What's the difference between $P[A \cap B]$ and $P[A | B]$ in this question?

Question

Problem: We roll a die. If we obtain a 6, we choose a ball from box A where three balls are white and two are black. If we do not obtain a 6, we choose a ball from box B where two balls are white and four are black.

(a) What is the probability of obtaining a white ball?

The answer on my notes went something like this: $P[\text{white ball}] = P[\text{get 6} \cap \text{white ball}]$ + $P[\text{not 6} \cap \text{whiteball}]$.

Why not $P[\text{white ball}] = P[\text{white ball} | \text{get 6}]$ + $P[\text{white ball} | \text{not 6}]$?

What's the difference here?

Answer 1

$P(\text{white ball}|\text{get 6}) + P(\text{white ball}|\text{not 6})$ just summates the probability of obtaining a white ball in the scenario you obtained six with the probability of obtaining a white ball in the scenario you did not obtain six.

You can see that this expression is incorrect by two ways. First, it is not a valid probability. For example, if the two boxes had only white balls, this expression would be 2. You might suggest to solve this issue by replacing the summation with averaging: $\frac{1}{2}P(\text{white ball}|\text{get 6}] + \frac{1}{2}P(\text{white ball}|\text{not 6})$

However, here comes the second issue with this expression - you must take into account $P(\text{get 6})$ and $P(\text{not 6})$. if you just average the conditional probabilities, you implicitly assume that $P(\text{get 6})$=$P(\text{not 6})=0.5$.

The correct solution using conditional probabilities requires multiplying them with the prior probabilities: $P(\text{white ball}|\text{get 6})P(\text{get 6}) + P(\text{white ball}|\text{not 6}]P(\text{not 6})$. This solution is equal to your first solution since $P(A \cap B)=P(A|B)P(B)$.