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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

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.

Re: Probability of at least one outcome [#permalink]

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04 Apr 2013, 08:19

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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 \(\frac{91}{216}\)
_________________

It is beyond a doubt that all our knowledge that begins with experience.

Re: Probability of at least one outcome [#permalink]

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04 Apr 2013, 09:26

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.

Re: Probability of at least one outcome [#permalink]

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04 Apr 2013, 10:27

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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
_________________

It is beyond a doubt that all our knowledge that begins with experience.

Re: Probability of at least one outcome [#permalink]

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12 Apr 2013, 02:28

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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

gmatclubot

Re: Probability of at least one outcome
[#permalink]
12 Apr 2013, 02:28

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