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# Probability n-sided dice

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Director
Joined: 23 Apr 2010
Posts: 583
Followers: 2

Kudos [?]: 35 [0], given: 7

Probability n-sided dice [#permalink]  03 Jun 2010, 00:51
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Question Stats:

50% (01:43) correct 50% (02:11) wrong based on 1 sessions
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:

[Reveal] Spoiler:
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

Official solution:

[Reveal] Spoiler:
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

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.
Intern
Joined: 28 May 2010
Posts: 4
Followers: 0

Kudos [?]: 3 [0], given: 1

Re: Probability n-sided dice [#permalink]  09 Jun 2010, 09:41
I got the same answer as you did. What is the source of this problem?
Intern
Joined: 28 May 2010
Posts: 4
Followers: 0

Kudos [?]: 3 [0], given: 1

Re: Probability n-sided dice [#permalink]  09 Jun 2010, 10:32
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.
Director
Joined: 23 Apr 2010
Posts: 583
Followers: 2

Kudos [?]: 35 [0], given: 7

Re: Probability n-sided dice [#permalink]  10 Jun 2010, 02:49
It's a text book called Probability and Statistics.
Re: Probability n-sided dice   [#permalink] 10 Jun 2010, 02:49
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