Coin flips

There are $2^{10}$ ways to flip 10 coins. And the number that have exacly 5 heads and 5 tails is $\binom{10}{5} = \frac{10!}{5!*5!}$ . So the answer is $\binom{10}{5}/(2^{10}) = \boxed{0.246}$

We can imagine the different ways to obtain 5 heads and 5 tails as a permutation of 10 elements (10 throws) with two sets of identical elements (5 heads and 5 tails):
$\frac {10!}{5! * 5!}={10 \choose 5}=252$
The total number of combinations of 10 coin throws (2 possibilities: head or tail) is
$2^{10}=1024$
So the probability is
$\frac {252}{1024}=\boxed{0.246}$

By the binomial distribution theorem: $\frac{10!}{5!(10-5)!} \times 0.5^2 \times 0.5^2 = 0.246$