I toss a coin 6 times and get a total of four heads (H). What is the probability my first roll was heads?
My thinking:
Let B be the event of 4H out of 6 tosses. I do this by binomial.
$P(B) = \left(\frac{1}{64}\binom{6}{4}\right) = \frac{15}{64}$
Let A be the event the first toss is H. I originally thought this be $P(A)=\frac{1}{2}$ however I was looking through my notes at a similar question and I believe
$P(A) = \left(\frac{1}{64}\binom{5}{3} + \binom{5}{4} +\binom{5}{5}\right) = \frac{16}{64}$
Thus by bayes theorem: $$ P(A \mid B) = \frac{P(B \mid A) \, P(A)}{P(B)} $$
where ${P(B \mid A) \, P(A)}$ = $P(A \cap B)$
$P(A \mid B) =\frac{15}{16}$
The answer in the book is $\frac{2}{3}$ so clearly my answer is wrong. Any guidance as to why I did not get $\frac{2}{3}$ for the probability the first role is a head? Thank you.