probability
probability
Probability Using Counting Techniques
MDM4U: Mathematics of Data Management
Some probabilities can be calculated using the three main
counting techniques covered earlier.
These include:
• Fundamental Counting Principle
Probabilities and Counting
• Permutations
Determining Probabilities Using Counting Techniques
• Combinations
J. Garvin
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probability
probability
Fundamental Counting Principle
Fundamental Counting Principle
Many probability calculations rely on computing the number
of choices for one item, followed by the number of choices for
a second, and so on.
Example
What is the probability that a randomly-selected Canadian
postal code contains no two identical letters or digits?
The FCP is a useful tool for these situations.
Solution: Let C be the event a postal code with unique
characters is selected.
n(S) = 26 × 10 × 26 × 10 × 26 × 10 = 17 576 000.
n(C ) = 26 × 10 × 25 × 9 × 24 × 8 = 11 232 000.
P(C ) =
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n(C )
n(S)
=
11 232 000
17 576 000
=
108
169 .
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probability
probability
Fundamental Counting Principle
Permutations
Your Turn
Some situations require specific orderings or rankings of
items.
What is the probability that a random 4-digit odd integer is a
palindrome?
Since order is important, permutations are used.
Solution: Let F be the event a 4-digit odd palindrome is
selected.
n(S) = 9 × 10 × 10 × 5 = 4500.
n(F ) = 1 × 1 × 10 × 5 = 50.
P(F ) =
n(F )
n(S)
=
50
4 500
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=
1
90 .
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probability
probability
Permutations
Permutations
Example
Your Turn
Nine students have a race. What is the probability of
successfully predicting the three fastest runners?
Four of seven students randomly line up against a wall. What
is the probability that Diana is included in the line-up?
Solution: Let R be the event the three fastest runners are
predicted.
Solution: Let D be the event Diana is included in the line-up.
n(S) = 9 P3 =
9!
(9−3)!
= 504.
n(R)
n(S)
=
7!
(7−4)!
n(D) = 4 × 6 P3 =
n(R) = 1 (the order is correct)
P(R) =
n(S) = 7 P4 =
P(D) =
1
504 .
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n(D)
n(S)
=
= 840.
6!
(6−3)!
480
840
= 480.
= 47 .
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probability
probability
Combinations
Combinations
Some probabilities are not influenced by ordering.
Example
For example, a question may ask about the probability of
having a three-card hand in cards, or forming a committee.
What is the probability that a committee of four, selected
from six doctors and five nurses, contains exactly two from
each group?
In these cases, combinations are applicable.
Solution: Let C the the event a valid committee is chosen.
n(S) = 11 C4 = 330.
n(C ) = 6 C2 × 5 C2 = 15 × 10 = 150.
P(C ) =
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n(C )
n(S)
=
150
330
Combinations
probability
Questions?
Your Turn
The same six doctors and five nurses will form a second
three-member committee, which must contain at least one
doctor. What is the probability that Nurse Jane is on this
committee?
Solution: For the sample space, eliminate all possible
committees composed entirely of nurses. So
n(S) = 11 C3 − 5 C3 = 155.
Let J be the event a committee with Jane is chosen. Then
1 C1 × 6 C1 × 9 C1
n(J) =
= 27.
2
The reason why we divide by 2 is because we overcount when
choosing 1 of 6 doctors, then 1 of the 9 remaining people.
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5
11 .
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probability
Therefore, P(C ) =
=
n(C )
n(S)
=
27
155 .
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