Why do you multiply probabilities even if they are dependent? I thought that if you have two events, A and B, and you want to find P(A n B), you do P(A) x P(B) ONLY IF they are independent events.
But, I'm not really confused. There's a question on red and yellow counters, where there are 4 red counters and 5 yellows in a bag. Two counters are removed, without being replaced. What is the probability of picking both counters yellow?
I know that the answer is 20/72 - because it's the probability of yellow for the first pick (5/9) x the probability of yellow for the second pick (4/8).
But surely this isn't allowed because the events are not independent?
The first event clearly affects the second event...
 A: We  have $$P(A \cap B) = P(B)P(A|B)$$
When they are independent, then $P(A|B)=P(A)$.
You are not multiplying by the probability of getting yellow for the second pick. You are multiplying by the probability of getting yellow for the second pick given that the first pick is yellow.
A: Further to Siong Thye’s answer:
Since the two events $Y_1$ and $Y_2$ are indeed dependent on each other, $P(Y_1\cap Y_2)=P(Y_1)\times P(Y_2)=\frac59\times\frac59$ would have been the wrong computation.
Instead, you computed $P(Y_1\cap Y_2)=P(Y_1)\times P(Y_2|Y_1)=\frac59\times\frac48,$ which is correct.
(If it helps, one might think of event $Y_1$ and conditional event $(Y_2|Y_1)$ as being independent of each other.)
Notice that in a probability tree (aka tree diagrams), the probability of a particular outcome simply equals the product of the individual probabilities along its branches, regardless of whether the trials are dependent or independent?
The same reason applies: any dependence among the trials has implicitly been taken into account, because by design, a probability tree displays its trial-outcome probabilities as conditional probabilities.
