you will land on 1 or 6 5/6 times
1-(5/6)^60 ≈ 99.99822529882%
the answer is the probability 6o =1/6 = 60%
the probability 6o =1/6 = 60%
The probability of getting a 2 is 1 - (1/6)60 = 1 - 2.05*10-47
With a fair die, you would expect it 60*(1/6) = 10 times.
1-(5/6)^60 ≈ 99.99822529882%
Pr(You roll a four in 60 rolls) = 1 - Prob(No fours in 60 rolls) = 1 - [Prob(Not 4 in a roll)]60 = 1 - (5/6)60 = 0.999982 ie 99.9982 per cent
60
You add the probabilities together. For each roll, it is 1/6; therefore, you will have 60 times 1/6, which is 60/6, which is 10. Since probability is only 1 at the most, which means that it WILL happen, I would say that the probability is 1 (or 100%).
I'm assuming you're looking for the probability that you roll either a one or six at least once. So the problem can be rewritten as: 1 - probability of rolling 60 times and never getting ones or sixes = 1 - (2/3)^60
the answer is the probability 6o =1/6 = 60%
the probability 6o =1/6 = 60%
The probability of rolling a 6 on each roll of an unbiased cuboid die is 1/6 If you mean at least one of the rolls shows a 6 then it is the same as 1 - pr(no roll shows a 6) = 1 - (5/6)⁶⁰ ≈ 1 - 0.0000177 = 0.9999823 If you mean that exactly one 6 is rolled then: Pr(exactly one 6) = 60 × 1/6 × (5/6)⁵⁹ ≈ 0.0002130
The probability of rolling a 3 on each roll of an unbiased cuboid die is 1/6 If you mean at least one of the rolls shows a 3 then it is the same as 1 - pr(no roll shows a 3) = 1 - (5/6)⁶⁰ ≈ 1 - 0.0000177 = 0.9999823 If you mean that exactly one 3 is rolled then: Pr(exactly one 3) = 60 × 1/6 × (5/6)⁵⁹ ≈ 0.0002130
-4
It is 60/100 = 0.6
The probability of getting a 2 is 1 - (1/6)60 = 1 - 2.05*10-47