Question 478361
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LET X BE A RANDOM VARIABLE WITH THE FOLLOWING PROBAILITY DISTRIBUTION
<pre>
    X       0    1    2    3
    P(X)  0.4  0.3  0.2  0.1
</pre>DOES X HAVE A BINOMIAL DISTRIBUTION ? JUSTIFY YOUR ANSWER.
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&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;It is a good problem.  It is good, because it is non-standard and is different from 

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;thousands other standard problems, that are usually offered in this area.


&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;Standard problems check if you know what is written in textbooks.

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;Non-standard problems check, in addition, if you able to think independently,

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;and motivate you to be creative.


&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;Tutor @Theo in his post wrote many words, but did not provide a direct solution.

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;Meanwhile, the direct solution is simple, but requires to find and to apply some fresh idea.

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;Therefore, it is educative and deserves your attention.



<pre>
                - - - S O L U T I O N  - - - 


Let's  {{{highlight(highlight(assume))}}}  for a minute that the distribution P(X) is a binomial.


Then the number of trials is 3, and should be some probability 'p' such that


    P(i) = {{{C[3]^i*p^(3-i)*(1-p)^i}}},  i = 0, 1, 2, 3.    (1)



According to (1), for i = 0 should be  P(0) = 0.4 = {{{C[3]^3*p^3*(1-p)^0}}} = {{{1*p^3*1}}} = {{{p^3}}}.

           It implies  p = {{{root(3,0.4)}}} = 0.7368063  (rounded).



Next, according to (1), for i = 3 should be  

    P(3) = 0.1 = {{{C[3]^3*p^0*(1-p)^3}}} = {{{1*1*(1-0.7368063)^3}}} = = {{{0.2631937^3}}} = 0.018231671.



Thus we got the  {{{highlight(highlight(CONTRDICTION))}}} : we got the number of 0.018231671 for P(3), different from the given value P(3) = 0.1.


It  {{{highlight(highlight(PROVES))}}}  that the given distribution is {{{highlight(highlight(NOT))}}}  a binomial.
</pre>

At this point, the problem is solved completely.