Success is: picking 2 e's in a row.
The probability of choosing a single E from GEESE is 3/5.
Once I've chosen an E, the probability of choosing a second E, from the remaining letters, is 2/4. One less letter available, and one less E available. That's 1/2.
The probability of doing both of those, in succession, is the first times the second. (Think of it this way: 3/5 of the time, I pick an E, and then of that 3/5, I pick another E 1/2 of the time. So, to get 1/2 of 3/5, I multiply.)
So, that's 3/5 * 1/2, which reduces to 3/10.
Now, let's do PLEASE the same way:
The probability of choosing a single E is 2/6 (or 1/3).
Once I've chosen an E, the probability of getting the remaining E, out of the remaining letters, is 1/5.
The probability of doing both of those, in succession, is 1/3 * 1/5, or 1/15.
Now, put together the geese and the please, the probability of getting 2 E's from geese AND the 2 E's from please:
3/10 * 1/15 = 3/150, or 1/50.
Again, 3/10 of the time, I get 2 E's from geese, and then of that 3/10, I get another 2 E's from please 1/15 of the time. 1/15 of 3/10 is 1/50.
Those are the situations where I get all 4 E's. To do the "everything but E's" scenario, I do the same thing, but only choosing the non-E's. The short version:
GEESE: 2/5 * 1/4 = 1/10
PLEASE: 4/6 * 3/5 = 2/5
No E's from both: 1/10 * 2/5 = 2/50
We now need all 4 E's or no E's. If these are completely separate possibilities, then I can just add them together. For an example of when they're not completely separate, think of a deck of cards. If I'm trying to draw a ten or a spade, there's actually a card that works for both: namely, the ten of spades. So, if I just add the probabilities together (the four tens, plus the thirteen spades in the deck), I just counted the ten of spades twice, and that's going to mean that I've overcounted my possible success by a card.
In this situation, "All 4 E's" and "no E's" don't mix, so 1/50 + 2/50 = 3/50.
Hope this helps! Maybe other tutors have a simpler way of explaining it, too...