probability counting
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Probability countingTRANSCRIPT
Quiz 3 Answer Key Question Set: Craps Firstly, let’s consider the sample space. We have a pair of dice being tossed and the outcome is the sum
of the two die. We can easily see the whole sample space using a table
Die 1
1 2 3 4 5 6
Die
2
1 2 3 4 5 6 7
2 3 4 5 6 7 8
3 4 5 6 7 8 9
4 5 6 7 8 9 10
5 6 7 8 9 10 11
6 7 8 9 10 11 12
This is basically a 6x6 addition table and as such, has 36 total cells (outcomes). That is, in the two
questions in this question set, the total number of outcomes is 36.
1. Outside Numbers. Using the sample space for two die on page 216 of your text book, what is the
probability of throwing “Outside Numbers” – the numbers 4, 5, 9 and 10 in the game of craps?
We assume, of course, that we’re throwing fair dice.
To obtain the desired outcomes, we count the number of times 4, 5, 9 and 10 occur in the
sample space.
Die 1
1 2 3 4 5 6
Die
2
1 2 3 4 5 6 7
2 3 4 5 6 7 8
3 4 5 6 7 8 9
4 5 6 7 8 9 10
5 6 7 8 9 10 11
6 7 8 9 10 11 12
From the highlighted cells, we can see there are 14 such outcomes. Thus, if we let E be the event
“Throwing outside numbers”, then
P(E) = #desired outcomes / total # of outcomes
= 14/36
2. Field Bet. Using the sample space for two die on page 216 of your text book, what is the
probability of winning a “Field Bet” – A bet on 2, 3, 4, 9, 10, 11 or 12 in the game of craps? We
assume of course, that we’re throwing fair dice.
To obtain the desired outcomes, we count the number of times, 2, 3, 4, 9, 10, 11 or 12 occur in
the sample space.
Die 1
1 2 3 4 5 6
Die
2
1 2 3 4 5 6 7
2 3 4 5 6 7 8
3 4 5 6 7 8 9
4 5 6 7 8 9 10
5 6 7 8 9 10 11
6 7 8 9 10 11 12
From the highlighted cells, we can see there are 16 such outcomes. Thus, if we let E be the event
“winning a field bet”, then
P(E) = #desired outcomes / total # of outcomes
= 16/36
Question Set: Alarm Systems At ACME alarms, the company has developed three different types of wireless home alarm systems:
Model X100 (aka Model “X”), Model Y200 (aka Model “Y”) and Model Z300 (aka Model “Z”). One day on
the assembly line, the boxes used to package the system were not labeled, i.e., they were all packed into
plain, unmarked boxes of the same size and shape, and therefore, by just looking at the box, you couldn't
tell whether it contained Model X100, Model Y200 or the Model Z300. It's not until you actually open the
box that you can tell which model was packed inside of it. On this day, hundreds of such boxes were
packaged and loaded into a truck. We don't know the exact number; the only thing we know for sure is
that an equal number of Model X100's, Model Y200's and Model Z300's were packed into the truck. The
plant manager asks you to pick three boxes out of the truck at random and place them in bins labeled “A,”
“B” and “C.”
3. Sample Space. If you place the boxes in bins as described, how many possible outcomes could
there be when you open the boxes? For example, one outcome could be that in bin A we had the
model X100, in bin B we had the model Y200 and in bin C we had the model Z300. This outcome
could be written in shorthand (by using the short name for each model) as XYZ.
We have three bins. In each bin we have 3 options (Model X100, Model Y200 or Model Z300).
3 3 3
Bin A Bin B Bin C
Thus, we have 3 * 3 * 3 = 27 total possible outcomes of the three models in the 3 bins.
These 27 outcomes are presented in the following table.
Bin A Bin B Bin C
X X X
X X Y
X X Z
X Y X
X Y Y
X Y Z
X Z X
X Z Y
X Z Z
Y X X
Y X Y
Y X Z
Y Y X
Y Y Y
Y Y Z
Y Z X
Y Z Y
Y Z Z
Z X X
Z X Y
Z X Z
Z Y X
Z Y Y
Z Y Z
Z Z X
Z Z Y
Z Z Z
4. Same Model. What is the probability that all three bins will contain the same model?
First, let’s enumerate the desired outcomes.
