More Probability

The Binomial and Geometric Distributions

What you’ll learn

• The conditions for a binomial setting

• Calculating binomial probabilities

• How to find the mean and standard deviation of a binomial setting

• The conditions for a geometric setting

• Calculating geometric probabilities

• How to find the mean of a geometric setting

Binomial vs Geometric

• In this lesson we will learn how to determine if a setting is Binomial or Geometric and how to use the correct distribution to calculate probabilities

• Binomial: A binomial setting looks at a fixed number of independent observations and counts the number of successes.

• Geometric: A geometric setting counts the number of observations until a success is observed. This distribution is sometimes called a waiting distribution.

The Quiz

• Consider the following: On a multiple-choice quiz, Joan guesses on

each of five questions. There are four possible answers for each question. Assuming that Joan did not study and must guess on each question,

What is the probability that she will get exactly 2 questions correct?

What is the probability that Joan passes the quiz?

• We can model this type of setting with a mathematical model called a Binomial Model.

Conditions for a Binomial Setting• A setting follows a binomial model if the

following is true:Fixed number of observations

Independent observations

Two possible outcomes (success/failure)

Same probability for success on each outcome

• So if the setting FITS, we can use a binomial model when finding probabilities.

What about our Quiz?• Let’s check the conditions

F There are 5 questions on our quiz, n=5

I We are assuming that Joan did not study and does not learn from each

question, so observations are independent

T There are two possible outcomes,

– Answers correctly…….success

– Answers incorrectly…..failure

S Since there are four possibilities with only 1 correct answer, the

probability of success is p = .25

What about our Quiz?• Let’s check the conditions

F There are 5 questions on our quiz, n=5

I We are assuming that Joan did not study and does not learn from each

question, so observations are independent

T There are two possible outcomes,

– Answers correctly…….success

– Answers incorrectly…..failure

S Since there are four possibilities with only 1 correct answer, the

probability of success is p = .25

Calculating probabilities• So how can we use the binomial model to

calculate probabilities? Let’s first consider the question:

• What is the probability that Joan answers 2 questions correctly ---- P(X=2)

Now Joan getting 2 questions correct can happen in several ways.

CCWWW WCCWW WWCCW WWWCC

CWCWW WCWCW WWCWC

CWWCW WCWWC

CWWWC

• In fact, there are 5C2 possibilites

• So what is 5C2 and how do we calculate it?

5C2 is short-hand notation for the number of ways we can choose two items from a group of 5.

The formula for calculating this number is

So:

Notice that the (n-k)! cancels with factors in the numerator. (That always happens—go ahead try it out!)

Finding combinations

)!(!

!

knk

nCkn

)123(12

1234525

C

• So what is 5C2 and how do we calculate it?

5C2 is short-hand notation for the number of ways we can choose two items from a group of 5.

The formula for calculating this number is

So:

Notice that the (n-k)! Cancels with factors in the numerator. (That always happens—go ahead try it out!)

Finding combinations

)!(!

!

knk

nCkn

)123(12

1234525

C

Finding Combinations

• Since that cancellation happens every time, we can short-cut our formula in the following way----

12

4525

C

)123(12

1234525

C

Notice that what is left in the numerator is 5*4. We can think about the numerator as starting at “n” and multiplying down “k” factors.

The denominator, after cancellations, consists of 2*1 which is simply “k!”

Finding Combinations on the TI

• To determine the number of combinations using your TI-83/84 use the following sequence of commands: “n” Math Prob 3: nCr

“k” Note: TIuses “r” insteadof “k”

Finding a probability• So, we have determined that on a 5 question quiz,

there are 10 ways in which Joan can get 2 problems correct.CCWWW WCCWW WWCCW WWWCC

CWCWW WCWCW WWCWC

CWWCW WCWWC

CWWWC

• Since multiplication is commutative, each of these combinations has the same probability Two successes p=.25 Three failures p=1-.25 = .75

• So the probability for each of these combinations is.252 (.75)3 = .02637

• And since there are ten of these, the probability that Joan gets exactly 2 questions correct is

10(.25)2(.75)3= .2637 or approx 26.37%

So what is the formula?• Let’s look at that probability again:

P(X=2) = 10 (.25)2 (.75)3

• Now let’s turn each piece into a formula:

• 10 -> is the number of ways the event can happen, i.e. the number of combinations (5C2)

• (.25)2 -> probability of a success raises to the number of success in our observations (p)2

• (.75) -> (1- probability of a success) raised to the number of failures in our observations (1-p)(5-2)

P(X=k) = nCk(p)k(1-p)(n-k)

What about multiple possibilities?

• What about our question: What is the probability that Joan passes the class?

• Of course this happens if she get 3,4, or 5 questions correct.

• Of course we could set up a probability distribution for each possibility and then use what we know about probability distributions to find the answer to that question

The Probability Distribution• Let X = the number of questions Joan guesses

correctly.

• Then X takes on the values 0,1,2,3,4,5

• We can then find the probability for each of these values using our binomial formula. P(X=0) = 5C0 (.25)0 (1-.25) (5-0) = .2373

P(X=0) = 5C1 (.25)1 (1-.25) (5-1) = .3955

P(X=0) = 5C0 (.25)0 (1-.25) (5-2) = .2637

P(X=0) = 5C0 (.25)0 (1-.25) (5-3) = .0879

P(X=0) = 5C0 (.25)0 (1-.25) (5-4) = .0147

P(X=0) = 5C0 (.25)0 (1-.25) (5-5) = .00098

X 0 1 2 3 4 5

P(x) .2373 .3955 .2637 .0879 .0147

.00098

Back to Multiple Possibilities• What is the probability that Joan passes the

test?• P(X ≥ 3) = P(X = 3) + P(X = 4) + P(X = 5)

• Now from our distribution• P(X≥3) = .0879 + .0147 + .00098 = .0135

X 0 1 2 3 4 5

P(x) .2373 .3955 .2637 .0879 .0147

.00098

Ok, so much for variables that have a small number of possibilities….

