Transcript (1/2) 3 x - danagoins

Chapter 7
Special Discrete
Distributions
Binomial Distribution B(n,p)
• Each trial results in one of two
mutually exclusive outcomes.
(success/failure)
• There are a fixed number of trials
• Outcomes of different trials are
independent
• The probability that a trial results in
success is the same for all trials
• The binomial random variable x is
defined as the number of successes out
of the fixed number
Are these binomial distributions?
1) Toss a coin 10 times and count the
number of heads
Yes
2) Deal 10 cards from a shuffled deck
and count the number of red cards
No, probability does not remain constant
3) Two parents with genes for O and A
blood types and count the number of
children with blood type O
No, no fixed number
Binomial example
Take the example of 5 coin tosses. What’s
the probability that you flip exactly 3 heads
in 5 coin tosses?
Binomial distribution
Solution:
One way to get exactly 3 heads: HHHTT
What’s the probability of this exact arrangement?
P(heads)xP(heads) xP(heads)xP(tails)xP(tails)
=(1/2)3 x (1/2)2
Another way to get exactly 3 heads: THHHT
Probability of this exact outcome = (1/2)1 x (1/2)3
x (1/2)1 = (1/2)3 x (1/2)2
Binomial distribution
In fact, (1/2)3 x (1/2)2 is the probability of each
unique outcome that has exactly 3 heads and 2
tails.
So, the overall probability of 3 heads and 2 tails
is:
(1/2)3 x (1/2)2 + (1/2)3 x (1/2)2 + (1/2)3 x (1/2)2
+ ….. for as many unique arrangements as
there are—but how many are there??
5
 
 3
5C3
ways to
arrange 3
heads in
5 trials
= 5!/3!2! = 10
Outcome
Probability
THHHT
(1/2)3 x (1/2)2
HHHTT
(1/2)3 x (1/2)2
TTHHH
(1/2)3 x (1/2)2
HTTHH
(1/2)3 x (1/2)2
HHTTH
(1/2)3 x (1/2)2
THTHH
(1/2)3 x (1/2)2
HTHTH
(1/2)3 x (1/2)2
HHTHT
(1/2)3 x (1/2)2
THHTH
(1/2)3 x (1/2)2
HTHHT
(1/2)3 x (1/2)2
10 arrangements x (1/2)3 x (1/2)2
The probability
of each unique
outcome (note:
they are all
equal)
P(3 heads and 2 tails) =
10 x (½)5=31.25%
5
 
 3
x P(heads)3 x P(tails)2 =
Binomial distribution
function:
X= the number of heads tossed in 5 coin
tosses
p(x)
0
1
2
3
4
number of heads
5
x
n!
n
n Cr  

 r  r ! n  r  !
Pascal’s Triangle!
1
11
121
1331
14641
1 5 10 10 5 1
1 6 15 20 15 6 1
1 7 21 35 35 21 7 1
1 8 28 56 70 56 28 8 1
1 9 36 84 126 126 84 36 9 1
Example 2
As voters exit the polls, you ask a representative
random sample of 6 voters if they voted for
proposition 100. If the true percentage of voters
who vote for the proposition is 55.1%, what is the
probability that, in your sample, exactly 2 voted
for the proposition and 4 did not?
Solution:
6
 
 2
ways to
arrange 2
Obama votes
among 6
voters
Outcome
YYNNNN
NYYNNN
NNYYNN
NNNYYN
NNNNYY
Probability
= (.551)2 x (.449)4
(.449)1 x (.551)2 x (.449)3 = (.551)2 x (.449)4
(.449)2 x (.551)2 x (.449)2 = (.551)2 x (.449)4
(.449)3 x (.551)2 x (.449)1 = (.551)2 x (.449)4
(.449)4 x (.551)2
= (.551)2 x (.449)4
.
.
15 arrangements x (.551)2 x (.449)4
P(2 yes votes exactly) =
6x
 
 2
(.551)2 x (.449)4 = 18.5%
n = number of trials
Binomial Formula:
1-p = probability
of failure
n  k
n k
P (x  k )    p 1  p 
k
 
Where:
K = # successes
out of n trials
p = probability
of success
n 
 n C k
k 
Toss a 3 coins and count the number of
heads
Find the discrete probability distribution
X
P(x)
0
1
2
3
.125
.375
.375
.125
Out of 3 coins that are tossed, what is
the probability of getting exactly 2 heads?
Out of 3 coins that are tossed,
what is the probability of
getting exactly 2 heads?
3 2
1
P (x  2)   0.5 0.5  .375
2
Binomial distribution: example

If I toss a coin 20 times, what’s the
probability of getting exactly 10 heads?
 20  10 10
 (.5) (.5)  .176
 10 
The number of inaccurate gauges in a
group of four is a binomial random
variable. If the probability of a defect
is 0.1, what is the probability that only
1 is defective?
 4 1
3


