### The binomial model

```Chapter 17:
The binomial model of probability
Part 2
AP Statistics
1
The binomial model:
jumping in with both feet
• There’s a close connection between tree
diagrams, combinations, expanding binomial
expressions, and the binomial model
• I’ve debated about giving you a lot of background
on it to show you the connections FIRST
• Instead, I think it will be more effective to show
you how to solve the problems, and then show
you the connections, which then may be more
meaningful.
2
The binomial model:
formulae
• So here they are, the three formulae you need
to recognize and use.
• Let’s do the easiest two first, mean and
standard deviation
– Check out the Math Box on p. 392 if you want to
see the derivations
• Mean:
• Standard deviation:
3
The binomial model:
probability of exactly k successes in n trials
• This is the key equation to recognize
(memorizing it is hard, at least at first)
• Remember that p=probability of success,
q=probability of failure, and the binomial
coefficient is the number of combinations that
satisfy k successes in n trials
• Here it is……wait for it…….
4
The binomial model:
how to dissect the binomial formula
• It’s the number of combinations you can get from
taking k samples from a population of n
• That, in turn, is equal to
• These are the calculations we’ll be working on
today
5
The binomial model/Exercise 13(d):
Starting with Exercise 13(d)
• We were not able to do Exercises 13(d)-(f) or
14(d)-(f) because we needed to apply the
binomial model.
what percentage of the populations has exactly 3
lefties.
• Before going on to the next slide, work for 2-3
minutes and see how many different
combinations you can make using three lefties
and two righties.
6
The binomial model/Exercise 13(d):
combinations of 3 lefties, 2 righties
• Here’s the distribution on
the right.
• It took me about 5
minutes to derive and
check it; YMMV.
• But even if you do it
faster than I did, what
about 3 out 10? Or 3 out
of 15? Or even 3 out of 8?
• It will take way too long.
• We need a better (faster)
method.
1. L
2. L
3. L
4. L
5. L
6. L
7. R
8. R
9. R
10. R
L
L
L
R
R
R
L
L
L
R
L
R
R
L
L
R
L
R
L
L
R
L
R
L
R
L
L
L
R
L
R
R
L
R
L
L
R
L
L
L
7
The binomial model/Exercise 13(d):
Using the binomial coefficient
• The binomial coefficient will save us from the
tedium of finding such distributions
• For n=5 and k = 3 (the conditions of our
problem), we can calculate the number of
combinations directly:
8
The binomial model/Exercise 13(d):
Applying the binomial coefficient to the probabilities
• We have the binomial coefficient, which is 10
for this problem.
• Now, how to apply the q and p probabilities?
• The best thing to do is to create a table. We’re
going to spend some time doing this, and it
should help to make everything clear.
9
The binomial model/Exercise 13(d):
Making a probability table
• Because the calculations are complicated,
we’re going to make a couple of tables to help
make things clear.
• Here’s the form of the tables to copy for this
part of the exercise. Make two of them:
5-k
k (5 k)
0
1
2
3
4
5
Totals:
q5-n
pn
(5 k)q5-npn Lefties
0
1
2
3
4
5
10
The binomial model/Exercise 13(d):
Filling in the 1st probability table with formulae
• Note that the table k=0 through k=5. We’ll
discuss why in a bit as we fill it in.
• Now, let’s fill in the formulas before you do
the calculations. You’ll do the calculations in
the 2nd table. Fill in the blanks as indicated:
5-k
5-0=5
5-1=4
k (5 k)
q5-n
0 5!/5!0! 0.87^5
1 5!/1!4! 0.87^4
2 5!/2!3!
3
4
5
Totals:
pn
0.13^0
0.13^1
(5 k)q5-npn Lefties
0
1
2
3
4
5
11
The binomial model/Exercise 13(d):
What the 1st table should look like when you’re done
5-k
5-0=5
5-1=4
5-2=3
5-3=2
5-4=1
5-5=0
k
0
1
2
3
4
5
Totals:
(5 k)
5!/5!0!
