Probability n-sided dice

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Probability n-sided dice [#permalink]

03 Jun 2010, 01:51

00:00

Question Stats:

50% (01:43) correct

50% (02:11) wrong

based on 4 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

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Re: Probability n-sided dice [#permalink]

09 Jun 2010, 10:41

I got the same answer as you did. What is the source of this problem?

Intern

Joined: 28 May 2010

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Re: Probability n-sided dice [#permalink]

09 Jun 2010, 11:32

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

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Re: Probability n-sided dice [#permalink]

10 Jun 2010, 03:49

It's a text book called Probability and Statistics.

Re: Probability n-sided dice [#permalink] 10 Jun 2010, 03:49

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Probability n-sided dice

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Probability n-sided dice

Probability n-sided dice

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