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If a coin has a $0.7$ chance of landing tails. What is the probability of getting EXACTLY $5$ heads in $10$ flips?

I know that the probability is $\frac{63}{256}$ if the coin is fair but I cannot work out how to do this problem. This is all i have so far: $$ \frac{(10!)}{2^{10}\cdot 5!\cdot5!} = \frac{252}{1024} = \frac{63}{256} $$

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  • $\begingroup$ Hint: Binomial distribution $\endgroup$ – celtschk Feb 14 '17 at 21:24
  • $\begingroup$ Refer this: en.m.wikipedia.org/wiki/Binomial_distribution $\endgroup$ – rookie Feb 14 '17 at 21:27
  • $\begingroup$ Are you looking for odds or probability? These two values are different. $\endgroup$ – amd Feb 14 '17 at 21:28
  • $\begingroup$ @amd meant probability, sorry $\endgroup$ – KeyboardLamp Feb 14 '17 at 21:36
  • $\begingroup$ @celtschk Not too sure on what binomial distribution is, but I read the wiki and got an answer in the region of 0.1029. Does this seem reasonable? $\endgroup$ – KeyboardLamp Feb 14 '17 at 21:40
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The number of ways you can arrange five heads and five tails (first toss to last) is $_{10}C_{5}$.

The probability of getting five of each, in one particular order (say all tails, then all heads), is $0.3^50.7^5$.

Can you take it from here?

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In the numerator: 5 heads, so $$\binom{10}{5}(1-0.7)^5(0.7)^{10-5}$$

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  • $\begingroup$ shouldn't i t be 0.3 in stead of the 0.7 in there? because the 0.7 represents the chance of landing a tail $\endgroup$ – KeyboardLamp Feb 14 '17 at 21:37
  • $\begingroup$ Yes, $1-0.7=0.3$ is the probability of head. So $5$ times we have heads, and the remaining $10-5=5$ times we have tails. The number of combinations is $\binom{10}{5}$. $\endgroup$ – msm Feb 14 '17 at 21:55

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