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

 Apr 19, 2015

Best Answer 

 #2
avatar+25569 
+5

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:

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

 Apr 20, 2015
 #1
avatar+249 
+5

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

 Apr 19, 2015
 #2
avatar+25569 
+5
Best Answer

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:

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

heureka Apr 20, 2015
 #3
avatar+111115 
0

These are both really good answers.  :))

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

 Apr 20, 2015

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