A standard six-sided dice is repeatedly rolled until it lands on a two. What is the probability that the first two is rolled after the first six rolls?

C bro how did you get that

To find the probability that the first two is rolled after the first six rolls, we need to consider the probability of rolling any number other than two in each of the first six rolls, followed by rolling a two as the seventh roll.

Since each roll of a standard six-sided dice is independent and has an equal probability of 1/6 for each outcome (in this case, rolling a two or any number from one to six), the probability of rolling any number other than two in any single roll is 5/6.

Therefore, the probability of rolling any number other than two in each of the first six rolls is:

(5/6) * (5/6) * (5/6) * (5/6) * (5/6) * (5/6) = (5/6)^6

Then, the probability of rolling a two on the seventh roll is:

1/6

To find the probability that the first two is rolled after the first six rolls, we multiply the probability of rolling any number other than two in each of the first six rolls by the probability of rolling a two on the seventh roll:

(5/6)^6 * 1/6 ≈ 0.0214

Therefore, the probability that the first two is rolled after the first six rolls is approximately 0.0214 or 2.14%.

To find the probability that the first two is rolled after the first six rolls, we need to consider the probability of not rolling a two in the first six rolls and then rolling a two on the seventh roll.

The probability of not rolling a two on any one roll is 5/6, as there are five other possible outcomes (1, 3, 4, 5, and 6) out of six total possibilities.

Therefore, the probability of not rolling a two in the first six rolls is (5/6) * (5/6) * (5/6) * (5/6) * (5/6) * (5/6) = (5/6)^6 ≈ 0.3349.

The probability of rolling a two on the seventh roll is 1/6, as there is only one possible outcome (rolling a two) out of six total possibilities.

To find the probability that the first two is rolled after the first six rolls, we multiply these probabilities:

(5/6)^6 * 1/6 ≈ 0.0562.

Therefore, the probability is approximately 0.0562 or 5.62%.

P(not 2) = 5/6 and P(2) = 1/6, so 5/6*5/6*5/6*5/6*5/6*5/6*1/6 = 0.0558 = 5.6%