Final answer:
To find the probability of rolling no more than two twos in 12 rolls of a die, compute the sum of individual binomial probabilities for rolling exactly 0, 1, and 2 twos.
Step-by-step explanation:
The question is asking to find the probability of rolling no more than two twos when a die is rolled 12 times. This can be considered as a binomial probability problem, where 'success' is rolling a two (with probability 1/6) and 'failure' is rolling any other number (with probability 5/6).
To solve this, we calculate the probabilities of rolling exactly 0, 1, and 2 twos and sum them up. The binomial probability formula P(X=k) = (n choose k) * (p^k) * (q^(n-k)) where 'n' is the number of trials, 'k' is the number of successful outcomes, 'p' is the probability of success, and 'q' is the probability of failure (q=1-p), can be applied here for k=0, 1, and 2.
Thus, the calculation for each case is as follows:
- For 0 twos: P(X=0) = (12 choose 0) * (1/6)^0 * (5/6)^12
- For 1 two: P(X=1) = (12 choose 1) * (1/6)^1 * (5/6)^11
- For 2 twos: P(X=2) = (12 choose 2) * (1/6)^2 * (5/6)^10
Finally, combine the probabilities of these three separate events to find the total probability of rolling no more than two twos:
P(X ≤ 2) = P(X=0) + P(X=1) + P(X=2)