5!/1!4!
5!/2!3!
5!/3!2!
5!/4!1!
5!/5!0!
5-n
q
0.87^5
0.87^4
0.87^3
0.87^2
0.87^1
0.87^0
n
5-n n
p
(5 k)q p Lefties
0.13^0
0
0.13^1
1
0.13^2
2
0.13^3
3
0.13^4
4
0.13^5
5
12
The binomial model/Exercise 13(d):
Starting the 2nd table
• Your second table will contain the calculations
using the formulas in the 1st table. As you
start, it should look like this:
5-k
k
(5 k)
0
1
2
3
4
5
Totals:
q5-k
pk
(5 k)q5-kpk Lefties
0
1
2
3
4
5
13
The binomial model/Exercise 13(d):
Completing the 2nd table
• Using the formulas in Table 1, do the calculations.
Be sure to fill in the calculations in the second
column from the right that I’ve grayed out.
• By the end, you should have a table of all the
values possible where there are 0 lefties, 1 lefty,
…., 4 lefties, and 5 lefties.
• It will probably take about 10 minutes to
complete this. Be sure to complete the totals for
the columns headed by (5 n) and by (5 k)q5-npn .
14
The binomial model/Exercise 13(d):
What the second table should like and its meaning
5-k
5
4
3
2
1
0
k (5 k)
0
1
1
5
2
10
3
10
4
5
5
1
Totals:
32
q5-k
0.4984209
0.5728976
0.6585030
0.7569000
0.8700000
1.0000000
pk (5 k)q5-kpk Lefties
1.0000000 0.4984209 0
0.1300000 0.3723834 1
0.0169000 0.1112870 2
0.0021970 0.0166291 3
0.0002856 0.0012424 4
3.71E-05 3.713E-05 5
1
15
The binomial model/Exercise 13(d):
Important points
1. (5 k)q5-kpk is the probability that you will get k
events in 5 trials.
1, which tells us that this is the TOTAL
probability model.
probability that we will get zero lefties.
4. If you have k trials, you will have k+1 entries
16
The binomial model/Exercise 13 (e):
Using the table to determine summed probabilities
• 13(e): at least 3 lefties in the group.
• That means there could be 3….or 4…..or 5.
• Find the total by adding the areas marked:
5-k
5
4
3
2
1
0
k (5 k)
0
1
1
5
2
10
3
10
4
5
5
1
Totals:
32
q5-k
0.4984209
0.5728976
0.6585030
0.7569000
0.8700000
1.0000000
pk
(5 k)q5-kpk Lefties
1.0000000 0.4984209 0
0.1300000 0.3723834 1
0.0169000 0.1112870 2
0.0021970 0.0166291 3
0.0002856 0.0012424 4
3.71E-05 3.713E-05 5
1
17
The binomial model/Exercise 13 (f):
Using the table to determine summed probabilities (2)
• 13(d): no more than 3 lefties in the group.
• That means there could be 3, 2, 1, or 0
• Find the total by adding the areas marked:
5-k
5
4
3
2
1
0
k (5 k)
0
1
1
5
2
10
3
10
4
5
5
1
Totals:
32
q5-k
0.4984209
0.5728976
0.6585030
0.7569000
0.8700000
1.0000000
pk
(5 k)q5-kpk Lefties
1.0000000 0.4984209 0
0.1300000 0.3723834 1
0.0169000 0.1112870 2
0.0021970 0.0166291 3
0.0002856 0.0012424 4
3.71E-05 3.713E-05 5
1
18
The binomial model:
Answers to Exercises 13(e) and 13(f)
• Exercise 13(e):0.982
– Does this answer make sense?
• Exercise 13(f): 0.0179
– Does THIS answer make sense?
