A fair coin is flipped 7 times. What is the probability that at least 5 of the flips come up heads?

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:

$$\small{\text{ $ \textcolor[rgb]{1,0,0}{1\cdot (\frac{1}{2})^7} + \textcolor[rgb]{1,0,0}{7\cdot (\frac{1}{2})^7} +\textcolor[rgb]{1,0,0}{21\cdot (\frac{1}{2})^7} =(1+7+21)\cdot (\frac{1}{2})^7 =29\cdot (\frac{1}{2})^7 = 29 \cdot 0.00781250000$}}\\ \small{\text{ $ =0.22656250000 =22.66\% $ }}$$

Use the binomial distribution where N=7 , P = 0.5, 1 - P = 0.5

Calculate the probability of getting 5 heads, 6 heads, 7 heads and then add them all up.

eg. probability of getting 5 heads is (7C5) x (0.5^5) x (0.5 ^2) = 0.164

These are both really good answers.  :))

Heureka, your presentation never fails to astound, but Brodude's answer is really good too :))