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Unformatted text preview: Chapter 11:
Chapter
Applications of Binomial Probabilities
Applications
For repeated, independent, experiments
For
with just two outcomes, success and
failure, where P (success) = p and
failure,
P (failure) = q = 1p, the probability of
1p the
gaining x successes out of n trials is
trials P( X = x) = ( ) p q
n
x x n− x , x = 0,1,....., n The P(X=x) is called a binomial probability,
and the whole list of P(x) for x=0,1,2,….,n
is called a binomial distribution.
We use the symbols
X ~ Bin (n, p)
Bin
to stand for the sentence,
to
“X is distributed like a binomial model with
number of trials n and probability of
success per trial p”
success The values of the binomial probabilities can be
computed directly.
They are also tabulated for certain selected values
of n and p in Table 2 which is an extract from a
much larger book for binomial distributions.
much
A: Straightforward Applications
Example 1(a):
A plays B in the final of a badminton tournament and each
performs independently from game to game.
Whoever wins two games out of three first gets the gold
medal.
In each game, A’s chance of winning is 0.45 while B’s
chance of winning is 0.55.
What is the probability that A gets the medal?
What Solution:
Solution
Assume that they play all the three games,
Assume
whether or not the match ends before the 3rd
whether
game.
game.
Let X = the number of games A wins in these
three
three
Then X ~ Bin( n = 3, p = 0.45)
From Table 2, one finds, for n=3 and p=0.45,
From
P(2)=0.3341 and P(3)=0.0911
P(2)=0.3341
Therefore,
Therefore,
P(A gets gold medal)
P(A
=P(X≥2)
=P(X=2)+P(X=3)= 0.3342 + 0.0911
=0.4252 Example 1(b)
A plays B in the final of a badminton
tournament.
Who ever wins three games first gets the
cup.
In each game, A’s chance of winning is 0.65.
Both perform independently from game to
game.
What is the probability that A gets the cup?
What Recommended Method:
X = the number of times A wins
X ~ Bin ( n=5, p=0.65)
P(A gets the cup)=P(X=3) + P(X=4) + P(X=5)
look up Binomial tables
= 0.3364 + 0.3124 + 0.1160
= 0.7648
This method is recommended by Prof Chiu and
Gabrielle! Let’s look at Example 1(b) in more detail:
The way of working out the answer on this page is a method that many students like to do.
This method works BUT is NOT recommended by Gabrielle or Prof Chiu because it is not efficient
(It’s correct but NOT fast!)
Who ever wins 3 games first!
Possibilities:
AA  A
A=3
BAA  A
ABA  A
AAB  A
BBAA  A
BABA  A
BAAB  A
ABBA  A
ABAB  A
AABB  A
Method B:
P(A gets the cup) B=0 A=3 B=1 A=3 = B=2 ( )(0.65) (0.65) + ( )(0.65) (0.35)(0.65) + ( )(0.65) (0.35) (0.65)
2
2 2 3
2 2 4
2 2 2 = (0.65) 2 (0.65) + 3(0.65) 2 (0.35)(0.65) + 6(0.65) 2 (0.35) 2 (0.65)
= (0.65)3[1 + 1.05 + 0.735]
= 0.7648 Let’s look at a much BETTER (Prof Chiu’s) method:
Who ever wins 3 games first!
Possibilities:
AA  A
A=3
B=0 BAA  A A=3 AAA BB
AAA AA
AAA BA
AAA AB B=1 BAA AB
BAA AA
ABA  A
A=3
B=1
ABA AA
ABA AB
AAB  A
A=3
B=1
AAB A A
AAB AB
And so on………………………………………………
Think of this like the MEN’S Tennis finals in Melbourne or Wimbledon
Recommended Method:
X = the number of times A wins
X ~ Bin ( n=5, p=0.65)
P(A gets the cup)=P(X=3) + P(X=4) + P(X=5) look up Binomial tables
This method is recommended and is MUCH FASTER than the other way Example 2:
Example
The probability that a soldier returns from a
battle uninjured is 0.85.
