Combinatorics - Probability

and or events

Let's consider the following set of 7 books from which we can win 2 books: one at the time:
Books = {Physics1, Physics2, Physics3, Math1, Math2, History, Philosophy}

Let's consider:
A is the event associated to "win a Physics book", and B is the event associated to "win a Math book"

These two events are mutually exlusive (incompatible or disjoint) . That is the two events (win a book) cannot occur at the same time: we win only a book at a time.

1. The first win

The probability to win, at first, a Physics book, is:
P(A) = 3/7

The probability to win, at first, a Math book, is:
P(B) = 2/7

The probability to win a Physics book AND a Math book is:
P(A ∩ B ) = P(Φ) = 0 (since "A" and "B" are incompatible).

The probability to win a Physics book OR a Math book is: P(A ∪ B ) = P(A) + P(B) - P(A ∩ B) = 3/7 + 2/7 - 0 = 5/7

2. The second and last win

Given the fact that one Physics book is already won, the probability to win, a Math book, is:
P(B/A) = 2/(7 - 1) = 2/6 = 1/3

The event "B" is dependent of the event "A" because it changes its probability (from 2/7 to 2/6). The event "A" influences the event "A".

The probability to win a Physics book AND a Math book is:
P(A ∩ B) = P(B/A) x P(A) = 1/3 x 3/7 = 1/7


The probability to win a Physics book OR a Math book is: P(A ∪ B ) = P(A) + P(B) - P(A ∩ B) = 3/7 + 2/7 - 1/7 = 4/7

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P(A ∪ B ) =
P(A) + P(B) - P(A ∩ B)

P(A ∩ B ) =
P(B/A) x P(A)

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3. Let's get money instead

Now, let's assume that once we win a book, we replace it and get its price in dollars. In this case, the two events "A" and "B" remain mutually exlusive, but become independent. That is the event "A" does not influences the event "B". Hence P(B) remains equal to 2/7. In other words, P(B/A) = P(B) = 2/7 (the fact that the event "A" has occured does not affect the event "B" at all).
Then:
P(A) = 3/7
P(B) = 2/7
P(A ∩ B ) =
P(B/A) x P(A) = P(B) x P(A) = 2/7 x 3/7 = 6/49, and
P(A ∪ B ) =
P(A) + P(B) - P(A ∩ B) =
3/7 + 2/7 - 6/49 = 8/49

4. Winning two books at a time

Now, we consider the case of we win two books at a time
Winning two books can happen from the following set of 21 pairs:
Books x Books = {(Physics1, Physics2), (Physics1, Physics3), (Physics1, Math1), (Physics1, Math2), (Physics1, History), (Physics1, Philosophy), (Physics2, Physics3), (Physics2, Math1), (Physics2, Math2), (Physics2, History ), (Physics2, Philosophy), (Physics3, Math1), (Physics3, Math2), (Physics3, History ), (Physics3, Philosophy ), (Math1, Math2 ), (Math1, History), (Math1, Philosophy), (Math2, History ), (Math2, Philosophy), (History, Philosophy)}

The number of paires from the set "Books" is the number of combinations of two books of the set "Books". It is C(2,7) = 7!/2!(7 - 2)! = 6 x 7 /2 = 21

Then:
P(A ∩ B ) = 6/21 = 2/7, and
P(A ∪ B ) = 20/21 (= 1 - P(History, Philosophy))

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incompatible = do not occur at the same time independent = do not affect each other
P(A ∪ B ) = P(A) + P(B) - P(A ∩ B)
P(A ∩ B ) = P(B/A) x P(A)
P(event) <= 1

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