Calculating conditional lottery probabilities - and example from DeGroot.

Question

I'm trying to understand this example in Probability and Statistics in DeGroot:

https://i.stack.imgur.com/b23Sr.jpg

"You learned that the event B = {one of the numbers drawn is 15} has occurred. You want to calculate the probability of the event A that your ticket is a winner [given B]."

It suggests that the probability of P(AnB) = P(A) but how can this be if there exist outcomes in A that aren't in B? Surely it should be equal to P(B) if anything?

Thanks.

Answer 1

The event $A$ is that you have won the lottery, presumably meaning that all your chosen numbers were "good."

The event $B$ is that at least one of your numbers is good. There are many more events in $B$ than in $A$, since there are many events in $B$ in which only the $15$ was "good," or only a couple of numbers including $15$ were good.

So the event $A$ is a subset of $B$, and therefore $A\cap B=A$. Having all good and at least one good is the same as having all good.

Answer 2

The probability of $A$ given $B$ denoted $P(A|B)$ will not be equal to $P(A)$

We do have a formula for conditional probability of $A$ given $B$ as shown below.

$P(A|B) = \frac{P(A \cap B)}{P(A)}$