Chapter 7.4: Counting Techniques and Probability
•
In many probability problems, the sample space
S
is too large to count
simply by listing all sample points.
•
If this is the case, we may use our different counting techniques (i.e.
permutations, combinations, etc.) to find
n
(
S
) and
n
(
E
).
Example:
In the lottery BC49, a player selects six different numbers
from 1
,
2
,
3
,...,
49. On the day of the lottery drawing, the lottery official
randomly selects the six winning numbers. If the player has exactly three
of these numbers, then the player wins a $10 prize. What is the probability
of a player winning the $10 prize with exactly one ticket?
Let
E
be the event that exactly three numbers match.
We need to count
n
(
S
) and
n
(
E
).
First, the sample space consists of all possible selections of six numbers.
The total number of ways in which 6 numbers can be selected from 49 is
parenleftbigg
49
6
parenrightbigg
. Thus,
n
(
S
) =
parenleftbigg
49
6
parenrightbigg
.
To count
n
(
E
), there are two tasks: selecting three of the six winning
numbers (which can be done
parenleftbigg
6
3
parenrightbigg
ways) and selecting three of the 43
remaining nonwinning numbers (which can be done
parenleftbigg
43
3
parenrightbigg
ways).
Thus,
n
(
E
) =
parenleftbigg
6
3
parenrightbiggparenleftbigg
43
3
parenrightbigg
. This gives us:
P
(
E
) =
n
(
E
)
n
(
S
)
=
parenleftbigg
6
3
parenrightbiggparenleftbigg
43
3
parenrightbigg
parenleftbigg
49
6
parenrightbigg
≈
0
.
0176
The probability of winning the $10 prize is approximately 0.0176.
Example:
If all six numbers match, then the player wins the grand
prize (usually at least $1 000 000). What is the probability that a player
wins the grand prize with one ticket?
Let
E
be the event that all six numbers match on the ticket.
1
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Since the experiment has not changed, the sample space is the same,
and
n
(
S
) =
parenleftbigg
49
6
parenrightbigg
.
To count
n
(
E
) there is only one task: out of six winning numbers, select
all six. This can be done
parenleftbigg
6
6
parenrightbigg
ways. Thus,
n
(
E
) =
parenleftbigg
6
6
parenrightbigg
. This gives
us:
P
(
E
) =
n
(
E
)
n
(
S
)
=
parenleftbigg
6
6
parenrightbigg
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 Spring '11
 stephenlang
 Calculus, Permutations, Counting, Probability, Probability theory, Pen

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