But what if our event is looking a large number of observations?

Binomial Probabilities and the TI

• Let’s find out how to use the calculator for our probabilities.

• We can use the built in binomial distribution to find probabilities.

• Let’s look at our first question again. P(X=2)

P(X=2) TI Style

• 2nd VARS (this is the distribution menu)

• 0:binompdf(n,p,k)• This distribution answers the

Questions P(X=k)

• n=5 (the number of questions)

• p=.25 (probability of success)

• k=2(the value we are interested in)

Cumulative Probabilities on the TI

• What about probabilities for more than 1 possibility, P(X ≤ 2 ) (This is the probability that Joan fails the quiz)

• 2nd VARS

• A:binomcdf(n,p,k) (this answers the question P(X ≤ k)

We can verify from our table that this is in fact the P(X≤ 2) = P(X=0) + P(X=1) + P(X=2)

= .2373 + .3955 + .2637

= .8965

What about “greater than”

• So, what about the probability that Joan passes the quiz? P(X ≥ 3)

• Since our calculator only finds probabilities “less than” we first need to change the probability and use complements P(X ≥ 3) = 1 – P(X ≤ 2)

Mean and Standard Deviation of the Binomial Distribution

• We can find the mean and standard deviation of a binomial distribution in the same way we found means and standard deviations of any other discrete distribution.

• However, we can also find the mean of a binomial distribution by

• μ(x) = np 5(.25) = 1.25– So, we would expect, on average, for Joan to answer 1.25

questions correctly.

• We can find the standard deviation

• σ(x) = 9682.)25.1)(25(.5)1( pnp

Summarizing the Binomial Distribution

• Remember to be a binomial setting, it must

–FITS

• To find individual probabilities use:• P(X=k) = nCk(p)k(1-p)(n-k)

• The mean of a binomial setting:• μ(x) = np

• The standard deviation of a binomial

• σ(x) = )1( pnp

Geometric Distribution

• Now let’s move on to our Geometric (or waiting distribution).

• The initial setting may be the same On a multiple-choice quiz, Joan guesses on

each of five questions. There are four possible answers for each question. Assuming that Joan did not study and must guess on each question,

• It’s the question that is different. What is the probability that the first

correct answer Joan gets is on the 3rd question?

Conditions for a Geometric Distribution

• A setting follows a geometric model if the following is true:Waiting for a success

Independent observations

Two possible outcomes (success/failure)

Same probability for success on each outcome

• So if the setting WITS, we can use a binomial model when finding probabilities.

What about our Quiz?• Let’s check the conditions

W We are waiting for the first correct answer

I We are assuming that Joan did not study and does not learn from each

question, so observations are independent

T There are two possible outcomes, – Answers correctly…….success– Answers incorrectly…..failure

S Since there are four possibilities with only 1 correct answer, the

probability of success is p = .25

Finding Geometric Probabilities• Since our conditions have been met, we can

use a geometric model to find probabilities.

• Our question: What is the probability that the first correct answer is the 3rd question?

• That means our sequence is– WWC

• Now the probability that the answer is correct is p=.25 and the probability that the answer is wrong is (1-.25)=.75

• So, our probability is .75(.75)(.25) =.752(.25) = .1406

So, the formula is……

• Let’s find a general formula for the geometric distribution.

• Now for this type of distribution, there will always be 1 less failure than the number, k, we are interested in. i.e.---if we want the probability that the

first success comes on the 5th question, that means we have had 4 failures.

• In general, where k is the first success, p=probability of a success… P(X=k) = (1-p)(k-1)(p)

Geometric on the TI

• Like the binomial, the TI-83/84 has built-in distributions geometric settings.

• 2nd VARS (this is the distribution menu)

• D: geometpdf(p,k)• This distribution answers the

Question P(X=k)

• p=.25 (probability of success)

• k=3(the value we are interested in)

Multiple Probabilities TI-style

• 2nd VARS

• E: geometcdf(p,k) (this answers the question P(X ≤ k)

• So, for P(X ≤ 3)

Multiple Probabilities

• What if our question was: What is the probability that the first correct

answer is one of the first 3 questions?

• P(X≤ 3)= P(X=1) + P(X=2) + P(X=3)

= .750 (.25) + .751 (.25) + .752 (.25)

= .25 + .1875 + .1406

= .5781

And Greater than????

• Just like for binomial, we want to turn a greater than question into 1 – “less than or equal to”

• For example, what is the probability that it takes more than 2 questions to get a correct answer?

• P(x > 2) = 1- P(X ≤ 2), then with technology

Note: be careful when rewriting.

P(X > k) 1 – P(X ≤ k)

P(X ≥ k) 1 – P (X ≤ (k-1))

What about the Mean?

• To find the average number of observations it will take to get 1 success use the following:

Μ(x) =

p

1

Note:

We do not calculate a standard deviation for a geometric distribution.

Binomial vs Geometric

• Binomial: This is a counting

distribution. We have a fixed number of observations and are counting the number of successes

• P(X=k)=>n Ck pk(1-p)(n-k)

• M(x) = np

• σ(x) = √(np(1-p))

• Geometric This is a waiting

distribution. We are waiting for the first success.

P(X=k) => (1-p)(k-

1)p M(x) => 1/p

Additional Resources

• The Practice of Statistics—YMM Pg 415 - 452

• The Practice of Statistics—YMS Pg 438 - 483