Find $P(1 le x < 3)$ if $X$ ~ $Binomial (100, 0.01)$












2












$begingroup$


Here's my approach:



$n = 100 , $ p = 0.01



To find $P(1 le x < 3)$, do I just find $P(X < 3) - P(1 le X)$?






probability probability-theory probability-distributions







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$endgroup$












  • $begingroup$
    no, you have to find $P(X<3)-P(X=0)$ which is $$P(Xleq 2)-P(X=0)=(P(X=0)+P(X=1)+P(X=2))-P(X=0)=P(X=1)+P(X=2)$$
    $endgroup$
    – Fakemistake
    Dec 1 '18 at 7:22
















2












$begingroup$


Here's my approach:



$n = 100 , $ p = 0.01



To find $P(1 le x < 3)$, do I just find $P(X < 3) - P(1 le X)$?






probability probability-theory probability-distributions







share|cite|improve this question











$endgroup$












  • $begingroup$
    no, you have to find $P(X<3)-P(X=0)$ which is $$P(Xleq 2)-P(X=0)=(P(X=0)+P(X=1)+P(X=2))-P(X=0)=P(X=1)+P(X=2)$$
    $endgroup$
    – Fakemistake
    Dec 1 '18 at 7:22














2












2








2





$begingroup$


Here's my approach:



$n = 100 , $ p = 0.01



To find $P(1 le x < 3)$, do I just find $P(X < 3) - P(1 le X)$?






probability probability-theory probability-distributions







share|cite|improve this question











$endgroup$




Here's my approach:



$n = 100 , $ p = 0.01



To find $P(1 le x < 3)$, do I just find $P(X < 3) - P(1 le X)$?






probability probability-theory probability-distributions




probability probability-theory probability-distributions






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 29 '18 at 17:43







Yolanda Hui

















asked Nov 29 '18 at 15:55









Yolanda HuiYolanda Hui

14810




14810












  • $begingroup$
    no, you have to find $P(X<3)-P(X=0)$ which is $$P(Xleq 2)-P(X=0)=(P(X=0)+P(X=1)+P(X=2))-P(X=0)=P(X=1)+P(X=2)$$
    $endgroup$
    – Fakemistake
    Dec 1 '18 at 7:22


















  • $begingroup$
    no, you have to find $P(X<3)-P(X=0)$ which is $$P(Xleq 2)-P(X=0)=(P(X=0)+P(X=1)+P(X=2))-P(X=0)=P(X=1)+P(X=2)$$
    $endgroup$
    – Fakemistake
    Dec 1 '18 at 7:22
















$begingroup$
no, you have to find $P(X<3)-P(X=0)$ which is $$P(Xleq 2)-P(X=0)=(P(X=0)+P(X=1)+P(X=2))-P(X=0)=P(X=1)+P(X=2)$$
$endgroup$
– Fakemistake
Dec 1 '18 at 7:22




$begingroup$
no, you have to find $P(X<3)-P(X=0)$ which is $$P(Xleq 2)-P(X=0)=(P(X=0)+P(X=1)+P(X=2))-P(X=0)=P(X=1)+P(X=2)$$
$endgroup$
– Fakemistake
Dec 1 '18 at 7:22










2 Answers
2






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0












$begingroup$

Hints
Notice that
$$P(X=k)=binom{100}{k}(0.01)^k(0.99)^{100-k}$$
and that
$$
P(1leq X<3)=P(X=1)+P(X=2).
$$






share|cite|improve this answer











$endgroup$





















    0












    $begingroup$

    To find $P(1 le x < 3)$, you do $P(X < 3) - P(X<1)$ not $P(X < 3) - P(1 le X)$ because



    $$P(X<3) = P(X=0) + P(X=1) + P(X=2)$$



    $$P(X<1) = P(X=0)$$



    $$P(1 le X) = P(X=1) + P(X=2) + P(X=3) + P(X=4) + ... + P(X=100)$$



    $$therefore, P(X < 3) - P(X<1) = P(X=1) + P(X=2)$$



    and $P(X < 3) - P(1 le X) < 0$ which is impossible if $P(X < 3) - P(1 le X)$ is equal to the probability of some event which you guessed is $P(1 le X < 3) = P(X < 3) - P(1 le X)$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Yeah thats a typo, thanks.
      $endgroup$
      – Yolanda Hui
      Dec 1 '18 at 10:04











