Transcript 2.1 Discrete and Continuous Variables

Chapter 6 Some Special Discrete
Distributions
6.1 THE BERNOULLI DISTRIBUTION
6.2 THE BINOMIAL DISTRIBUTION
6.3 THE GEOMETRIC DISTRIBUTION
6.4 THE POISSON DISTRIBUTION
6.1 THE BERNOULLI
DISTRIBUTION
 6.1.1
Bernoulli Trials
 Example 1
– The calculation of a payroll check may be correct or
incorrect. We define the Bernoulli random variable for
this trial so that X = 0 corresponds to a correctly
calculated check and X = 1 to an incorrectly calculated
one.
 Example 2
– A consumer either recalls the sponsor of a T.V. program
(X = 1) or does not recall (X = 0)
 Example 3
xp
– In a process for manufacturing spoons each spoon may
either be defective(X
X = = 1) or not (X = 0).
1

X= 
0
for a success,
for a
failure.
And, the probability distribution is
X
1
0
P(X = x)
p
1-p
1
x
2
P( X  x) 
x 0
 Mathematically, the Bernoulli distribution is given by
P(X = x) = px ( 1- p)1 - x for x = 1,0
 To calculate the mean and variance,
1
Mean,  = E(X) =
 x  P( X  x)
x 0
1
 Variance,
2
=
E(X2)
-=
2
x
 P( X  x) 
x 0
 Note: Since Bernoulli distribution is determined by the value of p, p is
the parameter of this distribution
6.2 THE BINOMIAL DISTRIBUTION
 6.2.1 The Probability Functions
– Consider an experiment which has two possible
outcomes, one which may be termed ‘success’ and the
other ‘failure’. A binomial situation arises when n
independent trials of the experiment are performed , for
example
– Toss a coin 6 times;
– Consider obtaining a head on a single toss as a success
and obtaining a tail as a failure;
– Throw a die 10 times;
– Consider obtaining a 6 on a single throw as a success,
and not obtaining a 6 as a failure.
 Example
 A coin a biased so that the probability of obtaining a
head is . The coin is tossed four times. Find the
probability of obtaining exactly two heads.
 Example
 An ordinary die is thrown seven times. Find the
probability of obtaining exactly three sixes.
 Example
 The probability that a marksman hits a target is p and
the probability that he misses is q, where q = 1 – p.
Write an expression for the probability that, in 10
shots, he hits the target 6 times.
If the probability that an experiment results in a successful
outcome is p and the probability that the outcome is a failure
is q, where q = 1 – p, and if X is the random variable ‘the
number of successful outcomes in n independent trials’, then
the probability function of X is given by
P(X = x) = for x = 0,1,2,…,n
 Example
 If p is the probability of success and q = 1- p is the probability of
failure, find the probability of 0,1,2,…,5 successes in 5
independent trials of the experiment. Comment your answer.
In general:
The values P(X =x) for x = 0,1,…,n can be obtained by considering
the terms in the binomial expansion of (q + p)n, noting that q + p = 1
C0 q p 0  C1n q n1 p1  C2n q n2 p 2  ...  Crn q nr p r  ...  Cnn q 0 p n
n =n
(q + p)n
1 = P(X = 0) + P(X =1) + P(X = 2) +…+ P(X= r) +….+P(X = n)
 If X is distribution in this way, we write
X  Bin(n,p) where n is the number of independent trials
and p is the probability of a successful outcome in one trial
– n and p are called the parameters of the
distribution.
 Sometimes, we will use b(x; n,p) to
represent the probability function when
n n x x
XBin(n,p). i.e. b(x; n,p) = P(X = x) = C x q p
 Example
– The probability that a person supports Party A is 0.6. Find the
probability that in a randomly selected sample of 8 voters there are
(a) exactly 3 who support Party A,
(b) more than 5 who support Party A.
 Example
– A box contains a large number of red and yellow tulip bulbs in the
ratio 1:3. Bulbs are picked at random from the box. How many
bulbs must be picked so that the probability that there is at least
one red tulip bulb among them is greater than 0.95?
 6.2.2 Mean and Variance of the Binomial
Distribution
–
 = np
2 = npq
– Proved it by yourself. :^)
 Example
– If the probability that it is find day is 0.4, find the
expected number of find days in a week, and the
standard deviation.
 Example
– The random variable X is such that X  Bin(n,p) and
E(X) = 2, Var(X) = . Find the values of n and p, and
P(X = 2).
 6.2.3 Application
 C.W. Applications of Binomial distributions
Throughout this unit, daily lift examples and discussions are the
essential features.
– Binomial Distribution
Q1The probability that a salesperson will sell a magzine subscription to
someone who has been randomly selected from the telephone
directory is 0.1. If the salesperson calls 6 individuals this evening, what
is the probability that
(I) There will be no subscriptions will be sold?
(II) Exactly 3 subscriptions will be sold?
(III) At least 3 subscriptions will be sold?
(IV) At most 3 subscriptions will be sold?
6.3 THE GEOMETRIC
DISTRIBUTION
 6.3.1
The Probability Function
 A geometric distribution arises when we have a sequence of
independent trials, each with a definite probability p of success and
probability q of failure, where q = 1 – p. Let X be the random variable
‘the number of trials up to and including the first success’.
 Now,

