SOLUTION: A dodecahedral die has 12 faces, numbered 1-12. If the die is weighted in such a way that 2 is twice as likely to land facing up as 1, 3 is three times as likely to land facing up

Question 243506: A dodecahedral die has 12 faces, numbered 1-12. If the die is weighted in such a way that 2 is twice as likely to land facing up as 1, 3 is three times as likely to land facing up as 1, and so on, what is the probability distribution for the face landing up?
Outcome 1 2 3 4 5 6 7 8 9 10 11 12
Probability ? ? ? ? ? ? ? ? ? ? ? ?
Answer by oberobic(2304) (Show Source): You can put this solution on YOUR website!
Since each of the 12 faces is weighted, as stated, we need to find the sum of the weights to create the probability distribution.
1+2+3...+12 = ?
Well, we know that the sum of consecutive digits from 1 to k = (k*(k+1))/2. (Gauss' solution)
.
So the sum of the weights is: (12*13)/2 = 6*13 = 78.
.
x....P(x)
--------
1....1/78
2....2/78
3....3/78
4....4/78
5....5/78
6....6/78
7....7/78
8....8/78
9....9/78
10...10/78
11...11/78
12...12/78
---------
.
Done.
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