Question
Suppose that we flip a fair coin 4 times.
(a) What is the probability of at least two consecutive heads are flipped given that the first flip is a heads?
(b) Are the events in the above part independent?
My attempt
(a) H will denote heads, and T denote tails. Since our first flip is already a H, there are $2^3=8$ different ways that the remaining flips can go. Of these, there are four ways we can get two or more consecutive heads: $HHHH$, $HTHH$, $HHTT$, and $HHTH$. I'm not sure if there is a more systematic way to approach this other than listing them out.
Since there are 4 ways that we can get two or more consecutive heads out of the 8 total, the probability is $\frac{4}{8}=\frac{1}{2}$.
(b) The two events are (A) a flip turns out heads (B) that there are because the coin is fair, and the probability of (B) I found by again listing out the ways we can get two or more consecutive heads $HHHH$, $HHHT$, $HHTT$, etc of a total $2^4=16$ possibilities two coins could take. There are 8 ways to get two consecutive heads. Again, I'm not sure if there's a more systematic approach here.
Two events are independent if their product is equal to the intersection of the events. $P(A)*P(B)= \frac{1}{2}*\frac{1}{2} = \frac{1}{4}$, which is not equal to the probability we found in part (a), $P(A\cap B) = \frac{1}{2}$. So the events are not independent.