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What is the probability of flipping a coin 8 times and getting heads 3 times? Round your answer to the nearest tenth of a percent. A.21.9% B.10.9% C.3.1% D.27.3%

Mathematics
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If you get head3 3 times, you also get tails 5 times. So there are only two possible outcomes. Each of the two outcomes has possibility 1/2. Say X is the number of heads when flipping 8 times. Then X is binomially distributed. The distribution function for such a process is:\[P(X=3)=\left(\begin{matrix}8 \\ 3\end{matrix}\right)p^3(1-p)^{5}=\left(\begin{matrix}8 \\ 3\end{matrix}\right)\left(\frac{1}{2}\right)^3\left(\frac{1}{2}\right)^5=\left(\begin{matrix}8 \\ 3\end{matrix}\right)\left(\frac{1}{2}\right)^8\]If you find this a little intimidating, you're right ;) It looks more complicated than it actually is. One possible outcome would be: hhttthtt. The chance of getting precisely this outcome is:\[\left( \frac{ 1 }{ 2 }\right)^3 \cdot \left( \frac{ 1 }{ 2 }\right)^5 = \left( \frac{ 1 }{ 2 }\right)^8 \]But this is not the only way go get 3 heads and 5 tails. Any combination of 3xh and 5xt would be ok. That is what the 8 above 3(between brackets) is about. It is a binomial coefficient. It is calculated as follows:\[\left(\begin{matrix}8 \\ 3\end{matrix}\right)=\frac{ 8! }{ 3!5! }=\frac{ 6 \cdot 7 \cdot 8 }{ 2 \cdot 3? }=56\]So there are 56 ways to get 3xh and 5xt. If you now multiply 56 and the outcome of (1/2)^8, you'll have the answer. Multiply that with 100 to get a percentage!
I can't figure it out
\[56 \cdot \left( \frac{ 1 }{ 2 } \right)^8=56 \cdot \frac{ 1 }{ 2^8 }=\frac{ 56 }{ 256 }=0.21875\]So times 100 and rounding off gives 21.9%

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