Bin A Bin B Bin C
For the first bin, we have 3 options.
3
Bin A Bin B Bin C
Once we’ve made a choice, the other two bins have only one option: the one we picked for the
first bin.
3 1 1
Bin A Bin B Bin C
Thus, our # of desired outcomes is 3 x 1 x 1 = 3. From the first question, our total # of possible
outcomes is 27. Thus, if we let E be the event “All three bins contain the same model” then
P(E) = # desired outcomes / total # of outcomes
= 3/27
REMEMBER: DO NOT REDUCE your fractions!
See the table below that identifies the desired outcomes with a highlighter.
Bin A Bin B Bin C
X X X
X X Y
X X Z
X Y X
X Y Y
X Y Z
X Z X
X Z Y
X Z Z
Y X X
Y X Y
Y X Z
Y Y X
Y Y Y
Y Y Z
Y Z X
Y Z Y
Y Z Z
Z X X
Z X Y
Z X Z
Z Y X
Z Y Y
Z Y Z
Z Z X
Z Z Y
Z Z Z
5. Model Z300 in bin B. What is the probability that bin “B” will contain the Model Z300?
First, let’s enumerate the desired outcomes.
Bin A Bin B Bin C
For the bin B, we have 1 option (the Model Z300):
1
Bin A Bin B Bin C
For the other two bins, we have 3 options:
3 1 3
Bin A Bin B Bin C
Thus, the # of desired outcomes is 3 x 1 x 3 = 9. From the first question, our total # of possible
outcomes is 27. Thus, if we let E be the event “Bin B contains the Model Z300” then
P(E) = # desired outcomes / total # of outcomes
= 9/27
REMEMBER: DO NOT REDUCE your fractions!
See the table below that identifies the desired outcomes with a highlighter.
Bin A Bin B Bin C
X X X
X X Y
X X Z
X Y X
X Y Y
X Y Z
X Z X
X Z Y
X Z Z
Y X X
Y X Y
Y X Z
Y Y X
Y Y Y
Y Y Z
Y Z X
Y Z Y
Y Z Z
Z X X
Z X Y
Z X Z
Z Y X
Z Y Y
Z Y Z
Z Z X
Z Z Y
Z Z Z
6. Bin A and Bin C contain the same model. What is the probability that bin A and bin C contain
the same model?
First, let’s enumerate the desired outcomes.
Bin A Bin B Bin C
For the bin A, we have 3 options:
3
Bin A Bin B Bin C
Once we've made a choice for Bin A, we have but one choice for Bin C (the same one that was
selected for Bin A):
3 1
Bin A Bin B Bin C
Finally, for Bin B we have 3 options:
3 3 1
Bin A Bin B Bin C
Thus, the # of desired outcomes is 3 x 3 x 1 = 9. From the first question, our total # of possible
outcomes is 27. Thus, if we let E be the event “Bin A and Bin C contain the same model” then
P(E) = # desired outcomes / total # of outcomes
= 9/27
REMEMBER: DO NOT REDUCE your fractions!
See the table below that identifies the desired outcomes with a highlighter.
Bin A Bin B Bin C
X X X
X X Y
X X Z
X Y X
X Y Y
X Y Z
X Z X
X Z Y
X Z Z
Y X X
Y X Y
Y X Z
Y Y X
Y Y Y
Y Y Z
Y Z X
Y Z Y
Y Z Z
Z X X
Z X Y
Z X Z
Z Y X
Z Y Y
Z Y Z
Z Z X
Z Z Y
Z Z Z
7. All three are different. What is the probability that all three bins will contain different models?
First, let’s enumerate the desired outcomes.
