
Probability 


The probability of two
independent events 
The probability of two
dependent events 
Calculating
probabilities, examples 





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.

The probability that two independent events
A
and B
both occur is given by 

P
(
A Ç B
)
= P (A) · P (B)



The probability of any set of
independent events
occurring equals the product of their individual probabilities.


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.

The probability that two
dependent events A
and B both occur
is given by 

P
(
A Ç B
)
= P (A) · P (BA) = P (B)
· P (AB) 



Calculating
probabilities, examples 
Example:
What is the probability when rolling a dice to get the result divisible by 2 or 3?

Solution:
The probability that an event
E
will occur, 

If rolling a
dice we get numbers, 2, 4 or 6, fulfilled is condition for the event
A,
and if we get numbers, 3 or 6, fulfilled is condition for the event B.

Notice that number 6 satisfy both events, so this outcome we
should include in the event A
or B,
thus "or” probability
of mutually exclusive events 


Example:
By rolling three dice at once find probability that the product of all three numbers that come up
is divisible by 50.

Solution:
The required condition (the product of all three numbers should be divisible by 50) will satisfy
triples,
events when three dice show up: 6, 5, 5 and 5, 5, 2. 
To calculate the number of favorable outcomes, each of the triples brings, use the formula for permutations of 
n
elements (objects), some groups, r,
s,
t, . . . of which, are the
same, 


So, each triple
( 6, 5, 5 and 5, 5,
2) gives 

permutations,
where 3 stands for the three shown figures, two 

of which are the
same. Therefore, 

where
the total number of outcomes 6^{3
}equals the
number of variations
with repetition the three dies, with six sides each, show up. 

Example:
What is the probability that in two throws of a die the sum of the numbers that come up is 5 or 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,
the event E_{1}:
(1, 4), (2, 3), (4, 1) and (3, 2). 
The second event (product is 4) is determined by pairs,
the event E_{2}:
(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
E_{1}
or E_{2}. 


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 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., 

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 threemember 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 threemember 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
E_{1}
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 E_{2
} occurs is determined by pairs: 
(3, 6), (4, 5), (5, 4) and (6, 3),
the pairs whose sum is 9. 









Combinatorics
and probability contents 



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