Counting Rules and Binomial Probabilities

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Counting Rules and Binomial Probabilities

• Stat 509 Lecture (E. Pena)
• February 12, 2001

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Fundamental Theorem of Counting
Suppose you have an experiment which could
be performed in k steps, so the outcome of this
experiment will be of form
(O1, O2, , Ok)
where Oi is the outcome from the ith step of the
experiment.
If on the ith step there are ni distinct possible
outcomes, then the total number of possible
(ordered) outcomes in the experiment is n1n2nk.
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Some Examples
1. How many possible outcomes are there when a
coin is tossed 25 times? Solution (2)(2)(2)(2)
225
2. Suppose that a man and a woman is to be chosen
from a group of people where 10 are men and 15
are women. How many possible choices can be
made? Solution Step 1 is to choose the man, and
there are 10 ways of doing this. Step 2 is to
choose the woman and there are 15 ways of doing
this. Therefore, the number of possible outcomes
is (10)(15) 150 ways.
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3. How many possible subsets could be formed from
a set with 10 elements, if we also include the
empty subset and the whole set as a
subset? Answer 210 1024. Why?
4. Suppose that you have n distinct objects. In
how many ways could these n objects be arranged
in a sequence? For example, suppose that there
are 20 students in this class, and there are
available 20 chairs. In how many ways could you
be seated?
Solution The first object can be chosen in n
ways the next in (n-1) ways then (n-2) ways, ,
until the last object has only one way to go.
Therefore, the number of ways is
(n)(n-1)(n-2)(2)(1) n! with 0! 1.
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Factorial Notation
For a positive integer, n, the factorial of n is
defined via n! (n)(n-1)(n-2)(2)(1).
Examples
4! (4)(3)(2)(1) 24 By convention, 0! 1 7!
(7)(6!) (7)(6)(5!) In general, n! n(n-1)!
n(n-1)(n-2)! etc.
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Permutations
5. Suppose that you have n distinct elements. In
how many ways could you choose r of these
elements if sampling is without replacement, and
the order in which the elements are chosen is
important, that is, the samples are
ordered? Solution nPr number of permutations
of n objects taken r at a time. This number is
given by
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Combinations
6. Suppose again that you have n distinct
objects. In how many ways could you form a sample
of size r if sampling is without replacement and
the order in which the objects are being included
in the sample does not matter?
Answer nCr number of combinations of n objects
taken r at a time. This is defined by
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Some Properties of Combinations
1. nC0 1 and nCn 1 2. nC1 n and nCn-1
n 3. nCr nCn-r 4. nCr are called the binomial
coefficients since they appear as the
coefficients in the binomial expansion
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Applications of Counting Rules
1. A batch of 20 electronic items (e.g., computer
chips) has 5 defective and 15 good items. Suppose
that a sample of size 4 is taken without
replacement from this batch, with the order in
which they are taken being unimportant.
A) How many possible samples are
possible? Solution 20C4 20!/4!(20-4)!
4845 samples.
B) How many samples are possible which has
exactly 2 defectives and 2 good items? Solution
(5C2)(15C2) (10)(105) 1050.
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C) If sampling is performed randomly, what will
be the probability of getting a sample with
exactly 2 defectives and 2 good items? Solution
P(2D,2G) N(A)/N(S) 1050/4845 0.2167.
D) What is the probability of getting a sample
with at least 2 defective items? Solution P(at
least 2 D) P(2D,2G) P(3D,1G) P(4D,0G)
(5C2)(15C2) (5C3)(15C1) (5C4)(15C0)/20C4
1050 150 5/4845 1205/4845 0.2487.
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2. Suppose that you have 8 (indistinguishable)
objects of type I, and 12 (indistinguishable)
objects of type II, and you have 20 positions on
which to place these 20 objects. In how many ways
could you place these 20 objects in these 20
positions? Solution You may place these objects
by first choosing the 8 positions out of the 20
positions on which to put the type I objects. The
remaining 12 positions will then be occupied by
the 12 type II objects. Therefore, the total
number of ways equals the number of ways of
choosing 8 positions out of 20, without regards
to the order in which the positions are chosen.
Therefore, the total number of ways is 20C8
20C12 20!/8!(20-8)! 125,970 ways.
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3. If a coin is tossed 25 times, in how many ways
could you get an outcome with exactly 15 heads
and 10 tails? Solution The number of ways would
be the number of ways of choosing the 15
positions out of the 25 tosses in which the heads
should appear. The other 10 positions will have
the tails. Thus, the total number of ways of
getting exactly 15 heads and 10 tails is 25C15
3,268,760.
Question What will be the probability of getting
exactly 15 heads in 25 tosses of a fair
coin? Answer P(15H,10T) 25C15/225 25C15
(1/2)15 (1/2)10 3268760/33554432 0.097, or
about 10.
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4. Consider the experiment of rolling a fair die
25 times. How many possible outcomes are there in
this experiment? Solution By the Fundamental
Theorem of Counting, there will be (6)(6)(6)
625 possible outcomes. B) How many possible
outcomes are there with exactly 15 6s and 10
non-6s? Solution Any configuration with 15
6s and 10 non-6s could occur in (115)(510)
possible ways. There are 25C15 configurations
with 15 6s and 10 non-6s. Therefore, the total
number of ways of getting 15 6s and 10 non-6s
is 25C15 (1)15(5)10.
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Binomial Probabilities
In the preceding example of rolling a fair die 25
times, the probability of obtaining exactly 15
6s and 10 non-6s is therefore
P(15 6s and 10 non-6s) 25C15 (1/6)15 (5/6)10.
This is an example of a binomial probability.
Definition A discrete random variable X taking
values in 0,1,2,,n is said to have a binomial
probability function with parameters n and p (0 lt
p lt 1) if its probability function is given by
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Binomial Probability Models
A random experiment that gives rise to binomial
random variables typically has the following
characteristics
1. The experiment consists of n trials or
steps. 2. Each trial or step has only two
possible outcomes labeled a success (S) or a
failure (F). 3. The probability of getting a
success (S) at a given trial is equal to p. 4.
The outcome on any given trial is unaffected by
the outcomes of the other trials (this is the
assumption of independent trials).
If X is the random variable denoting the number
of successes out of the n trials, then it has a
binomial probability distribution with parameters
n and p.
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Binomial Probability Shapes
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Computing Binomial Probabilities

