Transcript + The Practice of Statistics, 4 th edition

+
Chapter 6: Random Variables
Section 6.3
Binomial and Geometric Random Variables
The Practice of Statistics, 4th edition – For AP*
STARNES, YATES, MOORE
+
Quick Review
 What
is the difference between a discrete RV
and a continuous RV?
 What
is the table called that shows the
outcomes of a discrete RV and the
respective probabilities?
 How
 Is
do you graph a discrete RV?
the normal RV discrete or continuous?
+
Special Kinds of Discrete RVs
 You
need to train yourself to recognize the
difference between these two types of
scenarios:
 A couple
is going to have children until they have a
girl.

Here, the random variable is how many children will it take
to get a girl.
 A couple
is going to have 3 children and we’ll count
how many are girls.

Here, the random variable is how many girls there are out
of the 3 children.
 Both
of these situations have binary
outcomes. Others include coin tosses,
shooting free throws, etc.
A binomial setting arises when we perform several independent trials of the
same chance process and record the number of times that a particular
outcome occurs. The four conditions for a binomial setting are
B
• Binary? The possible outcomes of each trial can be classified as
“success” or “failure.”
I
• Independent? Trials must be independent; that is, knowing the result
of one trial must not have any effect on the result of any other trial.
N
• Number? The number of trials n of the chance process must be fixed
in advance.
S
• Success? On each trial, the probability p of success must be the
same.
+
Binomial and Geometric Random Variables
Definition:
+
Example
 Blood
type is inherited. If both parents in a couple
have the genes for the O and A blood types, then
each child has probability 0.25 of getting two O
genes and thus having type O blood. Is the number
of O blood types among this couple’s 5 children a
binomial distribution?
 If so, what are n and p?
 If not, why not?
Today
 Now,
we’ll learn about the basis for the binomial
calculations: the binomial formula.
n k
P( X  k )    p (1  p )n k
k 
Notice the = mark.
This is a combinatorial. It is
read “n choose k.”
p stands for the probability of
success. n represents the
number of observations. k is
the value of x of which you’re
asked to find the probability.
Combinatorials
 A combination
(or combinatorial) helps us find out how
many ways there are to choose k objects from n total
objects.
 For
example, let’s revisit the blood type example. If
we’re trying to find the probability that a couple has 2
type O children out of 5, then the coefficient of the
binomial formula is
 5  = 5!
 
2!3!
 2
n!
k!(n-k)!
Completing the Formula
 In
the previous example, we were trying to find out
the probability that a couple who has 5 children has 2
type 0 blood children.
 The
formula, then, is
n k
P( X  k )    p (1  p ) n k
k 
5
P( X  2)    (0.25) 2 (1  0.25)3
 2
Inheriting Blood Type
+
 Example:
Each child of a particular pair of parents has probability 0.25 of having blood
type O. Suppose the parents have 5 children
(a) Find the probability that exactly 3 of the children have type O blood.
Let X = the number of children with type O blood. We know X has a binomial
distribution with n = 5 and p = 0.25.
5
P(X  3)   (0.25) 3 (0.75) 2  10(0.25) 3 (0.75) 2  0.08789
3
(b) Should the parents be surprised if more than 3 of their children have
type O blood?
To answer this, we need to find P(X > 3).

P(X  3)  P(X  4)  P(X  5)
5 
5
4
1
  (0.25) (0.75)   (0.25) 5 (0.75) 0
4 
5
 5(0.25) 4 (0.75)1  1(0.25) 5 (0.75) 0
 0.01465  0.00098  0.01563
Since there is only a
1.5% chance that more
than 3 children out of 5
would have Type O
blood, the parents
should be surprised!
+
Try This…
 The
number of switches that fail inspection
follows a binomial distribution with n = 10
and p = 0.1. Find the probability that no
more than 1 switch fails. Hint: 0! = 1.