Chapter 17
Probability Models
The Binomial Distribution
1
Practice Problems

Determine the type of distribution and solve:
You roll a fair six-sided die:
• What is the probability that you will get the first 6 on the 9th
roll?

•
What is the probability that you will get the first 6 within 9
rolls?

•
This is a geometric distribution (since we wish to determine the
first success on the 9th roll). It satisfies all of the conditions, so we
will use geometpdf(1/6, 9) ≈ .0388
This is still a geometric distribution (since we wish to determine
the first success in any of the first 9 roll). It satisfies all of the
conditions, so we will use geometcdf(1/6, 9) ≈ .8062
The die is rolled 9 times. What is the probability that you will
get 6 exactly 4 times?

This is a binomial distribution (since it has a fixed number of
trials). It satisfies all of the conditions, so we will use
binompdf(9, 1/6, 4) ≈ .0391
2
Practice Problems

Determine the type of distribution and solve:
•
•
You pick a card from a standard 52-card deck and replace the
card. You do this 5 times. What is the probability that you will
get exactly 3 face cards? (Aces are not face cards)
 This is a binomial distribution (since it has a fixed number of
trials). It satisfies all of the conditions, so we use
binompdf(5, 12/52, 3) ≈ .073
 If we pick 5 cards from a standard 52-card deck, we expect
to see exactly 3 face cards about 7.3% of the time.
You pick a card from a standard 52-card deck and replace the
card. How many cards will you pick before you get a face card?
 This is a geometric distribution (since we are interested in
getting our first face card). It satisfies all of the conditions,
so we use 1/p = 1/(12/52) ≈ 4.333
 We expect to see the first face card in about 4.333 picks.
3
Practice Problems

Determine the type of distribution and solve:
•
You pick 5 card from a standard 52-card deck
without replacement. What is the probability that
you will get exactly 3 face cards?


This problem is not a Bernoulli Trial since it violates two
of our rules: the probabilities are not the same for each
card and each card is not independent. So, we need to
solve this using another method.
We want to determine the probability of getting 3 face
cards and two non-face cards and multiply it by the
number of ways of getting 3 face cards out of five:
 5  12  11  10  40  39 
        (10)(.0066)  .066
 3  52  51 50  49  48 
4
Practice Problems

Determine the type of distribution and solve:
•
•
•
A new sales gimmick has 30% of the M&M’s covered
with speckles. These “groovy” candies are mixed
randomly with the normal M&M’s as they are put
into the bags for distribution and sale. You buy a bag
and remove candies one at a time looking for the
speckles. What’s the probability that we’ll find two
speckled ones if you grab five candies?
Since we are looking at finding 2 speckled candies
out of five, this is a binomial distribution.
What’s the first thing you should do before you try to
answer this question?
5
Practice Problems
Check the conditions:
•
Two-Outcome Assumption

•
Consistent Probability Assumption

•
There are two outcomes:
 success = speckles
and failure = ordinary
The probability of success, based on the information from
the candy company, is 30%.
Independence Assumption

We want to assume that the trials are independent – there is
no reason to believe that finding a speckled candy reveals
anything about whether the next one out of the bag will
have speckles. However, we are actually drawing M&M’s
without replacement from a finite population, so the Independence
Condition is actually violated; the probability that the next one is
speckled depends on how many speckled ones we have removed
from the pool so far. So what condition will allow me to continue?
6
Practice Problems

Recall that the 10% Condition states that if
independence is violated, it is acceptable to proceed as
long as the sample is smaller than 10% of the
population.
•

The Independence Assumption is violated whenever we
sample without replacement, but is overridden by the 10%
Condition (as long as we don’t drain off more than 10% of the
population, the probabilities don’t change enough to matter).
Back to the M&M problem:
•
•
Obviously our sample (the small bag we bought) is such a tiny
subset of the vast number of M&Ms in the world that it
doesn’t really affect the probabilities much, and that’s the
essence of the 10% Condition.
So, we can apply the binomial distribution: P(X = 2) ≈ 0.3087
In the TI-84: binompdf(5, .3, 2)
7
Practice Problems



