Transcript Students Matter. Success Counts.

Section 4.5
Combining Probability and Counting
Techniques
With some added content
by D.R.S., University of Cordele
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Example 4.34: Calculating Probability Using
Combinations
A group of 12 tourists is visiting London. At one
particular museum, a discounted admission is given to
groups of at least ten.
a. How many combinations of tourists can be made for
the museum visit so that the group receives the
discounted rate?
b. Suppose that a group of the tourists does get the
discount. What’s the probability that it was made up
of 11 of the tourists?
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Example 4.34: Calculating Probability Using
Combinations (cont.)
a. “How many combinations …?”
a. The key words for this problem are at least. If at
least 10 are required, then the group will get the
discount rate if 10, 11, or 12 tourists go to the
museum. We must then calculate the number of
combinations for each of the three possibilities.
Note that we are counting combinations, not
permutations, because the order of tourists chosen
to go to the museum is irrelevant.
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Tourists group discount, part (a)
• Ways to take 12 tourists 10 at a time? 12 C 10
• Ways to take 12 tourists 11 at a time? 12 C 11
• Ways to take 12 tourists 12 at a time? 12 C 12
• But of course that’s “1”.
• Use TI-84 nCr to find these values, not the primitive
formula.
Example 4.34: Calculating Probability Using
Combinations (cont.)
As we noted, the question implies that the group will
get the discount rate if 10, 11, or 12 tourists attend.
Remember, the word “or” tells us to add the results
together. So, to find the total number of groups, we
add the numbers of combinations together.
66 + 12 + 1 = 79
There are 79 different groups that can be formed to
tour the museum at the discounted rate.
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Example 4.34: Calculating Probability Using
Combinations (cont.)
b. the probability that it was made up of 11 of the
tourists?
Think of the sentence as reading, “Given that a
group of tourists received the discount, what’s the
probability that it was a group of 11?” So, to find the
probability, we need the total number of ways a
group of 11 can be chosen from the group of 12
divided by the number of ways that a group could
receive the discount. Since both of these were
calculated in part a., we can simply substitute these
numbers in our fraction.
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Example 4.34: Calculating Probability Using
Combinations (cont.)
Therefore, we calculate the probability as follows.
12
P 11 tourists discount 
79
 0.1519
• Because there are 12 diff groups of eleven tourists.
• Because there are 79 diff groups qualify for the
discount.
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Example 4.35: Calculating Probability Using
Permutations
Jack is setting a password on his computer. He is told
that his password must contain at least three, but no
more than five, characters. He may use either letters or
numbers (0–9).
a. How many different possibilities are there for his
password if each character can only be used once?
b. Suppose that Jack’s computer randomly sets his
password using all of the restrictions given above.
What is the probability that this password would
contain only the letters in his name?
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Example 4.35: Calculating Probability Using
Permutations (cont.)
Solution – How many passwords can be formed?
a. First, notice the key words at least and no more
than. These words tell us that Jack’s password can
be 3, 4, or 5 characters in length. Also note that for
a password, the order of characters is important, so
we are counting permutations. There are 26 letters
and 10 digits to choose from, so we must calculate
the number of permutations possible from taking
groups of 3, 4, or 5 characters out of 36 possibilities.
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Password Permutations, continued
• It’s a lot like the tourists.
• How many ways to take 36 characters 3 at a time?
• How many ways to take 36 characters 4 at a time?
• How many ways to take 36 characters 5 at a time?
• How many ways in total to get 3, 4, or 5 characters?
Example 4.35: Calculating Probability Using
Permutations (cont.)
To find the total number of possible passwords, once
again we need to add all three of the above numbers of
permutations together since the solution contains an
“or” statement. Therefore, we obtain 46,695,600
passwords.
42,840  1,413,720  45, 239,040  46,695,600
(Imagine how many possibilities there would be if he
could use up to eight or nine characters, and mixed
upper/lowercase and punctuation characters too!)
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Example 4.35: Calculating Probability Using
Permutations (cont.)
Using only “J”, “A”, “C”, and “K”:
b. To find the probability that a randomly chosen
password would include only the four letters from
Jack’s name, we need the total number of possible
passwords calculated in part a. as well as the
number of possible passwords made from the four
letters in Jack’s name. (That is, the classic fraction,
n(outcomes for this event) / n(Sample space) ). The
numerator is the number of permutations of 4
things from a set of 4 (ways to arrange J, A, C, K)
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Example 4.35: Calculating Probability Using
Permutations (cont.)
4!
4 P4 
 4  4 !
4  3  2  1 24