A group of five is sent out.
What is the probability that at least four of
them will return uninjured?
them
(Assume the independence of probabilities)
Solution:
Solution:
Let s = number of soldiers who return
uninjured.
Then s ~ Bin (n=5, p=0.85)
Then Table 2 does not give
binomial probabilities
for p more than 0.5.
We can make use of the
block for n=5 and
n=5
p=0.15,
p=0.15
but downside up.
downside
Thus, P(s≥4)
Thus,
=P(4)+P(5)
=0.3915 +0.4437
=0.8352
=0.8352 Note :
“At least 4” means 4 or 5 soldiers returning
UNINJURED.
UNINJURED.
Thus, an alternative method is to look at
Thus,
r, which represents the number of returning
which
INJURED and for which,
r ~ Bin (n=5, p=0.15)
Bin
Therefore, P (r ≤ 1) = P(0) + P(1)
Therefore,
= 0.4437 + 0.3915
= 0.8352
0.8352 Example 3
Example
A large batch of apples is delivered to a customer, who adopts a twostage sampling inspection.
Stage 1:
He takes a sample of 10 apples.
He
If NO defective apple is found, he accepts the batch.
If two or more defectives are found, he rejects the batch.
If
If only one defective is found, he takes a 2nd sample, i.e. he goes to
stage 2.
stage
Stage 2:
He takes another sample of 14 apples.
He
If 0 or 1 defective apple is found, he accepts the batch.
Otherwise, he rejects the batch.
Otherwise,
Suppose the proportion of the batch that is actually defective is 0.15.
Suppose
What is the probability that it will be accepted?
What Draw a diagram to summarize this question
Draw Solution:
Solution
Stage 1:
Let X = the number of defectives out of 10
Let
then X ~ Bin (n=10, p=0.15)
P (batch Accepted at stage 1)
(batch
= P(X=0)
= 0.1969
from Table 2
0.1969
P(2nd sample needed)
= P(X=1)
= 0.3474
from Table 2
0.3474 Stage 2:
Stage
Let Y = the number of defectives out of 14.
Let
then Y ~ Bin (n=14, p=0.15)
P (Accepts batch at stage 2)
= P(Y=0) + P(Y=1)
P(Y=0)
= 0.1028 + 0.2539
= 0.3567
from Table 2
0.3567 Combining the analyzed results of
Stages 1 and 2, we have
Stages
P (customer accepts batch)
(customer
= P(X=0) + P(X=1)[P(Y=0)+P(Y=1)]
P(X=0)
=0.1969 + (0.3474 x 0.3567)
=0.3208
Note:
Acceptance sampling is an important topic
Acceptance
in Quality Control.
in Example 4
Example
Draw the graph for each of the following
Draw
distributions:
distributions
(a) Bin(6,0.1) (b) Bin(6,0.5) (c) Bin(6,0.85)
From Table 2, we have these figures from
From
which bar charts and histograms can be
drawn:
drawn: B: The Mean and Variance of a Binomial
Distribution
Distribution
Whenever one is given a set (i.e. a distribution) of data,
Whenever
whether it is presented in a frequency table like Chapter 5
Table 6,
or in a proportion table like Chapter 6 Example 1,
one’s job is to summarize information into two quantities,
mean, μ ,and variance, σ²
(or equivalently, standard deviation, σ)
Now we are studying binomial distributions.
We have, for each distribution, a proportion (i.e.
probability) table.
Our next task would be to find its mean and variance.
Our Example 5:
Example
For the distribution X ~ Bin (n=4,p=0.4), find
For
the mean and variance.
the
Note:
Note:
Solution:
Here, we have scores and
Table 2 gives
probabilities so we
Bin (n=4, p=0.4) distn:
Bin
should use the formulas
from Chapter 6
from µ = ∑xi pi
σ = ∑ x . pi − µ
2 2
i 2 With calculator:
With
ON, MODE, COMP, MODE, SD 0 SHIFT ; 0.1296 DATA
1 SHIFT ; 0.3456 DATA
………..
