[Puzzle] A Puzzle a Day #19 - Back to the Classics (Math)

Math puzzles haven't really been all that popular around here. It makes sense - I mean, who wants to get off from class just to do more work? Just kidding. Do a search for math and you'll see tons of random math quizzes all around the place. Well, when there's a prize involved, I suppose that's enough of an incentive. Chour has involved math with several of his puzzles, but I dont believe any of his quizzes are entirely math based. There are plenty of other users that have made math puzzles as well, and I'm not going to lay them all down here. I have noticed, however, that none of the quizzes ever get more complicated than simple calculus. I think one day I'll go into complex analysis, but today is not that day.

Instead, take some simple probabilities

Rules:
Do not share answers, hints, giveaway links, the name of the giveaway or generally spoil the puzzle in any form.

If you solve at least half (15) of the puzzles, you will get an invite to a private giveaway when the A Puzzle a Day series concludes. I'll try to pick something that no one has already, but no guarantees. Please make a parent post in the giveaway to be counted. Also, to be invited to the bonus giveaway, post in A Puzzle a Day #30 with the puzzles you have solved. You are not required to solve the puzzles before the giveaways end. Solving them at any point in time will count towards the 15 required puzzles.

Solution:

1. When you enter, there are 100 entries total, so you have a 1/100 chance to win = 1.0000%

2. Let P(W) be the chance you win and P(L) be the chance you lose. The chance to win at least one is defined as
   P(At least one) = 1- P(L)^3 = 5.88080%

3. The chance to win exactly one is defined as the chance to win giveaway #1 but not #2 or #3 plus the chance to win giveaway #2 but not #1 or #3 plus the chance to win giveaway #3 but not #1 or #2.

   P(exactly one) = P(W)P(L)P(L) + P(L)P(W)P(L) + P(L)P(L)P(W) = (P(W)P(L)P(L))^3 = 5.76240%

4. Chance that you will win all three is defined as

   P(all three) = P(W)^3 = 0.00080%

5. This is similar to question 3, but only take the first term

   P(only first) = P(W)P(L)P(L) + P(L) = 1.92080%

6. Let H be the event that you enter and N be the event that you don't. Then let D, S, and W be the event that the giveaway is real, the giveaway is deleted by the mods, and the giveaway is deleted by the giveaway owner, respectively. We are given the following:

   P(H) = .9, P(N) = .1
   P(D|N) = .8, P(S|N) = .2, P(W|N) = 0
   P(D|H) = .65, P(S|H) = .2, P(W|H) = .15

   The probability that that the giveaway is real is given by

   P(real) = P(D|H)P(H) + P(D|N)P(N) = 66.50000%

   And thus the probability that the giveaway is deleted is

   P(deleted) = 1 - P(real) = 33.50000%

7. This is just a simple application of Bayes Theorem

   P(N/D) = P(D/N)P(N) / P(D) = 12.00000%