On a practice test with no available solutions I was asked the following two-part question:
1) If a coin is tossed until three consecutive heads are shown, what is the probability that one tail is shown?
2) Given that the number of tosses is less than or equal to 4, what is the probability that the coin is tossed 3 times?
My approach to the first part was to "or" (add) together the probabilities of the three possible scenarios which could result in three consecutive heads preceded by a single tail.
$$ P(1tail)=P(THHH)+P(HTHHH)+P(HHTHHH) =\frac{1}{2^4}+\frac{1}{2^5}+\frac{1}{2^6} =\frac{7}{64} $$
This didn't match any of the available solutions, but it's very close to $\frac{3}{32}$, which was one of them. Is my approach incorrect?
For the second part, let's say $N$ is the number of tosses. This is really looking for $P(N=3|N\le4)$. The probability of having only three tosses is equal to that of three consecutive heads, or $P(N=3)=P(HHH)=\frac{1}{8}$, which remains true given any sequence longer than $3$, so $P(N=3|N\le4)=\frac{1}{8}$. But this also is not an choice, and my reasoning feels shady.
I feel like I'm thinking about this all wrong, but can't find my error.
Your reasoning for the first part is correct. We do need to know that with probability $1$, three consecutive heads will occur, but this is intuitively clear.
For the second, let $A$ be the event $N=3$ and $B$ be the event $N\le 4$. Then $\Pr(A|B)=\Pr(A\cap B)/\Pr(B)$. Note that $\Pr(A\cap B)=\Pr(A)=1/8$ and $\Pr(B)=1/8+1/16$. The ratio simplifies nicely.