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What is the probability that, on three rolls of a number cub

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What is the probability that, on three rolls of a number cub [#permalink] New post 04 Apr 2013, 07:09
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Hi,

This question is actually from the Manhattan Number Properties book,

I am trying to solve this, but am getting it wrong.

What is the probability that, on three rolls of a number cube with faces numbered 1 to 6, at least one of the rolls will be a 6?

The way I am solving it:

Quote:
Since I need to find the probability that at least one of the rolls is a 6, there are three outcomes possible:

Outcome 1:
Exactly one roll is a 6:
This would mean one roll is a 6, while the remaining two are not.
The probability would be: \frac{1}{6}*\frac{5}{6}*\frac{5}{6}
= \frac{25}{216}

Outcome 2:
Two rolls are 6s, one is not
The probability would be: \frac{1}{6}*\frac{1}{6}*\frac{5}{6}
= \frac{5}{216}

Outcome 3:
All three rolls are 6s
The probability would be: \frac{1}{6}*\frac{1}{6}*\frac{1}{6}
= \frac{1}{216}

Therefore, probability that at least one roll is a six would be Outcome 1 OR Outcome 2 OR Outcome 3

\frac{25}{216}+\frac{5}{216}+\frac{1}{216} = \frac{31}{216}



But the right answer is \frac{91}{216}. The book has given this example to explain the 1-x probability trick, ie. calculating the probability of not getting even one 6 and subtracting it from 1.

I was trying to solve it the normal way, to understand it.

I seem to get 91/216 if I consider WHICH ROLL is a 6 in the first two outcomes, and add their probabilities.
eg, for Outcome 1: First roll is a 6 AND the roll 2 and Roll 3 are not OR the Second roll is a 6 and the other two are not OR the third roll is a 6 and the other two are not

This would give me \frac{25}{216}+\frac{25}{216}+\frac{25}{216} = \frac{75}{216} for Outcome 1
and \frac{5}{216}+\frac{5}{216}+\frac{5}{216} = \frac{15}{216} for Outcome 2, Outcome 3 would be \frac{1}{216} for a total of \frac{91}{216}

But I don't see why WHICH ROLL is a 6 is important for this problem.

What am I thinking wrong?
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Re: Probability of at least one outcome [#permalink] New post 04 Apr 2013, 07:19
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What you say is correct, but not complete.

Case 1: only one 6. you here analyze one case (6,N,N) correctly, but you forget to analyze the other 2 cases : (N,6,N) and (N,N,6)
so the probability of case one is 3*\frac{25}{216}

The same happens for case 2: two 6s. Here again we have to multiply your result by 3 in order to get all the combinations: (N,6,6)(6,N,6)(6,6,N)
3*\frac{5}{216}

The probability of the third case is correct. Now we sum our results and obtain \frac{91}{216}
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Re: Probability of at least one outcome [#permalink] New post 04 Apr 2013, 08:26
Zarrolou wrote:
What you say is correct, but not complete.

Case 1: only one 6. you here analyze one case (6,N,N) correctly, but you forget to analyze the other 2 cases : (N,6,N) and (N,N,6)
so the probability of case one is 3*\frac{25}{216}

The same happens for case 2: two 6s. Here again we have to multiply your result by 3 in order to get all the combinations: (N,6,6)(6,N,6)(6,6,N)
3*\frac{5}{216}

The probability of the third case is correct. Now we sum our results and obtain \frac{91}{216}


Thanks Zarrolou.

I'm having trouble understanding the concept of probability.
Im trying to frame this question as a Combinatorics question. Please help me out.


Lets just consider Case 1, Only one 6 in three rolls.

So, for three rolls, all the possible outcomes are:

6 outcomes of first roll AND 6 of second AND 6 of third = 216

So, the right way to frame the question now would be : What are the different ways of getting exactly one 6 in three rolls of a dice?

Using the Slots method, that would be:

1 AND 5 AND 5 OR 5 AND 1 AND 5 OR 5 AND 5 AND 1.

That would be 75.

So, probability of getting exactly one 6 would now be, \frac{Ways of getting desired outcome}{All possible outcomes} = \frac{75}{216}

I am sorry for taking this roundabout rote, but am just trying to get it clear for myself.
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Re: Probability of at least one outcome [#permalink] New post 04 Apr 2013, 09:27
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rooker wrote:
Thanks Zarrolou.

I'm having trouble understanding the concept of probability.
Im trying to frame this question as a Combinatorics question. Please help me out.


Lets just consider Case 1, Only one 6 in three rolls.

So, for three rolls, all the possible outcomes are:

6 outcomes of first roll AND 6 of second AND 6 of third = 216

So, the right way to frame the question now would be : What are the different ways of getting exactly one 6 in three rolls of a dice?

Using the Slots method, that would be:

1 AND 5 AND 5 OR 5 AND 1 AND 5 OR 5 AND 5 AND 1.

That would be 75.

So, probability of getting exactly one 6 would now be, \frac{Ways of getting desired outcome}{All possible outcomes} = \frac{75}{216}

I am sorry for taking this roundabout rote, but am just trying to get it clear for myself.


Absolutely correct! The probability of getting one 6 and three N6s is: \frac{1}{6}*\frac{5}{6}*\frac{5}{6}
The number of possible outcome is which this can happen is 3. So 3*\frac{25}{216} is the magic number
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Re: Probability of at least one outcome [#permalink] New post 12 Apr 2013, 01:28
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Hi there,

This is my way of approach:
The probability of not getting a 6 in 3 throws is (5/6)X(5/6)x(5/6).
Therefore the probability of getting at least one 6 is 1- [P(no 6 in 3 throws) = (5/6)X(5/6)x(5/6)] = 91/216
Re: Probability of at least one outcome   [#permalink] 12 Apr 2013, 01:28
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