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A coin is flipped five times. For each of the events described below, express the event as a set in roster notation. Each outcome is written as a string of length 5 from{H, T}, such as HHHTH. Assuming the coin is a fair coin, give the probability of each event.

(c) The first flip comes up tails and there are at least two consecutive flips that come up heads.

When I did this via brute force, I came up with the following event of size 8:

{THHHH, THHHT, THHTH, THHTT, TTHHH, TTHHT, THTHH, TTTHH}. Since there are $2^5=32$, possibilities for the 5 coin flips, we have a probability of $\frac{8}{32} = \frac{1}{4}$.

This makes sense to me, the issue is that I tried to sanity check myself by doing this via complement. The idea being that p(consecutive heads)+p(non-consecutive heads)=1. Since the first flip must be tails, we're really looking at the ways to achieve consecutive and non-consecutive heads in 4 flips.

With that, there is only 1 way to see 0 heads, TTTTT, so $p(0heads)=\frac{1}{16}$.

In the case of 1 non-consecutive heads flip, there are 4 positions for it to be, thus $p(1heads) = \frac{4}{16}$.

Lastly, for the case of 2 non-consecutive heads flips, there are only two possibilities HTHT and THTH. Thus $p(2heads) = \frac{2}{16}$.

Using the sum rule we end up with $p(non-consecutive\, heads) = \frac{7}{16} \Rightarrow p(consecutive\, heads) = \frac{9}{16}$. Then using the product rule, since the first flip has to be heads, we end up with a probability of $\frac{1}{2} \cdotp \frac{9}{16} = \frac{9}{32}$.

I can't find the error in my logic for either of these methods but clearly there is one somewhere... If you can find it please let me know and thank you in advance.

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For two heads you missed $HTTH$. With the starting tail, that is the $\frac 1{32}$ that is missing.

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  • $\begingroup$ Ah! That was a bonehead move. Thanks for catching it. $\endgroup$ – L. Gomez Apr 3 at 3:35

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