• Use the binomial tables in the textbook. Provides
  probabilities of form P(X lt k), so P(X k) P(X
  lt k) - P(X lt k-1).
• Use the binomial function in your calculator.
• Use Minitab using the Probability Distributions
  option in the Calc Menu.
• Calculate it by hand and calculator.

Example Assume that the probability of a newly
manufactured computer of a certain brand having a
defect is 0.10. Suppose that 20 of these newly
manufactured computers are tested for defects.
Let X denote the number out of these 20 computers
which has defects. Then X has a binomial
distribution with n 20 and p .10.
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Some Questions Pertaining to the Preceding
Situation

• What is the probability that none of the 20
  computers will have defects?

Answer P(X0) 20C0 (.1)0(.9)20-0 (0.9)20
0.1216.

• What is the probability that at least one of
  these 20 computers will be found to have a defect?

Answer P(X gt 1) 1 - P(X0) 1 - 0.1216
0.8784.

• What is the probability that the number of
  defective computers will be at most 2?

Answer P(X lt 2) P(X0) P(X1) P(X2)
.1216 .2702 .2852 .6769 using Minitab.
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Mean and Standard Deviation of a Binomial Random
Variable
If X is a random variable whose distribution is
binomial with parameters n and p, then
Mean m np
Variance s2 np(1-p)
Standard Deviation s np(1-p)1/2.
Example In the preceding example, m (20)(.1)
2 and s 20(.1)(1-.1)1/2 (1.8)1/2 1.34.
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Computational Limitations when Dealing with
Binomial Distributions
Situation Suppose that you toss a fair coin 6
million times, and you denote by X the number of
heads that comes up. Then X will be binomial with
n 6,000,000 and p 0.5. Consequently,
On average, we should get m (6,000,000)(.5)
3,000,000 heads in this experiment and
The standard deviation will be s
(6000000)(.5)(.5)1/2 1224.74.
But, what will be the probability that X will
differ from 3 million by at most 500? This is
tough computationally, so later we will simply
approximate this probability! Relate this
situation to the Florida election!!