The owner of a small convenience store is trying to
decide whether to discontinue selling magazines. He
suspects that only 5% of the customers buy a magazine
and thinks that he might be able to use the display space
to sell something more profitable. Before making a final
decision he decides that for one day he’ll keep track of
the number of customers and whether or not they buy a
magazine.
What is the probability that exactly 2 of the first 10
customers buy magazines? Assume all the assumptions
are satisfied. What distribution should you use?
Since we know that we are looking at 10 customers, this
is a binomial distribution. P(X = 2) ≈ 0.0746
•
In the TI-84: binompdf(10, .05, 2)
8
Practice Problems

Using the same information as before, what is
the probability that at least 5 of his first 50
customers buy magazines?
•

P(X ≥ 5) = 1 – binomcdf(50, .05, 4) ≈ 0.1036
Using the same information as before, what is
the probability that 4 to 7 of his first 50
customers buy magazines?
•
•
Since we want 4 to 7, we want X = 4, 5, 6, or 7
SoP(4 ≤ X ≤ 7) ≈ 0.2364
=binomcdf(50, .05, 7) – binomcdf(50, .05, 3) ≈ 0.2364
9
Practice Problems

The store had 2752 customers that day. Assuming this
day was typical for his store, what would be the mean
and standard deviation of the number of customers
who buy magazines each day?
  np
  2752(.05)  137.6
  np(1  p)
  2752(.05)(.95)  11.4333

For simplicities sake, we often use q to represent (1 – p)
so   npq
10
Practice Problems

The store had 2752 customers that day. Using the
normal model, what is the approximate probability that
less than 125 people will buy magazines? What
condition do you need to check to solve this problem?
• The condition that allows us to use the normal
model is…
• Success/Failure Condition: np ≥ 10 and nq ≥ 10
 2752(.05) = 137.6 > 10 and 2752(.95)=2614.4 > 10,
since we expect to see more than 10 successes and
failures, it’s alright to use the normal model to
approximate the probabilities.
11
Practice Problems



The store had 2752 customers that day. Using the
normal model, what is the approximate probability that
less than 125 people will buy magazines? We know:
  137.6
  11.4333
If we use the z-score, we get:
125  137 .6
z
 1.102
11 .4333
Use the normal model to calculate the probability.
P(X < 50) = normalcdf (-E99, -1.102) ≈ 0.1352.
So, there is a 13.52% chance that less than 125 people
will buy a magazine on any given day.
•
Or we can simply use normalcdf(-E99, 125, 137.6, 11.4333)
12
What Can Go Wrong?

Be sure you have Bernoulli trials.
•
•


You need two outcomes per trial, a constant
probability of success, and independence.
Remember that the 10% Condition provides a
reasonable substitute for independence.
Don’t confuse Geometric and Binomial models.
Don’t use the Normal approximation with
small n.
•
You need at least 10 successes and 10 failures to use
the Normal approximation.
13
What have we learned?
•
Geometric model
 When
we’re interested in the number of Bernoulli
trials until the next success.
•
Binomial model
 When
we’re interested in the number of successes
in a certain number of Bernoulli trials.
•
Normal model
 To
approximate a Binomial model when we expect
at least 10 successes and 10 failures. Note: This
gives only a rough approximation.
14
What have we learned?
What are we
looking for?
E( X )   X
X
Geometric
The first or next
success
1
p
q
p
Binomial
Number of
success
np
npq
geometpdf(p, x) = probability the first success is on the xth trial
geometcdf(p, x) = probability the first success is within the 1st to the xth trial
binompdf(n, p, x) = probability of x successes in n trials.
binomcdf(n, p, x) = the sum of the probabilities of the first x success in n trials.
15
Assignment
Unit 4
Lesson:
Unit 4 Review
Read:
None
Problems:
p. 405 – 409: 1 – 37 (odd)
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