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An urn contains seven green balls and six white balls. A sample of four balls is selected at random from the urn.
(a) Find the probability that the four balls have the same color.
(b) Find the probability that the sample contains more green balls than white balls.
(a) The probability that the four balls have the same color is $◻$.
(Type an integer or a simplified fraction.)
(b) The probability that the sample contains more green balls than white balls is $◻$.
(Type an integer or a simplified fraction.)

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 ：$\boxed{{\displaystyle \frac{10}{143}}}$ 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 ：$\boxed{{\displaystyle \frac{112}{143}}}$ is the probability that the sample contains more green balls than white balls.