Suppose that $45 \%$ of all babies born in a particular hospital are boys. If 7 babies born in the hospital are randomly selected, what is the probability that at least 2 of them are boys?
Carry your intermediate computations to at least four decimal places, and round your answer to two decimal places.
(If necessary, consult a list of formulas.)

Step 1 ：We are given that 45% of all babies born in a particular hospital are boys. If 7 babies born in the hospital are randomly selected, we are asked to find the probability that at least 2 of them are boys.

Step 2 ：This is a binomial probability problem. The binomial distribution model is appropriate here because we have a fixed number of independent trials (7 babies), each trial results in one of two outcomes (boy or girl), and the probability of success (having a boy) is the same for each trial (45%).

Step 3 ：The probability of having at least 2 boys is equal to 1 minus the probability of having less than 2 boys (0 or 1 boy). We can calculate this using the binomial probability formula: \(P(X = k) = C(n, k) * (p^k) * ((1-p)^(n-k))\), where \(P(X = k)\) is the probability of k successes in n trials, \(C(n, k)\) is the number of combinations of n items taken k at a time, p is the probability of success on a single trial, n is the number of trials, and k is the number of successes.

Step 4 ：Let's calculate the probability of having 0 or 1 boy. We find that the probability is approximately 0.1024.

Step 5 ：Subtracting this from 1 gives us the probability of having at least 2 boys, which is approximately 0.9.

Step 6 ：Final Answer: The probability that at least 2 of the 7 babies born in the hospital are boys is \(\boxed{0.90}\).