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| Probability |
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Probability
definition, terms
and notation |
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Sample space, event |
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Probability formula |
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The probability of
mutually exclusive events |
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The probability of not
mutually exclusive events |
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Calculating
probabilities, examples |
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| 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, |
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| where, |
0
<
P (A) <
1, |
so
that, |
P
(not
A) = 1 -
P (A) |
and |
P
(A)
= 1 -
P (not A). |
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| The set of all possible outcomes
called the sample
space, |
S
= A
U not A. |
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| 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) |
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| The
probability of either events A
or B
(or both), occurring is written
P ( A
U B ). |
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| 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
)
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| The
probability of events A
and B
both occurring is written P
(
A Ç B
). |
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| 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?
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| 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. |
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| 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?
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| 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), |
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- 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 |
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| 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 |
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| 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?
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| 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., |
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| Example: What is the probability that in four consecutive throws of a
dice, come up four different numbers?
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| Solution: |
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| 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., |
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| 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, |
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| 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?
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| Solution:
From the group of three men a one man we can choose on |
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ways, |
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| and from four women we can
choose two women on |
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ways, |
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| so the number of favorable
outcomes, |
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| From the group of seven people a three-member group can be chosen on
n ways, i.e., |
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| Example: By rolling two dice at once find probability that the sum of numbers that come up is 6 or the sum
is 9.
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| 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. |
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- 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. |
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| Combinatorics
and probability contents |
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