4 SHIFT ; 0.0256 DATA
SHIFT SVAR 1.6 = µ x SHIFT SVAR xσ n x 2 0.96 = σ 2 The above results show that for
X ~ Bin (n, p) distribution: µ X = np,
Bin σ = npq
2
X In fact, they are always true.
In
You can verify these on every block of figures in
Table 2.
In fact, the above equations can be proven
mathematically, although we prefer to omit the
proof here.
These formulas are very convenient, for we do
NOT need to perform detailed computations
every time.
every Example 6:
Example
It is known that a binomial distribution has
It
mean 36 and variance 19.8
What are its two parameters, n and p?
parameters
Solution: µ = np = 36
σ 2 = npq = 19.8
npq 19.8
∴
=
np
36
q = 0.55
∴ p = 1 − q = 1 − 0.55 = 0.45
36
36
n=
=
= 80
p 0.45 More examples (a bit tricky)
More
A student takes an exam in which there are
student
5 hard questions and 5 easy questions.
The student answers the questions
independently.
Let X = the number of questions that the
student answers correctly.
Does X have a binomial distribution?
Does Solution:
Solution:
Let X=the number of correctly answered questions, in n=10
Let
trials.
Questions are answered independently
Questions
What about the value of p?
The wording of this question suggests that the probabilities
The
of answering each question correctly are NOT the
same, presumably being smaller for hard questions
than for easier questions.
than
Therefore, X does NOT have a binomial distribution.
Therefore,
NOT Another example:
Another
Players A and B are evenly matched in ability for
Players
playing handball.
They agree to continue to play games of handball
until player A wins a game.
Each game is played independently of all other
games.
games.
Let Y be the number of games they play.
Let
Does Y have a binomial distribution?
Does Solution:
Solution:
Since A and B are said to be evenly matched, we can
Since
assume that the probability that A wins any given
game is p=0.5.
Also, games are played independently.
However, the number of trials is NOT fixed
the
and
we are NOT counting successes.
NOT
In fact, the random variable Y is counting trials, NOT
successes.
Thus, Y does NOT have a binomial distribution.
NOT
binomial
The distribution of Y is actually GEOMETRIC with
parameter p=0.5
which we are NOT studying in this course.
which A very fishy example:
very
A pond is stocked with 100 fish of which 40 are
pond
of legal size.
You catch 3 fish.
Let X=the number of fish of legal size that you
catch.
catch. Does X have a binomial distribution? Answer: Maybe! It depends!
Answer
It matters whether you put back the first fish you catch
It
OR
you decide to take home and eat the first fish
you
You can think of the process of catching fish as if it were a
You
random sample of n=3 fish taken from the pond.
Each fish caught is a Bernoulli trial, with “success” being
when the fish is of legal size.
when
Note that a fish has probability p=of being legal size.
The only fact that remains is whether the trials (fish)
The
are independent.
IF each fish is RETURNED to the pond after it is caught
each
and a caught fish fails to learn from its experience, the
fish are being sampled WITH REPLACEMENT.
fish
In this case the trials are independent and X has a
independent
binomial distribution with parameters n=3 and p=0.4.
binomial
We write X ~ Bin ( n=3, p=0.4 )
We Fishy example cont’d
Fishy
On the other hand, most fishermen do NOT return caught fish to the pond; in
On
NOT
this case, the fish are sampled WITHOUT REPLACEMENT.
WITHOUT
If the first 2 fish you catch are of legal size, the third fish is sampled from a
population of 98 fish of whom 38 are of legal size.
Hence,
38
P(3rd fish is legal  first 2 fish legal ) = = 0.388 ≠ 0.4 = P(3rd fish is legal size)
P(3rd
98 So the trials (fish) are DEPENDENT.
So
DEPENDENT
Thus, X does NOT have a binomial distribution.
NOT
binomial
It has a HYPERGEOMETRIC distribution.
It
HYPERGEOMETRIC It’s really Fishy
It’s It has a HYPERGEOMETRIC distribution.
