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| Probability |
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The probability of two
independent events |
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The probability of two
dependent events |
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Calculating
probabilities, examples |
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| The probability of two
independent events |
| The
events are said to be
independent
if the occurrence of one event does not affect the probability of any of
the others.
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| The probability that two independent events
A
and B
both occur is given by |
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| The probability of any set of
independent events
occurring equals the product of their individual probabilities.
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| The probability of two
dependent events |
| The
events are said to be
dependent
if the occurrence of one event affects the outcome of the others.
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| The probability that two
dependent events A
and B both occur
is given by |
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P(
A Ç B
)
= P(A) · P(B|A) = P(B)
· P(A|B) |
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| Calculating
probabilities, examples |
| Example:
What is the probability that in two throws of a die the sum of the numbers that come up is 5 or
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| product is 4? |
| Solution:
Two throws of a die 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, |
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the event E1:
(1, 4), (2, 3), (4, 1) and (3, 2). |
| The second event (product is 4) is determined by pairs, |
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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 die the sum of the numbers that come up is 7 or
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| product is 10? |
| Solution:
The number of favorable outcomes that: |
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- the event E1
(the sum of the numbers is 7) occurs is determined by pairs: |
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(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
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| 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 |
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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 die, 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 die 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
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| 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
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| which consists
of one man and two women? |
| 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
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| 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. |
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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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