I roll twelve 4-sided dice. What is the probability that I roll at least one 2? What is the probability that I roll exactly three 2s?
i don't knowm sorry
the same question was answered by Economist at 4:59
To calculate the probability of rolling at least one 2 when you roll twelve 4-sided dice, you can use the concept of complementary probability.
The complementary probability states that the probability of an event occurring is equal to 1 minus the probability of the event not occurring.
So, to find the probability of rolling at least one 2, we need to find the probability of not rolling any 2s in all twelve dice rolls.
Each dice roll has four possible outcomes (1, 2, 3, or 4), so the probability of not rolling a 2 on one die is 3/4.
Since all twelve dice rolls are independent, we can multiply the probability of not rolling a 2 together for all twelve dice rolls:
(3/4) * (3/4) * (3/4) * ... * (3/4) (12 times)
This can be simplified as (3/4)^12.
Therefore, the probability of not rolling any 2s in twelve dice rolls is (3/4)^12.
To find the probability of rolling at least one 2, we subtract this probability from 1:
1 - (3/4)^12 ≈ 1 - 0.03125 ≈ 0.96875
So, the probability of rolling at least one 2 in twelve 4-sided dice rolls is approximately 0.96875, or 96.875%.
Now let's calculate the probability of rolling exactly three 2s.
There are twelve dice rolls, and we need exactly three of them to be 2s.
The probability of rolling a 2 on one die is 1/4, and the probability of not rolling a 2 is 3/4.
We use the binomial probability formula to calculate the probability of having exactly three 2s in twelve rolls:
P(x) = (nCx) * (p^x) * (q^(n-x))
Where:
P(x) is the probability of having exactly x successes,
n is the total number of trials (dice rolls),
x is the number of desired successes (in this case, three 2s),
p is the probability of a single success (rolling a 2),
q is the probability of failure (not rolling a 2),
and (nCx) is the number of combinations of n things taken x at a time.
Using this formula, we calculate the probability:
P(3) = (12C3) * ((1/4)^3) * ((3/4)^(12-3))
(12C3) = 12! / (3! * (12-3)!) = 220
(1/4)^3 ≈ 0.015625
(3/4)^9 ≈ 0.06505
Therefore, P(3) ≈ 220 * 0.015625 * 0.06505 ≈ 0.2211
So, the probability of rolling exactly three 2s in twelve 4-sided dice rolls is approximately 0.2211, or 22.11%.