It
HYPERGEOMETRIC
Let S=”success”= fish is legal size
Let =”success”=
legal
F=”failure”= fish is NOT legal size
=”failure”=
NOT
All the possibilities of catching 3 fish are:
SSS, SSF, SFS, FSS, SFF, FSF, FFS, FFF
We want only 2 successes, so the possibilities are:
We
successes
P ( X = 2)
SSF, SFS, FSS = ( 40 x 39 x 60 ) + ( 40 x 60 x 39 ) + ( 60 x 40 x 39 )
100 99 98
= 3 x0.0965
= 0.2895 100 99 98 100 99 98 or use the hypergeometric formula: C (40,2) xC (60,1)
C (100,3)
40! 60!
x
38!2! 59!1!
=
100!
97!3!
= 0.2895
P ( X = 2) = Past final exam questions on
Binomial Distribution
Binomial
Articles are produced by independent operations of a
Articles
machine.
At any one operation, the probability is 0.8 that the article is
acceptable.
The articles are placed in boxes as they are produced.
The
Each box contains 5 articles.
Each
(a) What is the probability that a randomly selected box
contains no acceptable articles?
contains
(b) What is the probability that a randomly selected box
(b)
contains more acceptable than defective articles?
contains
(c) Of 20 boxes examined, what is the probability that at most
(c)
one contains no acceptable articles?
one Solution:
Let X = number of acceptable articles in each box
(the number of nondefective articles)
X~ Bin (n=5, p=0.8)
Use the binomial formula : (a) P[X=x] = ( )p
n
x x (1 − p) n − x P [Box contains no acceptable articles]
= P[X=0] = 0.00032 You can use your calculator BUT it’s QUICKER to use table 2:
Look up n = 5 , p = 0.2 , x = 5 because
“no acceptable articles in each box” is the same as
“all 5 articles in each box are DEFECTIVE”
The probability that the articles are defective
equals 1 0.8 = 0.2 Solution: (b)
Solution:
P [Box contains more acceptable than
[Box
defective articles]
= P[X=3] + P[X=4] + P[X=5]
P[X=3]
= 0.2048 + 0.4096 + 0.32768
0.2048
= 0.94208
Using Table 2 is much quicker than the
calculator!
calculator! Solution: (c)
Solution:
Let Y = the number of boxes containing no
boxes
acceptable articles
acceptable
Y ~ Bin ( n=20, p = 0.00032) from (a)
Bin
P [Of 20 boxes examined , at most one contains no acceptable
articles]
articles]
= P[Y=0] + P[Y=1]
P[Y=0]
= 0.99362 + 0.006361
= 0.99998
0.99998
You CANNOT USE table 2 for this question.
You CANNOT
IT MUST be done with your calculator, so don’t forget to
practise with the 50FH! BINOMIAL and NORMAL Distributions in
the same question
the
The serum cholesterol level X in 14yrold boys has
The
approximately a normal distribution with mean µ= 170 and
170
variance =900.
variance
(a) Find the probability that a randomly selected 14yrold boy
(a)
has serum cholesterol level between 160 and 200.
has
(b) Find the probability that exactly 7 boys out of 10 randomly
(b)
selected 14yrold boys, have serum cholesterol levels
between 160 and 200.
between
(c) Usually high serum cholesterol levels are a cause for
(c)
concern.
Find the level c such that 1% of the population of 14yrold boys
have serum cholesterol levels higher than c.
have Solution: (a)
Solution:
Let X = (measuring) the serum cholesterol level in
Let
14yrold boys
14yrold
X ~ N( µ = 170 ,σ² = 900)
900)
P[ 160 < X < 200]
P[
= 160 − 170
200 − 170
P[ 30 <Z< 30 = P[0.33 < Z < 1]
P[0.33
= 0.1293 + 0.3413 (from table A)
0.1293
= 0.4706 Solution:(b)
Solution:(b)
Let Y = (counting) the number of 14yrold
Let
boys having serum cholesterol levels
between 160 and 200.
