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03 Jun 2010, 01:51
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
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Question Stats:
50% (00:43) correct
50%
(02:11)
wrong
based on 1
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n-sided dice is rolled n independent times. What is the probability that at least i will be observed on the i-th trial at least once?
My solution:
Official solution:
I think the official solution is wrong. Because during the first i trials we can observe any number except for i. But in rolls greater than i we can observe any number (even i).
Your thoughts would be greatly appreciated.
My solution:
We will solve it using a complement.
OMEGA = n^n
A' ... A' is an event that any number different from i appears on the i-th trial
A' = (n-1)^i * n^(n-i) ... (n-1)*(n-1)* ...*(n-1)*n*...*n
P(A) = 1 - A'/OMEGA
OMEGA = n^n
A' ... A' is an event that any number different from i appears on the i-th trial
A' = (n-1)^i * n^(n-i) ... (n-1)*(n-1)* ...*(n-1)*n*...*n
P(A) = 1 - A'/OMEGA
Official solution:
OMEGA = n^n
A' = (n-1)^n ... A' is an event that any number different from i appears on the i-th trial
P = 1 - A'/OMEGA
A' = (n-1)^n ... A' is an event that any number different from i appears on the i-th trial
P = 1 - A'/OMEGA
I think the official solution is wrong. Because during the first i trials we can observe any number except for i. But in rolls greater than i we can observe any number (even i).
Your thoughts would be greatly appreciated.
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Thank you for understanding, and happy exploring!
09 Jun 2010, 10:41
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I got the same answer as you did. What is the source of this problem?
09 Jun 2010, 11:32
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I feel your answer is correct. Even I got the answer as \(1- \frac{(n-1)^i}{n^n}\).
I am asking for the original source of this problem where you encountered this.
I am asking for the original source of this problem where you encountered this.
