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What is the probability that, on three rolls of a number cub
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04 Apr 2013, 08:09
04 Apr 2013, 08:09
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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:
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?
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
\(\frac{25}{216}+\frac{5}{216}+\frac{1}{216} = \frac{31}{216}\)
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?
Re: Probability of at least one outcome
[#permalink]
04 Apr 2013, 08:19
04 Apr 2013, 08:19
5
Kudos
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 \(\frac{91}{216}\)
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 \(\frac{91}{216}\)
General Discussion
Re: Probability of at least one outcome
[#permalink]
04 Apr 2013, 09:26
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 \(\frac{91}{216}\)
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.
