The manufacturer of a fertilizer guarantees that, with the aid of the fertilizer, $70\mathrm{\%}$ 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: $P(X=k)=C(n,k)\ast (p^k)\ast ((1-p{)}^{(}n-k))$, 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 $P(X=4)=C(6,4)\ast ({0.7}^4)\ast ((1-0.7{)}^{(}6-4))$.

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

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

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 $\boxed{{\displaystyle 0.74}}$.