Very simple probability question

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Rolling a six-sided die six times does not guarantee a roll of the number 1, as there is a chance of not rolling any 1's at all. The probability of rolling a 1 on a single roll is 1/6, while the chance of rolling any other number is 5/6. Since each roll is independent, the probability of not rolling a 1 in six attempts is (5/6)^6. Consequently, the probability of rolling at least one 1 in six rolls is calculated as 1 - (5/6)^6. This illustrates the concept of independent events in probability.
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This question has been bothering me for a while. If you roll a 6 sided die 6 times, is there a 1 probability that you with roll the number 1? What is the percent chance that you will roll the number 1? It clearly isn't 100%...
 
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piercebeatz said:
This question has been bothering me for a while. If you roll a 6 sided die 6 times, is there a 1 probability that you with roll the number 1?
No. You correctly intuit that there is a chance of not rolling any "1"'s at all.
What is the percent chance that you will roll the number 1? It clearly isn't 100%...
The probability of rolling a "1" is 1/6 for the die. Therefore the probability of rolling any other number is 5/6 ... since the rolls are independent, the probability of rolling anything but a 1 all six times is (5/6)^6 ... so the probability of getting at least one "1" in six rolls is 1-(5/6)^6.

That's the shortcut - you can also set it up as a binomial probability problem.
 
Simon Bridge said:
No. You correctly intuit that there is a chance of not rolling any "1"'s at all.The probability of rolling a "1" is 1/6 for the die. Therefore the probability of rolling any other number is 5/6 ... since the rolls are independent, the probability of rolling anything but a 1 all six times is (5/6)^6 ... so the probability of getting at least one "1" in six rolls is 1-(5/6)^6.

That's the shortcut - you can also set it up as a binomial probability problem.

Thanks a lot!
 
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