Question 1206643
Six different colored dice are rolled. Of interest is the number of dice that show a "1."
(a) In words, define the Random Variable X.
the outcome of the roll of the dice
the color of the dice
how many dice are rolled
how many dice show a "1"
(b) List the values that X may take on.
X = two, four, six
X = zero, one, two, three, four, five, six
X = 1
X = one, two, three, four, five, six
(c) Give the distribution of X. (Enter the probability as a fraction.)
(d) On average, how many dice would you expect to show a "1"?
(e) Find the probability that all six dice show a "1." (Round your answer to five decimal places.)
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(a)  

     the outcome of the roll of the dice
     the color of the dice
     how many dice are rolled              <<<---=== to answer correctly, it is enough to read the post.
     how many dice show a "1"


(b)  

     X = two, four, six
     X = zero, one, two, three, four, five, six   <<<---===  to answer correctly, it is enough to think 7 seconds
     X = 1
     X = one, two, three, four, five, six



(c)  To calculate the probabilities P(0), P(1), P(2), . . . , P(6), use the formulas of a binomial distribution

     P(0) = {{{(5/6)^6}}},

     P(1) = {{{C[6]^1*(1/6)*(5/6)^5}}} = {{{6*(1/6)*(5/6)^5}}},

     P(2) = {{{C[6]^2*(1/6)^2*(5/6)^4}}} = {{{15*(1/6)^2*(5/6)^4}}},

     . . . and so on . . . 

     P(5) = {{{C[6]^5*(1/6)^5*(5/6)}}} = {{{6*(1/6)^5*(5/6)}}},

     P(6) = {{{(1/6)^6}}}.



(d)  mean = 0*P(0) + 1*P(1) + 2*P(2) + . . . + 5*(P5) + 6*P(6) = 

     {{{0 + 1*6*(1/6)*(5/6)^5 + 2*15*(1/6)^2*(5/6)^4 + 3*20*(1/6)^3*(5/6)^3 + 4*15*(1/6)^4*(5/6)^2 + 5*6*(1/6)^5*(5/6) + 6*(1/6)^6}}} = 1.

     The mean is 1.00.



(e)  P(6) = {{{(1/6)^6}}} = 0.0000214335,  or 0.00002 rounded to 5 decimals.
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Solved in full.