If a coin is flipped $10$ times, what are the possibilities of getting at least $5$ H?

Question

If a coin is flipped $10$ times what are the possibilities of getting at least $5$H?

Is a pascal triangle really needed to do this type of questions?

Answer 1

Here's my approach to the problem:

If there are exactly $5$ heads,

This can be done in ${10\choose 5}$ ways. Of the remaining $2^{10}-{10\choose 5}$ cases, half of the time, there are more heads than tails and half of the time there are more tails than heads.

Thus, there are $\frac{2^{10}-{10\choose 5}}{2}$ ways in this case with more than 5 heads. Adding them, we get $\frac{2^{10}-{10\choose 5}}{2}+{10\choose 5}=2^9+\frac{{10\choose 5}}{2}$ ways with at least 5 heads.