Bin A Bin B Bin C
For the bin A we have 3 options:
3
Bin A Bin B Bin C
Once we made a choice, we have only two left for bin B:
3 2
Bin A Bin B Bin C
And now we have only one option for bin C:
3 2 1
Bin A Bin B Bin C
Thus, the # of desired outcomes is 3 x 2 x 1 = 6. From the first question, our total # of possible
outcomes is 27. Thus, if we let E be the event “All three bins have different models” then
P(E) = # desired outcomes / total # of outcomes
= 6/27
REMEMBER: DO NOT REDUCE your fractions!
See the table below that identifies the desired outcomes with a highlighter.
Bin A Bin B Bin C
X X X
X X Y
X X Z
X Y X
X Y Y
X Y Z
X Z X
X Z Y
X Z Z
Y X X
Y X Y
Y X Z
Y Y X
Y Y Y
Y Y Z
Y Z X
Y Z Y
Y Z Z
Z X X
Z X Y
Z X Z
Z Y X
Z Y Y
Z Y Z
Z Z X
Z Z Y
Z Z Z
8. Two bins are the same, the other is different. What is the probability that two bins will contain
the same model and the other is a different model?
First, let’s enumerate the desired outcomes.
Bin A Bin B Bin C
Case 1: Bin A and Bin B the same, Bin C different
For the bin A we have 3 options:
3
Bin A Bin B Bin C
One possibility is that Bin B has the same model, meaning it has only one choice:
3 1
Bin A Bin B Bin C
In this case, both Bin A and Bin B are the same model. For Bin C to be different than this one,
you have 2 options:
3 1 2
Bin A Bin B Bin C
Case 2: Bin A and Bin C the same, Bin B different.
Another possibility after having chosen the first one in Bin A, we have bin C be the same:
3 1
Bin A Bin B Bin C
And as before, this leaves 2 options for Bin B.
3 2 1
Bin A Bin B Bin C
Case 3: Bin B and Bin C the same, Bin A different.
2 3 1
Bin A Bin B Bin C
The # desired outcomes is the sum of these three cases: (3 x 1 x 2) + (3 x 2 x 1) + (2 x 3 x 1) = 6 +
6 + 6 = 18. From the first question, our total # of possible outcomes is 27. Thus, if we let E be the
event “Two bins are the same and the third is different” then
P(E) = # desired outcomes / total # of outcomes
= 18/27
REMEMBER: DO NOT REDUCE your fractions!
This question was, by far, the hardest one on this quiz.
See the table below that identifies the desired outcomes with a highlighter.
Bin A Bin B Bin C
X X X
X X Y
X X Z
X Y X
X Y Y
X Y Z
X Z X
X Z Y
X Z Z
Y X X
Y X Y
Y X Z
Y Y X
Y Y Y
Y Y Z
Y Z X
Y Z Y
Y Z Z
Z X X
Z X Y
Z X Z
Z Y X
Z Y Y
Z Y Z
Z Z X
Z Z Y
Z Z Z
9. Bin B contains the Model Y200. What is the probability that bin B will contain the Model Y200?
First, let’s enumerate the desired outcomes.
Bin A Bin B Bin C
For the bin B we have 1 option (Model Y200):
1
Bin A Bin B Bin C
For the other two bins, we have 3 options:
3 1 3
Bin A Bin B Bin C
Thus, the # of desired outcomes is 3 x 1 x 3 = 9. From the first question, our total # of possible
outcomes is 27. Thus, if we let E be the event “Bin B contains the Model Y200l” then
P(E) = # desired outcomes / total # of outcomes
= 9/27
REMEMBER: DO NOT REDUCE your fractions!
See the table below that identifies the desired outcomes with a highlighter.
Bin A Bin B Bin C
X X X
X X Y
X X Z
X Y X
X Y Y
X Y Z
X Z X
X Z Y
X Z Z
Y X X
Y X Y
Y X Z
Y Y X
Y Y Y
Y Y Z
Y Z X
Y Z Y
Y Z Z
Z X X
Z X Y
Z X Z
Z Y X
Z Y Y
Z Y Z
Z Z X
Z Z Y
Z Z Z