Transcript PowerPoint: Familiar Discrete Distributions.
Some Common Discrete Random Variables
Binomial Random Variables
Binomial experiment
• A sequence of each of which results in either a “success” or a “failure”.
n
trials (
called Bernoulli trials
), • The trials are independent and so the probability of success,
p
, remains the same for each trial.
• Define a random variable Y as the number of successes observed during the
n
trials.
• What is the probability p(
y
), for
y
= 0, 1, …,
n
?
• How many successes may we expect? E(Y) = ?
Returning Students
• Suppose the retention rate for a school indicates the probability a freshman returns for their sophmore year is 0.65. Among 12 randomly selected freshman, what is the probability 8 of them return to school next year?
Each student either returns or doesn’t. Think of each selected student as a trial, so
n
= 12.
If we consider “student returns” to be a success, then
p
= 0.65.
12 trials, 8 successes
• To find the probability of this event, consider the probability for just one sample point in the event.
• For example, the probability the first 8 students return and the last 4 don’t.
• Since independent, we just multiply the probabilities: ( 1
R
2 ( 1 ) ( 8 2 ) 4 (
R
8 8
R
9 ) ( 9 )
R
10
R
11
R
12 ) ( 12 )
12 trials, 8 successes
• For the probability of this event, we sum the probabilities for each sample point in the event.
• How many sample points are in this event?
• How many ways can 8 successes and 4 failures occur?
12
C C
8 4 4 , or simply
C
12 8 • Each of these sample points has the same probability. • Hence, summing these probabilities yields =
C
12 8 8 (0.65) (0.35) 4 0.237
Binomial Probability Function
• A random variable has a binomial distribution with parameters
n
and
p
if its probability function is given by
n
C p
y y
(1
p
)
Rats!
• In a research study, rats are injected with a drug. The probability that a rat will die from the drug before the experiment is over is 0.16. Ten rats are injected with the drug.
What is the probability that at least 8 will survive? Would you be surprised if at least 5 died during the experiment?
Quality Control
• For parts machined by a particular lathe, on average, 95% of the parts are within the acceptable tolerance.
• If 20 parts are checked, what is the probability that at least 18 are acceptable?
• If 20 parts are checked, what is the probability that at most 18 are acceptable?
Binomial Theorem
• As we saw in our Discrete class, the Binomial Theorem allows us to expand (
p
q
)
n
y n
0
n y y
C p q
• As a result, summing the binomial probabilities, where q = 1- p is the probability of a failure,
y
y
)
y n
0
n C p y y
(1
p
) (
p p
))
n
1
Mean and Variance
• If Y is a binomial random variable with parameters
n
and
p
, the expected value and variance for Y are given by
n p
n p
(1
p
)
Rats!
• In a research study, rats are injected with a drug. The probability that a rat will die from the drug before the experiment is over is 0.16. Ten rats are injected with the drug.
• How many of the rats are expected to survive?
• Find the variance for the number of survivors.
Geometric Random Variables
Your 1
st
Success
• Similar to the binomial experiment, we consider: • A sequence of • The independent Bernoulli trials.
probability of “success” equals
p
on each trial.
• Define a random variable Y as the number of the trial on which the 1 st success occurs. (
Stop the trials after the first success occurs
.) • What is the probability p(
y
), for
y
= 1,2, … ?
• On which trial is the first success expected?
S = success
• Consider the values of Y: y = 1: (S) y = 2: (F, S) S y = 3: (F, F, S) y = 4: (F, F, F, S) and so on… F p(1) =
p
p(2) = (
q
)(
p
) p(3) = (
q
2 )(
p
) p(4) = (
q
3 )(
p
) (S) S F (F, S) (F, F, S) S S (F, F, F, S) F ….
Geometric Probability Function
• A random variable has a geometric distribution with parameter
p
if its probability function is given by
where
q
q y
1
p p
, for
y
1, 2,...
Success?
• Of course, you need to be clear on what you consider a “success”. • For example, the 1 st success might mean finding the 1 st defective item!
(D) D (G, D) D G (G, G, D) D G G
Geometric Mean, Variance
• If Y is a geometric random variable with parameter
p
the expected value and variance for Y are given by
1
p
1
p
2
p
At least ‘
a’
trials? (#3.55)
• For a geometric random variable and
a
> 0, show P(Y >
a
) =
q a
• Consider P(Y >
a
) = 1 – P(Y < a) = 1 – p(1 + q + q 2 + …+ q a-1 ) = q a , based on the sum of a geometric series
“Memoryless Property”
• • For the geometric distribution P(Y > a + b | Y > a ) = q b = P(Y > b)
“at least 5 more trials?”
