Probability of two ones with 6 dice

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SUMMARY

The probability of rolling exactly two ones from a set of six dice is calculated using the binomial probability distribution. The correct formula is P = 6C2 * (1/6)^2 * (5/6)^4, where 6C2 represents the combinations of choosing 2 positions for the ones out of 6 dice. The initial calculation of 0.20 was incorrect due to a miscalculation of the combinations factor. The correct probability is derived from the combination of the two outcomes and their respective probabilities.

PREREQUISITES
  • Understanding of binomial probability distribution
  • Familiarity with combinations, specifically 6C2
  • Basic probability concepts, including independent events
  • Knowledge of factorial notation and calculations
NEXT STEPS
  • Study the binomial probability formula in depth
  • Practice calculating combinations using the formula nCr
  • Explore examples of probability with multiple dice
  • Learn about the implications of independent events in probability
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Students studying probability, educators teaching statistics, and anyone interested in understanding the mathematics behind dice games.

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Homework Statement



What is the probability of rolling exactly two ones from a set of six dice?

Homework Equations


Binomial probability distribution?


The Attempt at a Solution


I tried:

P = 6!/(2! x 4!) x (1/6)^2 x (5/6)^4 and got 0.20, but the book says that's a factor of 10 too high!
 
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relevant equation is correct... but, let's look at it this way:

You could get a 1,1,something else, something else, something else, something else.
Probability is 1/6 * 1/6 * 5/6 * 5/6 * 5/6 * 5/6
Or, you could get a
something else, 1, something else, something else, 1, something else
Probability is 5/6 * 1/6 * 5/6 * 5/6 * 1/6 * 5/6

Note, the probabilities of these two are the same; after a bit of rearranging,
(1/6)^2 * (5/6)^4

Now, how many different orders can you get 2 ones and 4 something elses?
The ones can be 1st and 2nd, 1st and 3rd, 1st and 4th, 1st and 5th, or 1st and 6th. Or, 2nd and 3rd, 2nd and 4th, etc. This is a combination of 2 elements out of 6, sometimes written as 6C2. (I don't know which way you've learned to write it.)

So, put these two parts together, and you should get your answer.
 
Whoever figured out the answers in your book is high on something. Your answer is correct.
 

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