Question 1201421
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You are going to draw 3 balls from the sack containing, 7 red balls,5 green balls, and 4 blue balls. 
If you REPLACE THE BALLS AFTER EACH DRAWING, what is the probability of drawing:

a) All red balls?
b) All green balls?
c) All blue balls?
d) A red ball, a blue ball and a green ball in THAT ORDER?
e) A red ball, a blue ball and a green ball in ANY ORDER?
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<pre>
Total balls is 7 + 5 + 4 = 16.


(a)  P = {{{(7/16)*(7/16)*(7/16)}}} = {{{7^3/16^3}}}.


(b)  P = {{{(5/16)*(5/16)*(5/16)}}} = {{{5^3/16^3}}}.


(c)  P = {{{(4/16)*(4/16)*(4/16)}}} = {{{4^3/16^3}}} = {{{1/4^3}}}.


(d)  P = {{{(7/16)*(5/16)*(4/16)}}} = reduce it.


(e)  P = {{{6*(7/16)*(5/16)*(4/16)}}} = reduce it.  

     Here 6 = 1*2*3 is the number of all possible permutations of 3 items.
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I answered all your questions.