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If a biased coin is flipped $10$ times, how can I calculate the probability of NOT having more than $5$ runs? A run or group is the maximal sequence of consecutive flips that are all the same. For instance HTTHHH has a total of three runs.

There is a similar thread here but the offered solution is unclear. Thanks

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    Can you calculate the complement? That is, can you calculate the probability of having more than $5$ runs? – 2017-02-06
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    Like The Count was saying, it's usually easier to find the compliment of an event occurring as opposed to finding the probability itself. – 2017-02-06
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    Do you know how to calculate the compliment? – 2017-02-06
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    @TheCount It is actually easier to calculate the probability of the event than the complement, in this case. – 2017-02-06

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$P(X=k)$ is the probability of $k$ groups.   For a fair coin, there are $2\binom {9} {k-1}$ outcomes in this event, among a sample space of $2^{10}$ equally probable outcomes.   Calculated by selecting $k-1$ from the $9$ 'spaces' between the flips where a change of group could happen.

Can you adjust this for a biased coin?

So you now can find $\mathsf P(X < 5)$