This is a binomial probability problem with n=11 (the number of drivers selected) and p=0.07 (the probability that a driver was involved in an accident).
To find the probability of getting 3 or more drivers who were involved in an accident, we need to find the probability of getting 3, 4, 5, ..., 11 drivers who were involved in an accident and then add those probabilities together.
We can use the binomial probability formula to calculate each of these individual probabilities:
P(X=k) = (n choose k) * p^k * (1-p)^(n-k)
where X is the number of drivers who were involved in an accident, k is the number of drivers we want to calculate the probability for, and (n choose k) is the binomial coefficient, which is calculated as:
(n choose k) = n! / (k! * (n-k)!)
Using this formula, we can calculate the probability of getting exactly k drivers who were involved in an accident.
For k=3, we have:
P(X=3) = (11 choose 3) * 0.07^3 * 0.93^8 = 0.1038
For k=4, we have:
P(X=4) = (11 choose 4) * 0.07^4 * 0.93^7 = 0.0286
For k=5, we have:
P(X=5) = (11 choose 5) * 0.07^5 * 0.93^6 = 0.0052
For k=6, we have:
P(X=6) = (11 choose 6) * 0.07^6 * 0.93^5 = 0.0007
For k=7, 8, 9, 10, 11, we have:
P(X=7) = (11 choose 7) * 0.07^7 * 0.93^4 = 0.0001
P(X=8) = (11 choose 8) * 0.07^8 * 0.93^3 = 0.0000
P(X=9) = (11 choose 9) * 0.07^9 * 0.93^2 = 0.0000
P(X=10) = (11 choose 10) * 0.07^10 * 0.93^1 = 0.0000
P(X=11) = (11 choose 11) * 0.07^11 * 0.93^0 = 0.0000
To get the probability of getting 3 or more drivers who were involved in an accident, we need to add up the probabilities for k=3, 4, 5, ..., 11:
P(X>=3) = P(X=3) + P(X=4) + P(X=5) + P(X=6) + P(X=7) + P(X=8) + P(X=9) + P(X=10) + P(X=11)
P(X>=3) = 0.1393
Therefore, the probability of getting 3 or more drivers who were involved in an accident out of 11 randomly selected drivers is 0.1393, or about 13.93%.