Chapter 5
Probability Theory
5.1 General Probability Rules
We have already met and used five rules in §4.2
• Rule 1 : The probability P(A) of any event A satisfies
0 < P(A) < 1
• Rule 2 : If S is the sample space in a probability model,
then P(S) = 1
• Rule 3 : The complement of any event A is the event that
A will not occur, written as Ac . The complement
rule states that P(AC) = 1 - P(A)
• Rule 4 : (Addition rule for disjoint events)
If A and B are disjoint, P(A or B) = P(A) + P(B)
• Rule 5 : (Multiplication rule)
If A and B are independent events, P(A and B) = P(A) P(B)
General Addition Rule
For any two events A and B,
P(A or B) = P(A) + P(B)-P(A & B)
Example
Two dice are thrown, what is the probability of
1) getting a total of 12 or an absolute difference of 4?
B: A Total of 12, C An absolute difference of 4
P(B or C)=P(B)+P(C)=(1/36)+(4/36)=5/36
2) getting at least one “6” on both dice or an absolute
difference of 4?
D: At least one 6
P(C or D)=P(C)+P(D)-P(C&D)
=(4/36)+(11/36)-(2/36)=13/36
3) getting an absolute difference of 4 or odd total?
P(C or D)=P(C)+P(D)-P(C&D)
=(4/36)+(18/36)-0=22/36
Caffeine Example
Common sources of caffeine are coffee, tea and cola
drinks. Suppose that:
55% of adults drink coffee,
25% of adults drink tea,
45% of adults drink cola
15% drink both coffee and tea,
5% drink all three beverages,
25% drink both coffee and cola
And 5% drink only tea
(a) Draw a Venn diagram marked with this notation.
(b) What percent of adults drink only cola?
(c) What percent drink none of these beverages?
5.2 The Binomial Distributions
The Binomial Setting:
1) There is a fixed number n of observations
2) The n observations are independent
3) Each observation falls into just one of two categories,
which, for convenience, we call “success” and “failure”.
4) The probability of a success, call it p, is the same for
each observation
Example: Toss a fair coin 10 times, and count the number of
heads which will appear. The random variable, X, is the
number of heads that appear. The probability of a success
is p = 0.5
This experiment satisfies the four requirements for the
Binomial Setting, so this is a Binomial Distribution.
Binomial Distributions
The distribution of the count X of successes in the binomial
setting is called the binomial distribution with parameters
n and p.
The parameter n is the number of observations
The parameter p is is the probability of a success on any
one observation.
The possible values of X are the whole numbers from 0
to n.
We say that X is B(n,p)
Binomial Distributions
Example: Imagine we have a big bunch of transistors, say
roughly 10,000. We will say that 1000 of them are bad.
Pull an SRS of size 10 without replacement.
Let X be the amount of transistors which are bad in the SRS.
Q: Is this the binomial setting?
A: No. The first transistor has a 1000/10,000 = 0.1 chance of
being bad. What about the second?
The second has either a 1000/9,999 = 0.10001 or
999/9,999 chance = 0.09991 chance of being bad
We can consider this a B(10,0.1)
Example
In each of the following cases, decide whether or not a
binomial setting is the appropriate model, and give your
reasons.
(a) Fifty students are taught about binomial distributions
by a television program. After completing their study,
all students take the same exam. The number of
students who pass is counted.
(b) A chemist repeats a solubility test 10 times on the
same substance. Each test is conducted at
temperature 10 higher than the previous test. She
counts the number of times that the substance
dissolves completely.
Example (cont.)
In each of the following cases, decide whether or not a
binomial setting is the appropriate model, and give your
reasons.
(c) A student studies binomial distributions using a
computer-assisted instruction. After the initial
instruction is completed, the computer presents 10
problems. The student solves each problem and
enters the answer; the computer gives additional
instruction between problems if the student’s answer
is wrong. The number of problems that the student
solves correctly is counted.
Binomial Probability Distribution
Function
n  k
n k
P (X  k)    p 1  p 
k 
n 
n!
  
 k  k! (n  k)!
n
p
k
=
=
=
sample size
probability of ‘success’
number of ‘successes’ in sample
(X = 0, 1, 2, ..., n)
Example 1
Toss 1 Coin 4 times in a row. Note # Tails. What’s the
Probability of 3 tails?
This is a binomial setting with n=4, p=0.5 i.e. B(4,0.5)
n k
nk
P ( X  k )    p 1  p 
k 
 4
4 3
P ( X  3)    0.53 1  0.5
3
 0.25
Using TI-83 Calculator under distributions menu
Binompdf(4,0.5,3)=0.25
Example 1
Evans is concerned about a low retention rate for
employees. On the basis of past experience,
management has seen a turnover of 10% of the
hourly employees annually. Thus, for any hourly
employees chosen at random, management
estimates a probability of 0.1 that the person will
not be with the company next year.
Choosing 3 hourly employees a random, what
is the probability that 1 of them will leave the
company this year?
Let:
p = .10, n = 3, x = 1
• Using the Binomial Probability Function
n!
f ( x) 
p x (1  p ) (n  x )
x !( n  x )!
3!
f (1) 
( 0.1)1 ( 0. 9 ) 2
1!( 3  1)!
= (3)(0.1)(0.81)
= .243
Binomial Mean & Standard Deviation
A: If a count X has the B(n,p) distribution, then :
X = np
X =
np (1-p)
Example: Refer back to Example 2
Expected Value
E(x) = 3(.1) = .3 employees out of 3
Variance
Var(x) = 3(.1)(.9) = .27
Standard Deviation
SD( x)    3(.1)(.9)  .52 employees
The Normal approximation to
Binomial distributions
• Suppose that a count X has the Binomial
distribution with n trials and success probability p.
When n is large, the distribution of X is
approximately Normal,

N np, np1  p

• As a rule of thumb, we will use the Normal
approximation when n and p satisfy
np  10 and n1  p   10.
Examples
• Read Pages 329-330
• Exercise 5.41, p. 334
a ) N 75 , 4.33  and P  X  70   P  Z  1.15   0.1251 .
• The chance that Jodi will score 70% or lower is
about 12.5%
b ) N 187.5, 6.85 and P  X  175  P  Z  1.82  0.0344.
• Slightly over 3% chance that Jodi will score 70% or lower.
5.3 The Poisson Distributions
•
The Poisson Setting:
1. The number of successes that occur in any
unit of measure is independent of the number
of successes that occur in any nonoverlapping unit of measure.
2. The probability that a success will occur in a
unit of measure is the same for all units of
equal size and is proportional to the size of
the unit.
3. The probability that 2 or more successes will
occur in a unit approaches 0 as the size of the
unit becomes smaller.
Poisson Probability Function
e   k
P (X  k ) 
k!
where:
P(X=k) = probability of x occurrences in an interval
µ = mean number of occurrences in an interval
e = 2.71828
Using the Poisson Probability Function
Patients arrive at the emergency room of
Mercy Hospital at the average rate of 6 per
hour on weekend evenings. What is the
probability of 4 arrivals in 30 minutes on a
weekend evening?
 = 6/hour = 3/half-hour, x = 4
34 (2.71828) 3
P (X  4) 
 .1680
4!
Example 5.15
Poisson Mean & Standard Deviation
• Mean =


• Variance =
• Standard Deviation =