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    2 Answers
    2






    active

    oldest

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    2 Answers
    2






    active

    oldest

    votes









    active

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    active

    oldest

    votes









    0












    $begingroup$

    Hints
    Notice that
    $$P(X=k)=binom{100}{k}(0.01)^k(0.99)^{100-k}$$
    and that
    $$
    P(1leq X<3)=P(X=1)+P(X=2).
    $$






    share|cite|improve this answer











    $endgroup$


















      0












      $begingroup$

      Hints
      Notice that
      $$P(X=k)=binom{100}{k}(0.01)^k(0.99)^{100-k}$$
      and that
      $$
      P(1leq X<3)=P(X=1)+P(X=2).
      $$






      share|cite|improve this answer











      $endgroup$
















        0












        0








        0





        $begingroup$

        Hints
        Notice that
        $$P(X=k)=binom{100}{k}(0.01)^k(0.99)^{100-k}$$
        and that
        $$
        P(1leq X<3)=P(X=1)+P(X=2).
        $$






        share|cite|improve this answer











        $endgroup$



        Hints
        Notice that
        $$P(X=k)=binom{100}{k}(0.01)^k(0.99)^{100-k}$$
        and that
        $$
        P(1leq X<3)=P(X=1)+P(X=2).
        $$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 1 '18 at 4:49









        farruhota

        19.7k2738




        19.7k2738










        answered Nov 29 '18 at 16:04









        Foobaz JohnFoobaz John

        21.6k41352




        21.6k41352























            0












            $begingroup$

            To find $P(1 le x < 3)$, you do $P(X < 3) - P(X<1)$ not $P(X < 3) - P(1 le X)$ because



            $$P(X<3) = P(X=0) + P(X=1) + P(X=2)$$



            $$P(X<1) = P(X=0)$$



            $$P(1 le X) = P(X=1) + P(X=2) + P(X=3) + P(X=4) + ... + P(X=100)$$



            $$therefore, P(X < 3) - P(X<1) = P(X=1) + P(X=2)$$



            and $P(X < 3) - P(1 le X) < 0$ which is impossible if $P(X < 3) - P(1 le X)$ is equal to the probability of some event which you guessed is $P(1 le X < 3) = P(X < 3) - P(1 le X)$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Yeah thats a typo, thanks.
              $endgroup$
              – Yolanda Hui
              Dec 1 '18 at 10:04
















            0












            $begingroup$

            To find $P(1 le x < 3)$, you do $P(X < 3) - P(X<1)$ not $P(X < 3) - P(1 le X)$ because



            $$P(X<3) = P(X=0) + P(X=1) + P(X=2)$$



            $$P(X<1) = P(X=0)$$



            $$P(1 le X) = P(X=1) + P(X=2) + P(X=3) + P(X=4) + ... + P(X=100)$$



            $$therefore, P(X < 3) - P(X<1) = P(X=1) + P(X=2)$$



            and $P(X < 3) - P(1 le X) < 0$ which is impossible if $P(X < 3) - P(1 le X)$ is equal to the probability of some event which you guessed is $P(1 le X < 3) = P(X < 3) - P(1 le X)$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Yeah thats a typo, thanks.
              $endgroup$
              – Yolanda Hui
              Dec 1 '18 at 10:04














            0












            0








            0





            $begingroup$

            To find $P(1 le x < 3)$, you do $P(X < 3) - P(X<1)$ not $P(X < 3) - P(1 le X)$ because



            $$P(X<3) = P(X=0) + P(X=1) + P(X=2)$$



            $$P(X<1) = P(X=0)$$



            $$P(1 le X) = P(X=1) + P(X=2) + P(X=3) + P(X=4) + ... + P(X=100)$$



            $$therefore, P(X < 3) - P(X<1) = P(X=1) + P(X=2)$$



            and $P(X < 3) - P(1 le X) < 0$ which is impossible if $P(X < 3) - P(1 le X)$ is equal to the probability of some event which you guessed is $P(1 le X < 3) = P(X < 3) - P(1 le X)$






            share|cite|improve this answer









            $endgroup$



            To find $P(1 le x < 3)$, you do $P(X < 3) - P(X<1)$ not $P(X < 3) - P(1 le X)$ because



            $$P(X<3) = P(X=0) + P(X=1) + P(X=2)$$



            $$P(X<1) = P(X=0)$$



            $$P(1 le X) = P(X=1) + P(X=2) + P(X=3) + P(X=4) + ... + P(X=100)$$



            $$therefore, P(X < 3) - P(X<1) = P(X=1) + P(X=2)$$



            and $P(X < 3) - P(1 le X) < 0$ which is impossible if $P(X < 3) - P(1 le X)$ is equal to the probability of some event which you guessed is $P(1 le X < 3) = P(X < 3) - P(1 le X)$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 1 '18 at 2:33







            user198044



















            • $begingroup$
              Yeah thats a typo, thanks.
              $endgroup$
              – Yolanda Hui
              Dec 1 '18 at 10:04


















            • $begingroup$
              Yeah thats a typo, thanks.
              $endgroup$
              – Yolanda Hui
              Dec 1 '18 at 10:04
















            $begingroup$
            Yeah thats a typo, thanks.
            $endgroup$
            – Yolanda Hui
            Dec 1 '18 at 10:04




            $begingroup$
            Yeah thats a typo, thanks.
            $endgroup$
            – Yolanda Hui
            Dec 1 '18 at 10:04


















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