Contents
 • one choice • many choices: permutations • many choices: combinations • many choices: combinations • independent events • dependent events • permutations without repettitions • permutations with identical elements • combinations • coditional probabilities • and or events • two outcomes • total probability theorem • Bayes rule • union sets and probability • fundamental counting principle

 Other exercices

 Combinatorics - Probability
Example5: Independent events

Independent events
combinations

Many choices: Independent events

Because we need 6 books, we want to choose 2 books from each the followings:
-- 5 books of Mathematics,
-- 4 books of Physics, and
-- 2 books of Biology.
How many possibilities do we have?
We have:
C(2,5) = 10 possibilities to choose 2 books among the 5 books of Mathematics.

C(2,4) = 6 possibilities to choose 2 books among the 4 books of Physics.
C(2,2) = 1 possibility to choose 2 books among the 2 books of Biology.

The total of possibilities is: 10 x 6 x 1 = 60 possibilities to choose 3 books among 11 books.

### Remarks:

1. If we have to choose 2 whatever books from the 11 books, the number of possibilities is : C(2,11) = 11!/2! (11 - 2)! = 11!/2! 9! = 10 x 11 /2 = 55 possibilities only.

2. Choosing books of Mathematics do not affect the choice of books of Physics or Biology. We say hat the three choices of books are independent events. Generally, the number of possibilities for independent events is the product of the possibilities for each event.

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