Lesson Problems on computing density probability distribution functions

Problems on computing density probability distribution functions

Problem 1

The words "with replacement" mean, in this context, that in every of 5 positions, from 1 to 5,
from left to right, any of 10 digits can appear independently of digits in other positions, 
with the probability   = 0.1.


For X, we have 6 possible values, X = 0, 1, 2, 3, 4, 5.


The probability density function of X is

    P(X=0) =  =  = 0.59049,

    P(X=1) =  =  = 0.32805,

    P(X=2) =  =  =  = 0.0729,

    P(X=3) =  =  =  = 0.0081,

    P(X=4) =  =  =  = 0.00045,

    P(X=5) =  =  = 0.00001.


CHECK.  0.59049 + 0.32805 + 0.0729 + 0.0081 + 0.00045 + 0.00001 = 1.   (! correct !)

Problem 2

(a)  X=1  means that one of the 7 men is in the first position,
          while the rest 6 men and 3 women are distributed in positions from 2 to 10 in arbitrary ways.

     The number of such possible outcomes is  = 7*9!,
     and to find the probability  P(X=1),  we should relate 7*9!  to  10!.


     It gives  P(X=1) =  =  = 0.7.     


     It is the  ANSWER to (a).



(b)  X=2 means that one of the 3 women is in the first position and one of the 7 men is in the second position,
         while the rest 3-1 = 2 women and 7-1 = 6 men are distributed in positions from 3 to 10 in arbitrary ways.

     The number of such possible outcomes is  = 3*7*8!,
     and to find the probability  P(X=2),  we should relate  3*7*8!  to  10!.


     It gives  P(X=2) = =  =  = .


     It is the ANSWER to (b).



(c)  X=3 means that two of the 3 women are in the first and in the second positions and one of the 7 men 
         is in the second position,
         while the rest 3-2 = 1 women and 7-1 = 6 men are distributed in positions from 3 to 10 in arbitrary ways.

     The number of such possible outcomes is   =  = 3*7*7!,
     and to find the probability  P(X=3),  we should relate  3*7*7!  to  10!.


     It gives  P(X=3) = =  =  = .


     It is the ANSWER to (c).



(d)  X=7  means that some man is in the 7th positions and the are no men in positions from 1 to 6.

          But this configuration is not possible (which is obvious), so

              P(X=7) = 0.


     It is the ANSWER to (d).


Thus, all the questions are answered, and the problem is solved completely.