 24
0!
1
So, the probability is calculated as follows.
P password containing only the letters J, A, C, K 
24

 0.0000005
46,695,600
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More with JACK
What if the question were “What is the probability that
the password will contain all four letters of his name?”
• What’s “new” here, besides the passwords we
counted in the previous problem?
• How to count the number of passwords that meet
the requirements?
• What is the probability now?
Example 4.36: Using Numbers of Combinations
with the Fundamental Counting Principle
Tina is packing her suitcase to go on a weekend trip.
She wants to pack 3 shirts, 2 pairs of pants, and 2 pairs
of shoes. She has 9 shirts, 5 pairs of pants, and 4 pairs
of shoes to choose from. How many ways can Tina pack
her suitcase? (We will assume that everything
matches.)
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Example 4.36: Using Numbers of Combinations
with the Fundamental Counting Principle (cont.)
Solution
Begin by thinking of this problem as a Fundamental
Counting Principle problem. Consider the act of packing the
suitcase to be a multistage experiment. There are three
stages—packing the shirts, packing the pants, and packing
the shoes. Thus, there are three suitcase “slots” for Tina to
fill—a slot for shirts, a slot for pants, and a slot for shoes. To
fill these slots, the combination formula must be used
because we are choosing some items out of her closet, and
the order of the choosing does not matter. Let’s first
calculate these individual numbers of combinations.
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Suitcase, continued
• Permutations or Combinations?
• ______ shirts, taken _____ at a time: __________
• ______ pants, taken _____ at a time: __________
• ______ pair of shoes, taken
_____ at a time: __________
Example 4.36: Using Numbers of Combinations
with the Fundamental Counting Principle (cont.)
The final step in this problem is different than the final
step in the last two examples, because of the word
“and.” The Fundamental Counting Principle tells us we
have to multiply, rather than add, the numbers of
combinations together to get the final result. There are
then
84  10  6  5040
different ways that Tina can pack
for the weekend.
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Example 4.37: Using Numbers of Combinations
with the Fundamental Counting Principle
An elementary school principal is putting together a
committee of 6 teachers to head up the spring festival.
There are 8 first-grade, 9 second-grade, and 7
third-grade teachers at the school.
a. In how many ways can the committee be formed?
b. In how many ways can the committee be formed if
there must be 2 teachers chosen from each grade?
c. Suppose the committee is chosen at random and
with no restrictions. What is the probability that 2
teachers from each grade are represented?
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Example 4.37: Using Numbers of Combinations
with the Fundamental Counting Principle (cont.)
Solution
In how many ways can the committee be formed?
a. There are no stipulations as to what grades the
teachers are from and the order in which they are
chosen does not matter, so we simply need to
calculate the number of combinations by choosing 6
committee members from the 8 + 9 + 7 = 24
teachers.
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Example 4.37: Using Numbers of Combinations
with the Fundamental Counting Principle (cont.)
24!
24 C 6 
6! 24  6 !
24!

6!18!
4

11
7
4
24  23  22  21  20  19  18  17
 6  5  4  3  2  1 18  17
2 1
2  1
 23  11  7  4  19
 134 ,596
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Example 4.37: Using Numbers of Combinations
with the Fundamental Counting Principle (cont.)
b. How many ways if we get two from each grade?
For this problem we need to combine the
Fundamental Counting Principle and the
combination formula. First, we see that there are
three slots to fill: one first-grade slot, one secondgrade slot, and one third-grade slot. Each slot is
made up of 2 teachers, and we must use the
combination formula to determine how many ways
these slots can be filled.
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Example 4.37: Using Numbers of Combinations
with the Fundamental Counting Principle (cont.)
Number of first-grade combinations:
8!
8 C2 
2! 8  2!
8!

