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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]

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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]

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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]

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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]

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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]

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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
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Re: What is the probability that, on three rolls of a number cub  [#permalink]

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New post 26 Feb 2018, 11:55
Hi All,

When it comes to probability questions, there are only 2 things that you can figure out:

1) What you WANT to have happen.
2) What you DON'T WANT to have happen.

If you add those two things together, then you get the number 1.

For some probability questions, it's pretty easy to figure out the exact thing that you want to have happen. In other questions though, it's easier to figure out what you DON'T WANT, then subtract that probability from 1 (and you'll have the probability of what you DO want).

In the example that you posted, we're asked for the probability of rolling 3 dice and getting AT LEAST one 6. That means getting one 6, two 6s or three 6s would fit the definition of what we WANT. That seems like a LOT of different possibilities to keep track of. It's easier to figure out what we DON'T WANT (which would be zero 6s).

With 3 dice rolls, the probability of getting zero 6s is....

(not a 6)(not a 6)(not a 6) = (5/6)(5/6)(5/6) = 125/216

1 - 125/216 = 91/216 = the probability of getting AT LEAST ONE 6.

As you deal with additional probability questions during your studies, you have to be clear on what the question asks you to figure out, then think about which of the two approaches would be easier.

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Re: What is the probability that, on three rolls of a number cub &nbs [#permalink] 26 Feb 2018, 11:55
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