A fair coin is flipped 7 times. What is the probability that at least 5 of the flips come up heads?
Flipping coin: we set h = head and t = tail
After 7 times we have the probabilities:
$$\small{
\begin{array}{c|c|l}
\text{heads} & \text{tails} & \text{probability}\\
\hline
&&\\
\textcolor[rgb]{1,0,0}{7} & 0 & \binom70 \cdot (\frac{1}{2})^7 = \textcolor[rgb]{1,0,0}{1\cdot (\frac{1}{2})^7}\\&& \\
\textcolor[rgb]{1,0,0}{6} & 1 & \binom71 \cdot (\frac{1}{2})^7= \textcolor[rgb]{1,0,0}{7\cdot (\frac{1}{2})^7}\\&& \\
\textcolor[rgb]{1,0,0}{5} & 2 & \binom72 \cdot (\frac{1}{2})^7= \textcolor[rgb]{1,0,0}{21\cdot (\frac{1}{2})^7}\\&& \\
4 & 3 & \binom73 \cdot (\frac{1}{2})^7= 35\cdot (\frac{1}{2})^7\\&& \\
3 & 4 & \binom74 \cdot (\frac{1}{2})^7= 35\cdot (\frac{1}{2})^7\\&& \\
2 & 5 & \binom75 \cdot (\frac{1}{2})^7= 21\cdot (\frac{1}{2})^7\\&& \\
1 & 6 & \binom76 \cdot (\frac{1}{2})^7= 7\cdot (\frac{1}{2})^7\\&& \\
0 & 7 & \binom77 \cdot (\frac{1}{2})^7= 1\cdot (\frac{1}{2})^7\\&& \\
\hline
\end{array}
}$$
The probabiliies of getting 7 heads and 6 heads and 5 heads is:
$$\small{\text{
$
\textcolor[rgb]{1,0,0}{1\cdot (\frac{1}{2})^7} + \textcolor[rgb]{1,0,0}{7\cdot (\frac{1}{2})^7}
+\textcolor[rgb]{1,0,0}{21\cdot (\frac{1}{2})^7}
=(1+7+21)\cdot (\frac{1}{2})^7
=29\cdot (\frac{1}{2})^7
= 29 \cdot 0.00781250000$}}\\
\small{\text{
$
=0.22656250000
=22.66\%
$
}}$$
.
