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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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    Hint: Binomial distribution – 2017-02-14
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    Refer this: https://en.m.wikipedia.org/wiki/Binomial_distribution – 2017-02-14
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    Are you looking for *odds* or *probability*? These two values are different. – 2017-02-14
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    @amd meant probability, sorry – 2017-02-14
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    @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? – 2017-02-14
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    @KeyboardLamp: Yes. – 2017-02-14
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    @celtschk Thanks man – 2017-02-14

2 Answers 2

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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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    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 – 2017-02-14
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    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 $\binom{10}{5}$. – 2017-02-14