P(X = 1) = P(success on the first trial) = p

P(X = 2) = P(failure on first trial, success on second) = q p

P(X= 3) = ___________________________

P(X = 4) = __________________________

….

….

P(X = x ) = __________________________
P(X = x) = qx – 1 p,
x = 1,2,3,……
where q = 1 – p.
 p is the parameter of the distribution.
 If X is defined in this way, we write
X  Geo(p)
 6.3.2 Mean and Variance
 = 1 and 2 = q
p2
p
 Proved it by yourself. 
 Example
– The probability that a marksman hits the bull’s
eye is 0.4 for each shot, and each shot is
independent of all others. Find
 the probability that he hits the bull’s eye for the first
time on his fourth attempt,
 the mean number of throws needed to hit the bull’s
eye, and the standard deviation,
 the most common number of throws until he hits the
bull’s eye.
 Example
– A coin is biased so that the probability of
obtaining a head is 0.6. If X is the random
variable ‘the number of tosses up to and
including the first head’, find
 P(X  4),
 P(X > 5),
 The probability that more than 8 tosses will be
required to obtain a head, given the more than 5
tossed are required.
 Example
– In a particular board game a player can get out
of jail only by obtaining two heads when she
tosses two coins.
 Find the probability that more than 6 attempts are
needed to get out of jail.
 What is the smallest value of n if there is to be at
least a 90% chance of getting out of jail on or before
the n th attempt.
 C.W. Application of Geometric Distribution
– 1)The probability that a student will pass a test on any
trial is 0.6. What is the probability that he will eventually
pass the test on the second trial?
– 2)Suppose the probability that Hong Kong Observatory
will make correct daily whether forecasts is 0.8. In the
coming days, what is the probability that it will make the
first correct forecast on the fourth day?
6.4 THE POISSON DISTRIBUTION
 [see textbook]
 Example

Verify that if XPo(), then X is a random variable.
 Example

If XPo() find (a) E(X), (b) E(X2), (c) Var(X).
 From above example , we can conclude that the MEAN
and VARIANCE of the Poisson distribution are  and 
respectively.
C.W. Application of Poisson
Distribution
–
1)The average number of claims per day
made to the Insurance Company for damage
or losses is 3.1. What is the probability that in
any given day




fewer than 2 claims will be made?
exactly 2 claims will be made?
2 or more claims will be made?
more than 2 claims will be made?
– 2) Based on past experience, 1% of the
telephone bills mailed to house-holds in Hong
Kong are incorrect. If a sample of 10 bills is
selected, find the probability that at least one bill
will be incorrect. Do this using two probability
distributions (the binomial and the Poisson) and
briefly compare and explain your result.