Transcript Document

Math 3680
Lecture #5
Important Discrete
Distributions
The Binomial Distribution
Example: A student randomly guesses at
three questions. Each question has five
possible answers, only once of which is
correct. Find the probability that she gets 0,
1, 2 or 3 correct. This is the same problem
as the previous one; we will now solve it by
means of the binomial formula.
Example: Recall that if X ~ Binomial(3, 0.2),
P(X = 0) = 0.512
P(X = 1) = 0.384
P(X = 2) = 0.096
P(X = 3) = 0.008
Compute E(X) and SD(X).
MOMENTS OF Binomial(n, p) DISTRIBUTION
E(X) = n p
SD(X) =
np (1  p )
Var(X) =
np (1  p )
Try this for the Binomial(3, 0.2) distribution.
Do these formulas make intuitive sense?
Example: A die is rolled 30 times.
Let X denote the number of aces that appear.
A) Find P(X = 3).
B) Find E(X) and SD(X).
Example: Three draws are made with replacement
from a box containing 6 tickets:
• two labeled “1”,
• one each labeled, “2”, “3”, “4” and “5”.
Find the probability of getting two “1”s.
0.4
0.3
0.2
0.1
1
2
3
The Hypergeometric Distribution
Example: Three draws are made without
replacement from a box containing 6 tickets:
• two labeled “1”,
• one each labeled, “2”, “3”, “4” and “5”.
Find the probability of getting two “1”s.
P(S1S2S3) = P(S1) P(S2 | S1) P(S3 | S1 ∩ S2) = 2  1  0  0
6 5 4
P(S1S2F3) = P(S1) P(S2 | S1) P(F3 | S1 ∩ S2) = 2  1  4  1
6 5 4 15
2 4 1 1
P(S1F2S3) = P(S1) P(F2 | S1) P(S3 | S1 ∩ F2) =   
6 5 4 15
2 4 3 3
P(S1F2F3) = P(S1) P(F2 | S1) P(F3 | S1 ∩ F2) =
  
6 5 4 15
4 2 1 1
P(F1S2S3) = P(F1) P(S2 | F1) P(S3 | F1 ∩ S2) =   
6 5 4 15
4 2 3 3
  
P(F1S2F3) = P(F1) P(S2 | F1) P(F3 | S1 ∩ F2) =
6 5 4 15
4 3 2 3
P(F1F2S3) = P(F1) P(F2 | F1) P(S3 | F1 ∩ F2) =   
6 5 4 15
4 3 2 3
  
P(F1F2F3) = P(F1) P(F2 | F1) P(F3 | F1 ∩ F2) =
6 5 4 15
0.6
0.5
0.4
0.3
0.2
0.1
1
2
3
The Hypergeometric Distribution
Suppose that n draws are made without replacement
from a finite population of size N which contains G
“good” objects and B = N - G “bad” objects. Let X
denote the number of good objects drawn. Then
 G  B 
  
g  b 

P( X  g ) 
,
N
 
n
where b = n - g.
Example: Three draws are made without replacement
from a box containing 6 tickets; two of which are
labeled “1”, and one each labeled, “2”, “3”, “4” and
“5”. Find the probability of getting two “1”’s.
MOMENTS OF
HYPERGEOMETRIC(N, G, n)
DISTRIBUTION
E(X) = n p
(where p = G / N)
SD(X) =
N n
np (1  p )
N 1
 N n
np (1  p )
Var(X) = 
 N 1 
REDUCTION FACTOR
The term
N n
is called the Small Population
N 1
Reduction Factor.
It always appears when we draw without replacement.
If the population is large (N > 20 n) , then the
reduction factor can generally be ignored (why?).
Example:
Thirteen cards are dealt from a well-shuffled deck.
Let X denote the number of hearts that appear.
A) Find P(X = 3).
B) Find E(X) and SD(X).
Example. A lonely bachelor decides to play the
field, deciding that a lifetime of watching “Leave
It To Beaver” reruns doesn’t sound all that
pleasant. On 250 consecutive days, he calls a
different woman for a date. Unfortunately,
through the school of hard knocks, he knows that
the probability that a given woman will
accept his gracious invitation is
only 1%.
What is the chance that he will land
three dates?