5-k
5
4
3
2
1
0
k (5 k)
0
1
1
5
2
10
3
10
4
5
5
1
Totals:
32
q5-k
0.4984209
0.5728976
0.6585030
0.7569000
0.8700000
1.0000000
pk
(5 k)q5-kpk Lefties
1.0000000 0.4984209 0
0.1300000 0.3723834 1
0.0169000 0.1112870 2
0.0021970 0.0166291 3
0.0002856 0.0012424 4
3.71E-05 3.713E-05 5
1
19
The binomial model/Exercise 14:
Preparing the chart
• Same as 13, except we now have 6 arrows, not
5 people
• Do the tables again, starting with a blank table
like this:
Go to the next slide when done.
6-k
1
2
3
4
5
6
7
TOTAL:
k
0
1
2
3
4
5
6
(6 k)
q6-k
pk
(6 n)q6-kpk bull's-eyes
0
1
2
3
4
5
6
20
The binomial model/Exercise 14:
The complete chart
• Here’s what you should have come up with.
• The answers are similar to what you did for
on next slide.
6-k
1
2
3
4
5
6
7
k
0
1
2
3
4
5
6
(6 k)
1
6
15
20
15
6
1
64
q6-k
pk
(6 n)q6-kpk bull's-eyes
0.000064
1 0.000064
0
0.00032
0.8 0.001536
1
0.0016
0.64 0.015360
2
0.008
0.512 0.081920
3
0.04
0.4096 0.245760
4
0.2 0.32768 0.393216
5
1 0.262144 0.262144
6
1
21
The binomial model/Exercise 14:
e) 0.0179 (adding numbers circled in red
f) 0.345 (adding numbers circled in green)
6-k
6
5
4
3
2
1
0
k
0
1
2
3
4
5
6
(6 k)
1
6
15
20
15
6
1
64
q6-k
pk
(6 n)q6-kpk bull's-eyes
0.000064
1 0.000064
0
0.00032
0.8 0.001536
1
0.0016
0.64 0.015360
2
0.008
0.512 0.081920
3
0.04
0.4096 0.245760
4
0.2 0.32768 0.393216
5
1 0.262144 0.262144
6
1
22
The binomial model:
Review of how to calculate combinations
• Remember the formula:
• k is the number of events you’re looking for in the
population n
• Be sure you put the k with the p, which, as you’ve
seen, may change from problem to problem.
• Don’t forget to calculate all three parts of the
equation and do the multiplication
23
The binomial model:
Practice: Exercise 19
• Similar to what we’ve done already
• Tennis player has successful first serve 70% of the
time.
• Assuming independence and 6 serves (n=6),
calculate the following probabilities:
a.
b.
c.
d.
All 6 serves go in
Exactly 4 serves go in
At least 4 serves go in
No more than 4 serves go in
• Spend 5-10 minutes calculating (a) through (c),
then advance to the next slide for the analysis.
24
The binomial model:
Practice: Exercise 19(a)
• All six serves go in.
• n=6, k=number of good serves=6, p=0.7, so q=0.3
• Calculate combinations first:
• Next, pkqn-k=(0.7)6(0.3)6-6=(0.7)6(0.3)0=(0.7)6=0.118
25
The binomial model:
Exercise 19(b)
• Exactly 4 serves go in.
• Similar to (a), but with different numbers: n
still 6, but k=4
• Binomial coefficient:
• Next, pkqn-k=(0.7)4(0.3)6-4 = (0.7)4(0.3)2
=(0.24)(0.09)=0.0216
• Multiply 15 times 0.0216 to get 0.324
26
The binomial model:
Exercise 19(c)
• At least 4 serves in means she got 4, 5 or 6
serves in. If we know probabilities, we can add
them.
• Already know 6 from (a) and 4 from (b)
• Calculate probabilities of 5:
• Calculate pkqn-k (continued on next slide)
27
The binomial model:
Putting all of 19(c) together
• Remember that the total probability of at
least 4 serves going in is P(4)+P(5)+P(6)
• From previous slides, we have
– P(4)= 0.324
– P(5)=0.301=6×0.0504 (from slide 27)
– P(6)=0.118
– TOT: 0.743
• Whenever you have multiple outcomes,