You flip a fair coin 3 times, let $B = \text{At least 2 flips T}$ and $A = \text{2nd flip T}$, find $P(A|B)$
I know their individual probability, which is $P(A) = 1/2$ and $P(B) = (3C2)(1/2)^3 + (1/2)^3 = 1/2$
We have $P(A|B) = P(AB)/P(B)$
These events are NOT independent right?
How would I calculate, $P(AB)$?
What is the sample space for your experiment {$TTT,HHH,TTH,THT,HTT,HHT,HTH,THH$}. Just see how many of the cases have atleast two tails.There are $4 $ cases. And once you know this, this reduces your sample size from $8$ to $ 4$ ,then find out how many cases are there with tails on the second flip , they are $3$ of them among the $4$ you shortlisted .
So your probability is $P(A|B)=\frac{3}{4}$
We can simply check the corresponding events for $A \cap B $ in the sample space which come out to be: $\{HTT, TTH, TTT\} $ giving us a probability of $\frac {3}{8} $. Thus, $$\boxed {P (A \mid B) = \frac {3}{8} \times \frac {2}{1} = \frac {3}{4}}$$ Hope it helps.
Hint -
Write sample spaces of A and B.
Then its easy to find $A \cap B$
How would I calculate, $P(AB)$?
It is the probability for the second flip showing a tail and at least one of the other flips also doing so (so there is at least two tails, total). That is, for the second flip to show a tail and not both of the others to show heads.
$\mathsf P(A\cap B) = \tfrac 12(1-\tfrac 12\tfrac 12)=\tfrac 38$
And since $\mathsf P(B)=\tfrac 12$ then $\mathsf P(A\mid B) = \tfrac 34$
This is the probability for at least two of the coin flips showing a tail, when given that the second flip does; which is also simply the probability for at least one of the first or third coin flips showing a tail.
These events are NOT independent right?
Well, it is usually best not to just assume some events are independent or not, without doing some calculation, because our intuition on this can very often be misleading.
However, indeed, since $\mathsf P(A)~\mathsf P(B)=\tfrac 12\times \tfrac 12$ we can now affirm that they are not independent events, since the product of their probabilities does not equal the probability of their intersection.
What we did know from the beginning was that the outcome of each individual flip was independent from either or both of the others. We knew this from past experience of how coin flips work.