This post is quite old, but I'm surprised that no one pointed out that in problem given by @Potato,
$P(A∩B)=\frac{1}{2}P(B)$
so
$P(A|B)=\frac{P(A∩B)}{P(B)}=\frac{\frac{1}{2}P(B)}{P(B)}=\frac{1}{2} $
One can see this by thinking about the sets of points involved. B is the set of points on the equator. A∩B is intersection of the set of points in the area of the Western Hemisphere with the set of points on the equator. In other words, A∩B is the set of points on the equator in the Western hemisphere, which is half the set of points on the equator. It doesn't matter that the chances of picking a point exactly on the equator are infinitesimally small. The chance of picking a point on the equator in the Western hemisphere is always exactly half that of picking a point anywhere on the equator. Basically you have to think about what the sets of points involved mean, how they are related, and not get caught up so much on the equator being an infinitesimally small portion of the globe. The ratio of probability relationships associated with halves and wholes remains the same, even in the presence of infinitesimals (or infinities). Assuming a uniform distribution of global point choices, one has to think in terms of "the probability of X for this half is the same as the probability of X for the other half."
It's also important to realize that this line of thinking is simple in this case, because the equator is orthogonal to the meridians that define East/West hemispheres, and both bisect the surface area into exact halves. If one were to use a different circumference, such as one at an angle to the equator, or if one chooses a different latitude than the equator, it gets more complicated.
As an aside, Bayes' Rule works to get the same answer:
$P(A|B)=\frac{P(B|A)P(A)}{P(B)} $
Then observe that
$P(B)=P(B|A)P(A)+ P(B|¬A)P(¬A) $
Which is to say that the probability of picking a point somewhere on the equator is equal to the sum of the probability of picking a point on the equator from the set of points comprising the Western hemisphere weighted by the probability of picking a point in the Western hemisphere from the global set of points and plus the probability of picking a point on the equator from the set of points that are not in the Western hemisphere weighted by the probability of picking a point that is not in the Western hemisphere from the global set of points. Making that substitution gives
$P(A|B)=\frac{P(B|A)P(A)}{P(B|A)P(A)+ P(B|¬A)P(¬A)} $
Next note that P(A) is just the probability of picking a point in the Western hemisphere from the global set of all points. Since the Western hemisphere is, by definition, half of the globe,
$P(A)=\frac{1}{2}$
and since the part of the globe that is not in the Western hemisphere is simply the other half,
$P(¬A)=\frac{1}{2}$
P(B|A) is the probability of a point being on the equator given that it is in the Western hemisphere. This probability is infinitesimally small, but we know that, as we have stated, the Western hemisphere is half the globe, and that exactly half of the equator lies in the Western hemisphere and the other half in the Eastern, we can say that this infinitesimally small probability is the same for both hemispheres, so
$P(B|A)=P(B|¬A)$
Making these substitutions gives
$P(A|B)=\frac{P(B|A)\frac{1}{2}}{P(B|A)\frac{1}{2}+ P(B|A)\frac{1}{2}}=\frac{1}{2} $