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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?

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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

#2
+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:

$$\small{ \begin{array}{c|c|l} \text{heads} & \text{tails} & \text{probability}\\ \hline &&\\ {7} & 0 & \binom70 \cdot (\frac{1}{2})^7 = {1\cdot (\frac{1}{2})^7}\\&& \\ {6} & 1 & \binom71 \cdot (\frac{1}{2})^7= {7\cdot (\frac{1}{2})^7}\\&& \\ {5} & 2 & \binom72 \cdot (\frac{1}{2})^7= {21\cdot (\frac{1}{2})^7}\\&& \\ 4 & 3 & \binom73 \cdot (\frac{1}{2})^7= 35\cdot (\frac{1}{2})^7\\&& \\ 3 & 4 & \binom74 \cdot (\frac{1}{2})^7= 35\cdot (\frac{1}{2})^7\\&& \\ 2 & 5 & \binom75 \cdot (\frac{1}{2})^7= 21\cdot (\frac{1}{2})^7\\&& \\ 1 & 6 & \binom76 \cdot (\frac{1}{2})^7= 7\cdot (\frac{1}{2})^7\\&& \\ 0 & 7 & \binom77 \cdot (\frac{1}{2})^7= 1\cdot (\frac{1}{2})^7\\&& \\ \hline \end{array} }$$

$$\small{\text{  {1\cdot (\frac{1}{2})^7} + {7\cdot (\frac{1}{2})^7} +{21\cdot (\frac{1}{2})^7} =(1+7+21)\cdot (\frac{1}{2})^7 =29\cdot (\frac{1}{2})^7 = 29 \cdot 0.00781250000}}\\ \small{\text{  =0.22656250000 =22.66\%  }}$$

Apr 20, 2015

#1
+249
+5

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

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

Apr 19, 2015
#2
+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:

$$\small{ \begin{array}{c|c|l} \text{heads} & \text{tails} & \text{probability}\\ \hline &&\\ {7} & 0 & \binom70 \cdot (\frac{1}{2})^7 = {1\cdot (\frac{1}{2})^7}\\&& \\ {6} & 1 & \binom71 \cdot (\frac{1}{2})^7= {7\cdot (\frac{1}{2})^7}\\&& \\ {5} & 2 & \binom72 \cdot (\frac{1}{2})^7= {21\cdot (\frac{1}{2})^7}\\&& \\ 4 & 3 & \binom73 \cdot (\frac{1}{2})^7= 35\cdot (\frac{1}{2})^7\\&& \\ 3 & 4 & \binom74 \cdot (\frac{1}{2})^7= 35\cdot (\frac{1}{2})^7\\&& \\ 2 & 5 & \binom75 \cdot (\frac{1}{2})^7= 21\cdot (\frac{1}{2})^7\\&& \\ 1 & 6 & \binom76 \cdot (\frac{1}{2})^7= 7\cdot (\frac{1}{2})^7\\&& \\ 0 & 7 & \binom77 \cdot (\frac{1}{2})^7= 1\cdot (\frac{1}{2})^7\\&& \\ \hline \end{array} }$$

$$\small{\text{  {1\cdot (\frac{1}{2})^7} + {7\cdot (\frac{1}{2})^7} +{21\cdot (\frac{1}{2})^7} =(1+7+21)\cdot (\frac{1}{2})^7 =29\cdot (\frac{1}{2})^7 = 29 \cdot 0.00781250000}}\\ \small{\text{  =0.22656250000 =22.66\%  }}$$

heureka Apr 20, 2015
#3
+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