Some Common Discrete Random Variables. Binomial Random Variables.

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 F

P 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

S

SF

FS

….

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 p

q 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

D

DG

G

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 Y

p 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 Y

p 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( ) 40

0.1E 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 y

y

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) 0.13385

4!

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 Y

N 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).