.



        $P(X=3) = 0.6$



        At the other extreme we can set up scenarios where at least one sample will always positive, but it is impossible for a 3 to be positive.



        $P(X=3) = 0$






        share|cite|improve this answer











        $endgroup$



        If your trials are independent



        $P(X=3)= 0.6^3$



        But, supposing you are biopsying some diseased tissue. There is a probability $p$ chance that the tissue has disease, and there is some probability $q$, that any sample will show disease if the tissue is diseased.



        All of the trials have equal likelihood of showing disease, but they are not independent.



        If $q$ is big... i.e. let $q = 1$



        Either all three samples show disease or none do.



        $P(X=3) = 0.6$



        At the other extreme we can set up scenarios where at least one sample will always positive, but it is impossible for a 3 to be positive.



        $P(X=3) = 0$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Nov 28 '18 at 19:48

























        answered Nov 28 '18 at 19:38









        Doug MDoug M

        44.4k31854




        44.4k31854






























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