Question 1029790
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A cube has 12 edges, 12 face diagonals, and 4 diagonals. These 28 objects are placed end to end to form a single line segment 
consisting of the sum of the 28 original line segments. A point is picked at random on the line segment that was formed. 
Find the probability that the point that was picked was a point on one of the original face diagonals. 
Express your answer as a decimal rounded to the nearest ten-thousandth. 

Answer:0.4727

So, I can visualize what is being asked; however, I was tripped up with the probability of the whole line segment/face diagonal. 
Could you please explain? Your help is most sincerely appreciated!!! Thank you so much!!!
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<pre>
1. Let "a" be the length of the cube edge.
   Then the length of the face diagonal is {{{a*sqrt(2)}}}.
   The length of the "true 3D" diagonal is {{{a*sqrt(3)}}}.

2. The length of the entire long segment consisting of all 28 elements is 
   
   {{{12a + 12*sqrt(2) + 4a*sqrt(3)}}} = {{{a*(12 + 12*sqrt(2) + 4*sqrt(3))}}}.     (1)

3. The length of 12 face diagonals is {{{12a*sqrt(2)}}}.   (2)

4. The probability you are asked for is the ratio (2) to (1), which is 

   {{{(12*sqrt(2))/(12 + 12*sqrt(2) + 4*sqrt(3))}}} = 0.4727.
</pre>