Question 1149314
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The first sentence means that repetitions are allowed in the code's digits.


The possible values for the random variable X are {0, 1, 2, 3, 4, 5}.



X= 0  means that there is NO zero among five digits of the code.

      The probability do not have zero in any one fixed/selected of 5 positions is  {{{9/10}}};

      The provability do not have zero in ALL        5 positions, is, therefore,  {{{(9/10)^5}}}.



X= 1  means that there is exactly one 0 among five digits of the code.

      The probability for it is  {{{C[5]^1*(9/10)^4*(1/10)}}} = {{{5*(9/10)^4*(1/10)}}}.



X= 2  means that there is exactly two 0 among five digits of the code.

      The probability for it is  {{{C[5]^2*(9/10)^3*(1/10)^2}}}.


The pattern is just clear.  The probability for X = k,  where  k= 0, 1, 2, 3, 4, 5,  is

            {{{C[5]^k*(9/10)^(5-k)*(1/10)^k}}}.   
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Solved.


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<U>CHECK</U>.  You may check that the sum of these 6 partial probabilities is 1.


<pre>
    Indeed, this sum is equal to  {{{((9/10) + (1/10))^5}}} = {{{1^5}}} = 1.
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