Finding the probability that in ten dice throws the digit six will appear at least five times 
Find the probability that in ten dice throws the digit six will appear at least five times.

My attempt:
Using complement, six will appear at most 4 times.
I'm pretty sure that $1-(\frac 5 6)^4$ isn't right since it doesn't account for six appearing 3, 2, 1, 0 times. So using inclusion exclusion:
Six not appearing: $(\frac 5 6 )^4$
Six appearing once: $\frac 1 6 \cdot  (\frac 5 6 )^3 $
Six appearing twice: $(\frac 1 6 \cdot \frac 5 6 )^2$
Six appearing three times: $(\frac 1 6)^3 \cdot \frac 5 6 $
Six appearing four times: $(\frac 1 6) ^4$
Calculating the above with inclusion exclusion: $1-( (5/6)^4 -(1/6)*(5/6)^3+(5/36)^2 - (1/6)^3*(5/6) + (1/6)^4 )=0.59799382716$
Now when I wrote this I noticed the complement and the given case are symmetric, the only difference between them is the case where six appearing exactly 5 times, so I tried to plug it into the calculator and got a very similar result: $1-(5/6)^5=0.59812242798$.
Are either of them correct? Why do we subtract the middle point from the general case in the symmetric approach?

There are $10$ independent experiments here that all have a chance $\frac16$ to succeed (i.e. a six is thrown). If $X$ denotes the number of successes then it has binomial distribution with parameters $n=10$ and $p=\frac16$. That determines its PMF (probability mass function): $$P(X=k)=\binom{10}{k}\left(\frac16\right)^k\left(\frac56\right)^{10-k}$$
To be found here is $$P(X\geq5)=P(X=5)+\cdots+P(X=10)$$
