some common discrete random variables
DESCRIPTION
Some Common Discrete Random Variables. Binomial Random Variables. Binomial experiment. A sequence of n trials ( called Bernoulli trials ), each of which results in either a “success” or a “failure”. - PowerPoint PPT PresentationTRANSCRIPT
Some Common Discrete Random Variables
Binomial Random Variables
Binomial experiment
• A sequence of n trials (called Bernoulli trials), each of which results in either a “success” or a “failure”.
• 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 2 8 9 10 11 12
1 2 8 9 12
8 4
(( , , , , , , , , , , , ))( )
( ) ( ) ( ) ( ) ( )
(0.65) (1 0.65)
P S S S S S S S S F F F FP R R R R R R R
P R P R P R P R P R
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 4 128 4 8, or simply C C C
• Each of these sample points has the same probability. • Hence, summing these probabilities yields
12 8 48
P(8 successes in trials)
= (0.65) (0.35) 0.237
n
C
Binomial Probability Function
• A random variable has a binomial distribution with parameters n and p if its probability function is given by
p( ) (1 )n y n yyy C p 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
• As a result, summing the binomial probabilities, where q = 1- p is the probability of a failure,
0
( )n
n n y n yy
y
p q C p q
0
( ) (1 ) ( (1 )) 1n
n y n y ny
y y
P Y y C p p p p
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
( ) and ( ) (1 )E Y n p V Y n p 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 1st Success
• Similar to the binomial experiment, we consider:• A sequence of independent Bernoulli trials.• The probability of “success” equals p on each trial.• Define a random variable Y as the number of the
trial on which the 1st 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)y = 3: (F, F, S)y = 4: (F, F, F, S)and so on…
S
SS
FF
S
….
F
(F, S)
(F, F, S)
(S)
(F, F, F, S)p(1) = pp(2) = (q)( p)p(3) = (q2)( p)p(4) = (q3)( p)
Geometric Probability Function
• A random variable has a geometric distribution with parameter p if its probability function is given by
1 p( ) where 1 , for 1,2,...
yy q pq p y
Success?
• Of course, you need to be clear on what you consider a “success”.
• For example, the 1st success might mean finding the 1st defective item!
D
DD
GG
G
(G, D)
(G, G, D)
(D)
Geometric Mean, Variance
• If Y is a geometric random variable with parameter p the expected value and variance for Y are given by
2
1 1( ) and ( ) pE Y V Yp p
At least ‘a’ trials? (#3.55)
• For a geometric random variable and a > 0,show P(Y > a) = qa
• Consider P(Y > a) = 1 – P(Y < a)
= 1 – p(1 + q + q2 + …+ qa-1) = qa , based on the sum of a
geometric series
“Memoryless Property”• For the geometric distribution
P(Y > a + b | Y > a ) = qb = P(Y > b)• “at least 5 more trials?”
We note P(Y > 7 | Y > 2 ) = q5 = 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 1st success, let Y be the number of the trial on which the rth success occurs. (Stop the trials after the rth success occurs.)
• For a given value r, the probability p(y) is
1, 1( ) (1 ) , , 1,...r y ry rp y C p p y r r
Negative Binomial
• To determine the probability the 4th success occurs on the 7th trial, we compute
4 36,3(7) (1 )p C p p
• Note this is actually just the binomial probability of 3 successes during the first 6 trials, followed by one more success:
3 36,3(7) (1 )p C p p p
“a success on 4th last trial”
Negative Binomial
• For the negative binomial distribution, we have
2
(1 )( ) and ( )r r pE Y V Yp p
• For example, if a success occurs 10% of the time (i.e., p = 0.1), then to find the 4th success, we expect to require 40 trials on average.
4( ) 400.1
E Y
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 if its probability function is given by
( )!
yep yy
where y = 0, 1, 2, …
We’ll see that is the “average rate” at which the events occur. That is, E(Y) = .
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.
4 66(4) 0.133854!ep
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)P Y P Y 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 the expected value and variance for Y are given by
( ) and ( )E Y V Y
Hypergeometric Random Variables
Sampling without replacement• When sampling with replacement, each trial
remains independent. For example,…• If balls are replaced, P(red ball on 2nd draw) =
P(red ball on 2nd draw | first ball was red).
Though for a large population of balls, the effect may be minimal.
• If balls not replaced, then given the first ball is red, there is less chance of a red ball on the 2nd draw.
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 N ry n y
Nn
C Cp y
C
Hypergeometric R. V.• A random variable has a hypergeometric
distribution with parameters N, n, and r if its probability function is given by
( )r N ry n y
Nn
C Cp y
C
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
( ) and ( )1
nr nr N r N nE Y V YN N N N
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).