We note P(Y > 7 | Y > 2 ) = q 5 = P(Y > 5).
That is, “knowing the first two trials were failures, the probability a success won’t occur on the next 5 trials” is identical to… “just starting the trials and a success won’t occur on the first 5 trials”
Negative Binomial Distribution
• Again, considering a independent Bernoulli trials with probability of “success”
p
on each trial… • Instead of watching for the 1 st success, let Y be the number of the trial on which the
r th
success occurs. (
Stop the trials after the r th success occurs
.) • For a given value
r
, the probability p(
y
) is
C y
1,
r
1
p r
(1
p
) ,
y
1,...
Negative Binomial
• To determine the probability the 4 th on the 7 th trial, we compute success occurs
p
(7)
C p
6,3 4 (1
p
) 3 • Note this is actually just the binomial probability of 3 successes during the first 6 trials, followed by one more success:
p
(7)
C p
6,3 3 (1
p
) 3 “a
success on 4 th last trial
”
Negative Binomial
• For the negative binomial distribution, we have
r
r
(1
p
)
p p
2 • For example, if a success occurs 10% of the time (i.e., p = 0.1), then to find the 4 th success, we expect to require 40 trials
on average
.
4 0.1
40
Intuitively, wouldn’t you expect 40 trials?
Poisson Random Variables
Number of occurrences
• Let Y represent the number of occurrences of an event in an interval of size
s
.
• Here we may be referring to an interval of time, distance, space, etc.
• For example, we may be interested in the number of customers Y arriving during a given time interval.
• We call Y a Poisson random variable.
Poisson R. V.
• A random variable has a Poisson distribution with parameter l if its probability function is given by l
y e
l
y
!
where
y
= 0, 1, 2, … We’ll see that l is the “average rate” at which the events occur. That is, E(Y) = l .
Queries
• If the number of database queries processed by a computer in a time interval is a Poisson random variable with an average of 6 queries per minute, find the probability that 4 queries occur in a one minute interval.
p
(4) 6 4
e
6 4!
0.13385
Fewer Queries
• As before, for the Poisson random variable with an average of 6 queries per minute… • find the probability there are less than 6 queries in a one minute interval: 6) 5) poissoncdf (6,5) 0.44568
Some PoissonVariables
• Number of incoming telephone calls to a switchboard within a given time interval; • Number of errors (incorrect bits) received by a modem during a given time interval; • Number of chocolate chips in one of Dr. Vestal’s chocolate chip cookies; • Number of claims processed by a particular insurance company on a single day; • Number of white blood cells in a drop of blood; • Number of dead deer along a mile of highway.
Poisson mean, variance
• If Y is a Poisson random variable with parameter l, the expected value and variance for Y are given by l l
Hypergeometric Random Variables
Sampling without replacement
• When sampling with replacement, each trial remains independent. For example,… • If balls are replaced, P(red ball on 2 nd draw) = P(red ball on 2 nd draw | first ball was red).
• If balls not replaced, then given the first ball is red, there is less chance of a red ball on the 2 nd draw.
Though for a large population of balls, the effect may be minimal.
n
trials,
y
red balls
• Suppose there are
r
red balls, and
N – r
other balls.
• Consider Y, the number of red balls in
n
selections, where now the trials may be dependent.
(for sampling without replacement, when sample size is significant relative to the population)
• The probability
y
of the
n
selected balls are red is
r C C y N r n y C n N
Hypergeometric R. V.
• A random variable has a hypergeometric distribution with parameters
N, n,
and
r
if its probability function is given by
r C C y N r n y C n N
where 0 <
y
< min(
n, r
).
Hypergeometric mean, variance
• If Y is a hypergeometric random variable with parameter
p
the expected value and variance for Y are given by
n r N
n r N
N
N r N N
n
1
Sample of 20
Suppose among a supply of 5000 parts produced during a given week, there are 100 that don’t meet the required quality standard. Twenty of the parts are randomly selected and checked to see if they meet the standard. Let Y be the number in the sample that don’t meet the standard.
a). Compute the probability exactly 2 of the sampled parts fail to meet the quality standard.
b). Determine the mean, E(Y).