Probability Probability definition, terms and notation
Sample space, event Probability formula The probability of mutually exclusive events The probability of not mutually exclusive events Calculating probabilities, examples
Probability definition, terms and notation
The probability is defined as a measure of the degree of confidence in the occurrence of an event, measured on a scale from 0 (impossibility) to 1 (certainty), expressed as the ratio of favorable outcomes of the event to the total number of possible outcomes.
Thus, the probability of an event A occurring, where, 0 <  P (A)  < 1, so that, P (not A) = 1 - P (A) and P (A) = 1 - P (not A).
 The set of all possible outcomes called the sample space, S = A U not A.
The probability of mutually exclusive events
The probability of mutually exclusive events, either A or B (or both) occurring is
 P ( A U B ) = P (A) + P (B)
The probability of either events A or B (or both), occurring is written P ( A U B ).
The probability of not mutually exclusive events
The probability of not mutually exclusive events, either A or B (or both) occurring is
 P ( A U B ) = P (A) + P (B) - P ( A Ç B )
The probability of events A and B both occurring is written P ( A Ç B ).
Calculating probabilities, examples
Example:   What is the probability that in two throws of a dice the sum of the numbers that come up is 5 or product is 4?
Solution:  Two throws of a dice we can consider as one throw of two dice. So, the number of favorable outcomes, that the first event occurs (i.e., the sum of the numbers that come up is 5) is determined by pairs, the event E1:  (1, 4), (2, 3), (4, 1) and (3, 2).
The second event (product is 4) is determined by pairs, the event  E2:  (1, 4), (2, 2), (4, 1).
Notice that pairs (1, 4) and (4, 1) appear in both events, so we should include them in the event E1 or E2. Example:  What is the probability that in two throws of a dice the sum of the numbers that come up is 7 or product is 10?
Solution:  The number of favorable outcomes that: - the event E1 (the sum of the numbers is 7) occurs is determined by pairs: (1, 6), (2, 5), (3, 4), (6, 1), (5, 2), (4, 3),
- the event E2 (the product of the shown numbers is 10) occurs is determined by pairs: (2, 5) and (5, 2).
Since the pairs (2, 5) and (5, 2) are already contained in the event E1 then, the probability is We get the same result by counting the two pairs in the event E2 (but then, we don't count them in the E1), so Example:  In a box there are 9 balls numbered from 1 to 9. If we draw from the box two balls at once, what is the probability that the sum of both numbers is odd and less than 8?
Solution:  According to stated conditions about the number of favorable outcomes m will give us the following pairs of numbered balls: (1, 2), (1, 4), (1, 6), (2, 3), (2, 5), and (3, 4).
The total number of possible outcomes is equal to the number of the combinations of the subset with k = 2 elements out of the set of n = 9 elements, i.e., Example:  What is the probability that in four consecutive throws of a dice, come up four different numbers?
 Solution: The number of favorable outcomes m corresponds to the number of permuted combinations (or variations) of the subset with k = 4 elements (four throws of one dice or one throw of 4 dice) out of the set of 6 elements (6 faces of the cube numbered from 1 to 6), i.e., The total number of possible outcomes n correspond to the number of permuted combinations with repetition of the subset with k = 4 elements out of the set of 6 elements, since each throw has 6 possible
 outcomes,  Example:  What is the probability that from a group of 3 men and 4 women we chose a three-member group which consists of one man and two women?
 Solution:  From the group of three men a one man we can choose on ways,
 and from four women we can choose two women on ways,
so the number of favorable outcomes, From the group of seven people a three-member group can be chosen on n ways, i.e.,  Example:  By rolling two dice at once find probability that the sum of numbers that come up is 6 or the sum is 9.
Solution:  - The number of favorable outcomes that the event E1 occurs is determined by pairs:
(1, 5), (2, 4), (3, 3), (4, 2) and (5, 1), the pairs whose sum is 6.
- The number of favorable outcomes that the event E2 occurs is determined by pairs:
(3, 6), (4, 5), (5, 4) and (6, 3), the pairs whose sum is 9.    Combinatorics and probability contents 