Step 1 :Calculate the total number of ways to select 4 balls out of 13 using the combination formula \(C(n, k) = \frac{n!}{k!(n-k)!}\). Here, \(n=13\) and \(k=4\). So, \(C(13, 4) = \frac{13!}{4!(13-4)!} = 715\).
Step 2 :Calculate the number of ways to select 4 green balls out of 7 using the combination formula. Here, \(n=7\) and \(k=4\). So, \(C(7, 4) = \frac{7!}{4!(7-4)!} = 35\).
Step 3 :Calculate the number of ways to select 4 white balls out of 6 using the combination formula. Here, \(n=6\) and \(k=4\). So, \(C(6, 4) = \frac{6!}{4!(6-4)!} = 15\).
Step 4 :Add the number of ways to select 4 green balls and 4 white balls to get the total number of favorable outcomes. So, \(35 + 15 = 50\).
Step 5 :Calculate the probability that the four balls have the same color by dividing the number of favorable outcomes by the total number of outcomes. So, \(\frac{50}{715} = \frac{10}{143}\).
Step 6 :\(\boxed{\frac{10}{143}}\) is the probability that the four balls have the same color.
Step 7 :Calculate the number of ways to select 3 green balls and 1 white ball. Here, \(n=7, k=3\) for green balls and \(n=6, k=1\) for white balls. So, \(C(7, 3) \times C(6, 1) = \frac{7!}{3!(7-3)!} \times \frac{6!}{1!(6-1)!} = 35 \times 6 = 210\).
Step 8 :Calculate the number of ways to select 2 green balls and 2 white balls. Here, \(n=7, k=2\) for green balls and \(n=6, k=2\) for white balls. So, \(C(7, 2) \times C(6, 2) = \frac{7!}{2!(7-2)!} \times \frac{6!}{2!(6-2)!} = 21 \times 15 = 315\).
Step 9 :Add the number of ways to select 4 green balls, 3 green balls and 1 white ball, and 2 green balls and 2 white balls to get the total number of favorable outcomes. So, \(35 + 210 + 315 = 560\).
Step 10 :Calculate the probability that the sample contains more green balls than white balls by dividing the number of favorable outcomes by the total number of outcomes. So, \(\frac{560}{715} = \frac{112}{143}\).
Step 11 :\(\boxed{\frac{112}{143}}\) is the probability that the sample contains more green balls than white balls.