What is the possibility of rolling a 6 on a die at least once in ten rolls?

Question

My answer is 84%. Am I right?

Thank you, Ronald, for the simplest answer. My real interest is in deriving a formula to determine the probability for achieving a desired outcome at least ONCE for a variable number of outcomes and "tries".

Where...
x = number of outcomes (as in x = number of sides on the die)
y = number of "tries" (rolls of the die)

Previously, I had a complicated formula, but the simplest formula I can create with Ronald's math is this:

Probability of getting desired outcome at least once = 1 - [(x-1)/x]^y

It seems to check out.

Answer 1

The probability of NOT rolling a 6 on a die in one roll is 5/6. The probablility of not rolling a 6 on ten rolls is (5/6)^10, so the probablility of rolling a die and getting at least one 6 is 1 - (5/6)^10 or 83.8%.

Answer 2

that would be a binomial experiment with
n = 10 and p = 1/6. You'd figure it out for k = 0 (no 6's) and then subtract this from 1. (This is an example of where it's a LOT easier to figure out by the compliments rule, than figuring out the probability of one or two or three or ... 10 sixes.)
84% seems a bit low, it seems like if the probability is 1 in 6, and the number of trials is more than 6, the outcome ought to be certain. So let's put this in the formula for binomial probability.
P(k=0) (n = 10, p = 1/6) is C(10,0) x (1/6)^0 x (5/6)^10
Since C(n,0) = 1 for all positive n, and p^0 = 1 for all nonzero p, this reduces to (5/6)^10, as others have said.

Answer 3

The probability is one minus the chance of not rolling a six in ten tries.

P(at least one 6) = 1 - (5/6)^10 ≈ .838 or 83.8%

So you are correct.

Answer 4

for each roll, the probability of not rolloing a 6 is 6/7. the probability of not rolling a 6 10 times in a row is
(6/7)^10=0.214 so the probability of rolling at least 1 6 is
1-(6/7)^10=0.786 or 78.6%

Answer 5

Anwer is still 1/6
possibility every roll is 1/6 , 6 possibilities for 6 rolls
total possibility for 6 rolls is 36
probability = 6/36 = 1/6 = 16.66%

Answer 6

I don't know how to use the keyboard to write the mathematical symbols, so, I will put it this way.
The probability equals to 1-(A/B)
Where B is 6X6 X6 X6 X6 X6 X6 X6 X6 X6=60466176 (all the combinations for ten rolls)
While A equals to 5X5X5X5X5X5X5X5X5X5= 9765625
A/B is the possibility of not having six at all =0.16150556
So the final probability is 1-1.6150556= 0.83849444 approximately %84

Answer 7

Yep, that's correct.

Figure out the odds of rolling no sixes on your ten rolls.
That's (5/6)*(5/6)*(5/6)*(5/6)*(5/6)*(5/6)*(5/6)*(5/6)*(5/6)*(5/6).

The odds of rolling at least one six are 1 - (5/6)^10
That's 1 - 0.1615, or 0.8385.
That rounds to 84%, which is the answer you got.

Answer 8

at least once is the complement of no sixes. The probability of no 6's is (5/6)^10 = .1615
1-.1615 = 83.8%, your answer looks good to me!

Answer 9

that is...

[5^9 + 5^8 + 5^7 + ... + 5^0]/6^10

compute in your calculator

Answer 10

The possibility is 100%
Probability???