A question about dice probability calculation

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Homework Help Overview

The discussion revolves around calculating the probability of the sum of two dice being under 9. The original poster expresses uncertainty about the calculation method and presents their manual approach of listing all possible outcomes. They also raise questions about the interpretation of events in probability calculations.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to calculate the probability by listing outcomes and questions the reasoning behind counting events differently for specific cases. Other participants explore the probabilities of sums greater than or equal to 9 and provide examples of possible outcomes.

Discussion Status

Participants are actively engaging with the problem, offering different perspectives on calculating probabilities. Some guidance has been provided regarding the interpretation of events, but there is no explicit consensus on the best approach to the original question.

Contextual Notes

The original poster's inquiry includes a potential misunderstanding of event counting in probability, and there is a mention of assumptions regarding the conditions of the sums being calculated.

TheNaturalStep
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A question about probability calculation
What is the probability that the sum from two dices are under 9, i don’t know a fast way to calculate that :(, instead i calculated it manually by listing all possible event.
Divided by the total number of outcomes 6*6=36

And something else that I can not understand,
p=probability that this happen

Case 1
dice (1,1), dice (1,1), <-> is one event <-> 2p
case 2
dice(1,2) and dice(2,1) <-> is two events <->2p

How come case one gets 1 event and 2p when case 2 gets two events and 2p ...

Kindly TNS ...
 
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If you get more than 9(assuming that you meant sum>9 and not equal to 9) you can get (6,3) or (6,4) or (6,5) or (6,6) or (5,5) or (5,4) <---- these gives you the possibilities to get a sum\geq9. NOTE: this follows the order, (dice1,dice2)

so to get (6,3)=\frac{1}{6}*\frac{1}{3}*2!=\frac{1}{9}

since all the possibilities will be similar to that it will simply be \frac{1}{9}*6 which is \frac{2}{3} but this is to find P(sum\geq9) so then to find what you want is simply 1-P(sum\geq9)
 
TheNaturalStep said:
And something else that I can not understand,
p=probability that this happen

Case 1
dice (1,1), dice (1,1), <-> is one event <-> 2p
case 2
dice(1,2) and dice(2,1) <-> is two events <->2p

How come case one gets 1 event and 2p when case 2 gets two events and 2p ...

Kindly TNS ...
If you were calculating the probability of getting a 2 on a pair of dice, there is only one way that can happen. Die "A" is a 1 and die "B" is a 1. Probability of rolling a 2 is 1/36.
If you were calculating the probability of getting a 3 on a pair of dice, there are two ways that can happen. Die "A" is a 2 and die "B" is a 1 or die "A" is a 1 and die "B" is a 2. Probability of rolling a 3 is 2/36= 1/18.
 
Thank you very much for your replies, i think it is clear now.

Kindly TNS
 

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