2!6!
4

8 7 65
 2  1 6  5
2 1
2  1
 4  7  28
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Example 4.37: Using Numbers of Combinations
with the Fundamental Counting Principle (cont.)
Number of second-grade combinations:
9!
9 C2 
2! 9  2!
9!

2!7!
4

9 8  76 5
 2  1 7  6  5
2 1
2  1
 9  4  36
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Example 4.37: Using Numbers of Combinations
with the Fundamental Counting Principle (cont.)
Number of third-grade combinations:
7!
7 C2 
2! 7  2!
7!

2!5!
3

7  6  5 4  32 1
 2  1  5  4  3  2  1 
 7  3  21
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Example 4.37: Using Numbers of Combinations
with the Fundamental Counting Principle (cont.)
Finally, the Fundamental Counting Principle says that
we have to multiply together the outcomes for all three
slots in order to obtain the total number of ways to
form the committee. Thus, there are
28  36  21  21,168
ways to choose the committee.
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Example 4.37: Using Numbers of Combinations
with the Fundamental Counting Principle (cont.)
c. P(a committee does have two from each grade)
The final part of this question asks us to find the
probability that if a committee was chosen at
random, it would meet the requirements given in
part b. That is, each grade would have two teachers
on the committee. We can combine our answers
from parts a. and b. to find the answer to this
probability question.
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Example 4.37: Using Numbers of Combinations
with the Fundamental Counting Principle (cont.)
P  committee has 2 teachers from each grade
Number of possible committees with
2 teachers from each grade

Number of possible committees of 6 teachers
21,168

 0.1573
134,596
In other words, if the members of the committee are
randomly selected, there is a 15.73% chance that the
committee will be made up of 2 teachers from each
grade.
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Menu planning example
150 guests are coming.
Select 5 appetizers, 4 entrees, 3 side dishes.
Available 18 appetizers, 9 entrees, 12 side dishes.
How many different menu plans are there?
Menu planning example
• Fundamental counting principle:
• How many ways to choose appetizers,
• times how many ways to choose entrees,
• times how many ways to choose side dishes
• What about the 150 guests?
Reading list example
There are 10 novels, 8 non-fiction, and 6 self-help
books on the shelf.
Choose 5 books to take on vacation.
How many different book choices?
Reading List Example
• Key insight – in this case, we’re just selecting books
• It doesn’t matter how many books of which types
Reading list example, modified
There are 10 novels, 8 non-fiction, and 6 self-help
books on the shelf.
Choose 5 books to take on vacation, and at least one
book must be a self-help book.
How many different book choices?
Reading list example, modified
There are 10 novels, 8 non-fiction,18 books besides the
self-help books, and 6 self-help books on the shelf.
Choose 5 books to take on vacation, and at least one
book must be a self-help book.
• How many ways to choose 1 self-help + 4 others
• + How many ways to choose 2 self-help + 3 others
• + How many ways to choose 3 self-help + 2 others
• + How many ways to choose 4 self-help + 1 other
• + How many ways to choose 5 self-help + 0 others
Reading list example, modified
There are 10 novels, 8 non-fiction,18 books besides the
self-help books, and 6 self-help books on the shelf.
Choose 5 books to take on vacation, and at least one
book must be a self-help book.
And… what if the order of the five books matters?
Reading list example, modified
There are 10 novels, 8 non-fiction,18 books besides the
self-help books, and 6 self-help books on the shelf.
Choose 5 books to take on vacation, and at least one
book must be a self-help book.
And… what if the order of the five books matters?
> Multiply the number of ways to choose the books
by 5!, since each of those ways has 5! possible orders.
Mixture of events of different natures
How many outcomes for the activity of
• Tossing a coin 5 times
• Drawing 2 cards without replacement
Mixing events of different natures
Fundamental Counting Principle, so multiply
• The coin tosses are independent events
• The card drawing is a combination
Tossing a coin 10 times
… and the number of heads is between 5 and 8
Tossing a coin 10 times
… and the number of heads is between 5 and 8
• 10 coin tosses … the sample space has 210
outcomes.
• How many ways to get 5 heads? 10 C 5
• How many ways to get 6 heads? 10 C 6
• How many ways to get 7 heads? 10 C 7
• How many ways to get 8 heads? 10 C 8