between
Y ~ Bin ( n=10, p= 0.4706) from (a)
n=10,
P[Y=7] = ( )(0.4706) (0.5294)
10
7 = 0.091
0.091 7 3 (with calculator) Solution: (c)
Solution:
Look up 0.49 in table B, gives z = 2.326
Look
2.326
Now
X − µ , find x=c z= c  170
170
c
∴c σ
= 30 x 2.326
= 30 x 2.326 + 170
30
= 239.78 C: Normal Approximation to Binomial
Distribution
Distribution
For X ~ Bin (n, p) the binomial probability is given by P( X = x) = ( )p q
n
x x n− x , x = 0,1,....., n () Now, when n is large, say, n≥30, the computation of n
Now,
the
x
is somewhat tedious.
We look for approximation methods.
There are two methods.
The first one is the case of n is large and p is somewhat
large
close to 0.5, the closer the better.
close
the
Actually, 0.1 < p < 0.9
Actually, Now, when p is close to 0.5, the distribution is
somewhat symmetric, see Case (b) of Example
4.
Whenever we have a symmetric distribution, we
think of normal distribution.
Now, we know from Example 5 that the formulas
for the mean and variance of X ~ Bin (n, p) are
μ= np and σ² = npqc, ~ N (np, npq )
npq respectively.
X
Therefore, we use
to match
Therefore,
X ~ Bin (n, p) and use Table 1 to find the
Bin
required probabilities.
required
Xc
= continuous version of X.
continuous
X. Normal Approximating Binomial Theorem
(NABT)
(NABT)
For X ~ Bin (n, p) where n is large
For
Bin
and p is not too remote from 0.5
in such a way that np ≥ 5, nq ≥ 5,
and 0.1 < p < 0.9,
and
X can be approximated by
X c ~ N (np, npq ) (Proof Omitted)
(Proof Example 7:
Example
A fair, ordinary dice is rolled 18 times. What
is the probability of getting at least 4 but
NOT more than 10 “red” results?
NOT
Solution:
On an ordinary fair dice, faces “1” and “4”
are red in colour.
are
Therefore, on each roll the chance of
Therefore,
getting “red” is 2/6=1/3.
getting Let X = the number of times “red” occurs when
rolling the dice
X ~ Bin (n =18, p =1/3) where
Bin
μ = np = 6 and σ² = npq = 6x2/3 = 4
Now by NABT, X is approximated by X c ~ N (np, npq)
X exact Xc ~ 1
Bin(n = 18, p = )
3 approx ~ N ( µ = 6, σ 2 = 4) Xc Now, we are required to find
P(4 ≤ X ≤ 10) = P(3.5 ≤ X c ≤ 10.5)
10)
That 4 ≤ X ≤ 10 has been changed into
That
3.5 ≤ X c ≤ 10.5 is called
the continuity correction.
continuity The rule is:
The
½(the last number included) +½ (tthe first number
he
½(
excluded)
excluded
1
1
×10 + ×11 = 10.5
2
2 which is equivalent to
which
for the right cutoff value
for
1
1
× 4 + × 3 = 3.5
2
2 and to
and for the left cutoff value. Now the Signal/Sigma ratio corresponding to 3.5
Now
and 10.5 are, respectively,
z1 = 3.5 − 6
= −1.25
2 and
and z2 = 10.5 − 6
= 2.25
2 Hence
Hence P (4 ≤ X ≤ 10)
= P (3.5 ≤ X c ≤ 10.5)
= P (−1.25 ≤ Z ≤ 2.25)
= 0.3944 + 0.487
= 0.8822 Note:
The two quantities, μandσ², characterizing a normal
²,
distribution, N(μ,σ²), are called parameters.
distribution,
²),
parameters.
The two quantities, n and p, characterizing a binomial
characterizing
distribution,
Bin (n, p), are also parameters.
Bin
parameters
See Example 6 above.
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This note was uploaded on 02/22/2010 for the course FBE STAT0302 taught by Professor Unknown during the Spring '10 term at HKU.
 Spring '10
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