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












3














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
















3














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
























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














3












3








3







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















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 at 17:43

























asked Nov 29 at 15:55









Yolanda Hui

15810




15810












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


















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
















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




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










2 Answers
2






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1














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































    0














    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





















    • Yeah thats a typo, thanks.
      – Yolanda Hui
      Dec 1 at 10:04











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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1














    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




























      1














      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


























        1












        1








        1






        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














        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 at 4:49









        farruhota

        19k2736




        19k2736










        answered Nov 29 at 16:04









        Foobaz John

        20.8k41250




        20.8k41250























            0














            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





















            • Yeah thats a typo, thanks.
              – Yolanda Hui
              Dec 1 at 10:04
















            0














            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





















            • Yeah thats a typo, thanks.
              – Yolanda Hui
              Dec 1 at 10:04














            0












            0








            0






            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












            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 at 2:33









            Jack Bauer

            1,3071631




            1,3071631












            • Yeah thats a typo, thanks.
              – Yolanda Hui
              Dec 1 at 10:04


















            • Yeah thats a typo, thanks.
              – Yolanda Hui
              Dec 1 at 10:04
















            Yeah thats a typo, thanks.
            – Yolanda Hui
            Dec 1 at 10:04




            Yeah thats a typo, thanks.
            – Yolanda Hui
            Dec 1 at 10:04


















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