Probability of getting exactly 5 heads in 10 flips of an unfair coin with 0.7 chance of tails

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: $$\frac{(10!)}{2^{10}\cdot 5!\cdot5!} = \frac{252}{1024} = \frac{63}{256}$$

• Hint: Binomial distribution – celtschk Feb 14 '17 at 21:24
• – rookie Feb 14 '17 at 21:27
• Are you looking for odds or probability? These two values are different. – amd Feb 14 '17 at 21:28
• @amd meant probability, sorry – KeyboardLamp Feb 14 '17 at 21:36
• @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? – KeyboardLamp Feb 14 '17 at 21:40

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 $0.3^50.7^5$.
In the numerator: 5 heads, so $$\binom{10}{5}(1-0.7)^5(0.7)^{10-5}$$
• 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 $\binom{10}{5}$. – msm Feb 14 '17 at 21:55