P (x  1)   0.1 0.9  .2916
1 
More than 1 is defective?
P (x  1)  1  (P (0)  P (1))  .0523
Practice problems
If the probability of being a smoker among
a group of cases with lung cancer is .6,
what’s the probability that in a group of 8
cases you have less than 2 smokers? More
than 5?
What are the expected value and variance of
the number of smokers?
Answer
X
0
1
2
3
4
5
6
7
8
P(X)
8
1(.4) =.00065
7
1
8(.6) (.4) =.008
6
2
28(.6) (.4) =.04
5
3
56(.6) (.4) =.12
4
4
70(.6) (.4) =.23
3
5
56(.6) (.4) =.28
2
6
28(.6) (.4) =.21
1
7
8(.6) (.4) =.090
8
1(.6) =.0168
1
11
121
1331
14641
1 5 10 10 5 1
1 6 15 20 15 6 1
1 7 21 35 35 21 7 1
1 8 28 56 70 56 28 8 1
Answer, continued
0 1 2 3 4 5 6 7 8
Answer, continued
P(<2)=.00065 + .008 = .00865
P(>5)=.21+.09+.0168 = .3168
0 1 2 3 4 5 6 7 8
E(X) = 8 (.6) = 4.8
Var(X) = 8 (.6) (.4) =1.92
StdDev(X) = 1.38
Calculator
• Binomialpdf(n,p,x) – this calculates
the probability of a single binomial
P(x = k)
• Binomialcdf(n,p,x) – this calculates
the cumulative probabilities from
P(0) to P(k) OR
P(X < k)
A genetic trait of one family manifests
itself in 25% of the offspring. If eight
offspring are randomly selected, find
the probability that the trait will
appear in exactly three of them.
P( X  3)  binomialpdf (8,.25,3)  .2076
Less than half?
P( X  4)  binomialcdf (8,.25,3)  .8862
At least 5?
P (X  5)  1  binomialcdf (8,.25,4)  .0273
In a certain county, 30% of the
voters are Republicans. If ten
voters are selected at random,
find the probability that no more
than six of them will be
Republicans.
P(x < 6) = binomcdf(10,.3,6) = .9894
What is the probability that at least 7
are not Republicans?
P(x > 7) = 1 - binomcdf(10,.7,6) = .6496
Binomial formulas for mean
and standard deviation
 x  np
 x  np 1  p 
In a certain county, 30% of the
voters are Republicans. How
many Republicans would you expect
in ten randomly selected voters?
What is the standard deviation for
this distribution?
 x  10(.3)  3 Republicans
x  10(.3)(.7)  1.45 Republicans
What happens to the shape
of the distribution as the
probability of success
increased?
•In L1 – seq(x,x,0,10)
•In L2 – binompdf(10, .1 ,L1)
•Sketch histogram on board
What happened to the
shape of the distribution
as the probability of
success increased?
As the probability of success
increases, the shape changes from
being skewed right to symmetrical
at p =.5 to skewed left.
•Calculate the mean and standard
deviations for each of the
probabilities
What do you notice?
As the probability of success increases,
•the means increase.
•the standard deviations increase to p = .5, then
decrease. Their values are also symmetrical.
Geometric Distributions:
• There are two mutually exclusive
So what are the
outcomesHow far
this go?
To will
infinity
possible values of X
• Each trial is independent of the
others
• The probability of success
remains constant for each trial.
1 2 variable
3 4 .x .is.the
•X
The random
number of trials UNTIL the
FIRST success occurs.
Differences between binomial
& geometric distributions
• The difference between
binomial and geometric
properties is that there is
NOT a fixed number of
trials in geometric
distributions!
Other differences:
•Binomial random variables
start with 0 while geometric
random variables start with 1
•Binomial distributions are
finite, while geometric
distributions are infinite
Geometric Formulas:
P (x )  p 1  p 
x 1
1
x 
p
1p
x 
2
p
Not on formula
sheet – they will
be given on quiz or
test
Count the number of boys in a
family of four children.
Binomial:
X
0
1
2
3
What are the
values for these
random
4
variables?
Count children until first son
is born
Geometric:
X
1
2
3
4
. . .
Calculator
• geometpdf(p,x) – finds the
geometric probability for P(X = k)
No “n” because
there is no
• Geometcdf(p,x) – finds the
fixed number!
cumulative probability for P(X < k)
What is the probability that the
first son is the fourth child born?
P (X  4)  geometricpdf (.5,4)  .0625
What is the probability that the first
son born is at most the fourth child?
P (X  4)  geometriccdf (.5,4)  .9375
What is the probability that the first
son born is at least the third child?
P (X  3)  1  geometriccdf (.5,2)  .25
A real estate agent shows a house to
prospective buyers. The probability that
the house will be sold to the person is
35%. What is the probability that the
agent will sell the house to the third
person she shows it to?
P (x  3)  geometricpdf (.35,3)  .1479
How many prospective buyers does she
expect to show the house to before
someone buys the house? SD?
1
1  .35
x 
 2.86 buyers  x 
 2.304 buyers
2
.35
.35
•In L1 – input the x-values
•In L2 – geometpdf(.1 ,L1)
•Sketch histogram on board
What happened to the
shape of the distribution
as the probability of
success increased?
As the probability of success
increases, the shape becomes more
& more strongly skewed right
•Calculate the mean and standard
deviations for each of the probabilities
What do you notice?
As the probability of success increase,
•the means decrease
•the standard deviations decrease.