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- Thread starter new_at_math
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pwsnafu

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I don't see the logic

how is it permuation formula(combination formula ).

What is wrong with my method?

- #4

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If each of these seems equally likely to you, then the answer is [itex]\frac1{36}[/itex].

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tiny-tim

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Say you roll a dice twice.You want to calculate the probablity of getting both dice to land on 3. Using the formula for combinations: the total number of combinations is 21.

no, you're completely misunderstanding what combinations are for

21 is the number of different results you can get from two

12 13 14 15 16 17 23 24 25 26 27 34 35 36 37 45 46 47 56 57 67 …

start again: write out the possible combinations for a 3 (you did mean 3-total?)

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I get it now it was a permutation with repetition;my bad.

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I don't believe any formula with the word "permutation"Say you roll a dice twice.You want to calculate the probablity of getting both dice to land on 3.

- #9

pwsnafu

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I get it now it was a permutation with repetition;my bad.

No. Permutation is a

Your example is a Bernoulli trial.

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HallsofIvy

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You roll aSay you roll a dice twice.

Another way of looking at this is that if "A" and "B" are independent events, then the probability of "A and B" is the probability of A times the probability of B.You want to calculate the probablity of getting both dice to land on 3. Using the formula for combinations: the total number of combinations is 21. so is the probabilty of getting a 3 on both dice 1 /21 ?

There are 6 faces on a die, one of which is a "3". As long as the faces are all equally likely to come up, the probability of a "3" is 1/6. The second roll of the die is independent of the first so the probability that both will come up "3" is (1/6)(1/6)= 1/36.

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