We toss a fair coin $k$ times and we set the following possible results:
$A$: both heads and tails appear at least 1 time. $B$: tails appears at least one time.
For which values of $k$ are $A$ and $B$ independent ?
Now i know that the intersection of $A$ and $B$ is $B$ itself and I want to use the Bayes rule but I cant figure out the probability of $A$.
So how to find $A$???
If $A \cap B = A$ then to show independence you would have to show $P[A] = P[A \cap B] = P[A]P[B]$, which implies that $P[A] = 1,0$. When, if ever, does this happen?
By the way, to compute $A$, let $X$ be the number of heads in $k$ flips. This has a binomial distribution with probability $p=1/2$. The set $A$ is the event $$ A = \bigcup_{i=1}^{k-1}\{ X = i\} $$ (if you have at least one head, but not $k$, there will also be at least one tail). You can compute $P[A]$ then as $$ P[A] = \sum_{i=1}^{k-1} \binom{k}{i} p^i(1-p)^{k-i} = \frac{1}{2^k}\sum_{i=1}^{k-1} \binom{k}{i} $$