What Is the Probability of Rolling At Least Two 6's in 13 Dice Rolls?

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Homework Help Overview

The problem involves calculating the probability of rolling at least two sixes when a die is rolled 13 times. The subject area pertains to probability theory and combinatorics.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the original poster's approach using combinations and the total number of outcomes. Some suggest considering the complement of the event to simplify calculations. Questions arise regarding the distinction between "at least two sixes" and "exactly two sixes."

Discussion Status

The discussion is ongoing, with participants providing guidance on alternative methods and clarifying concepts. There is an acknowledgment of different approaches, including calculating separate cases versus using the complement.

Contextual Notes

Participants are navigating the complexities of probability calculations and the implications of different interpretations of the problem statement. The original poster expresses uncertainty about the complement method.

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Homework Statement



A die is rolled 13 times. What is the probability of at least two 6's appearing? (Round your answer to four decimal places.)

2. The attempt at a solution
I know that the total number of outcomes is 6^13. I did (13nCr2)x(1x1x6x6x6x6x6x6x6x6x6x6x6)/6^13, but the answer isn't right. What am I doing wrong?
 
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Perhaps you should try explaining the theory behind your solution.
I think working on the complement would be much easier. That is, finding the number of times a 6 would appear 0 times and the number of times a six would appear 1 time, and then take the complement.
 
What is the difference between "at least two 6s appear" and "exactly two sixes appear?" which did you calculate?
VeeEight's suggestion is spot on.
 
statdad said:
What is the difference between "at least two 6s appear" and "exactly two sixes appear?" which did you calculate?
VeeEight's suggestion is spot on.


I see. I calculated the number of times exactly two sixes would appear. Not quite sure how to do the compliment though.
 
You can do separate cases (exactly two sixes, exactly three sixes, etc) and add them all up. It's just that taking the complement is less work
(a good exercise might be to do both to make sure they are the same)
 
The complement of "at least two sixes" is "0 or 1 sixes" and is easier to calculate sinced it involves only two cases rather than 5.
 
Thanks! Got it.
 

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