The manufacturer of a fertilizer guarantees that, with the aid of the fertilizer, $70 \%$ of planted seeds will germinate. Suppose the manufacturer is correct. If 6 seeds planted with the fertilizer are randomly selected, what is the probability that more than 3 of them germinate?
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 ：This problem is a binomial probability problem. The probability of success, which is the germination of a seed, is given as 0.7. We are asked to find the probability that more than 3 seeds germinate out of 6. This means we need to find the probability that 4, 5, or 6 seeds germinate.

Step 2 ：We can use the binomial probability formula to calculate this: \(P(X=k) = C(n, k) * (p^k) * ((1-p)^(n-k))\), 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, n is the number of trials, and k is the number of successes.

Step 3 ：We can calculate this for k=4, 5, and 6 and sum the results to get the final probability. Let's denote p as the probability of success (0.7), and n as the number of trials (6).

Step 4 ：Calculating for k=4, we get \(P(X=4) = C(6, 4) * (0.7^4) * ((1-0.7)^(6-4))\).

Step 5 ：Calculating for k=5, we get \(P(X=5) = C(6, 5) * (0.7^5) * ((1-0.7)^(6-5))\).

Step 6 ：Calculating for k=6, we get \(P(X=6) = C(6, 6) * (0.7^6) * ((1-0.7)^(6-6))\).

Step 7 ：Adding these probabilities together, we get the total probability that more than 3 seeds germinate.

Step 8 ：Final Answer: The probability that more than 3 of the 6 seeds germinate is \(\boxed{0.74}\).