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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 • Independentobservations • 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 • FThere are 5 questions on our quiz, n=5 • IWe are assuming that Joan did not study and does not learn from each question, so observations are independent • TThere are two possible outcomes, • Answers correctly…….success • Answers incorrectly…..failure • SSince 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

Finding combinations • 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 • 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 • Since that cancellation happens every time, we can short-cut our formula in the following way---- 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: TI uses “r” instead of “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

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 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) =

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) =

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 • Independentobservations • 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 • WWe are waiting for the first correct answer • IWe are assuming that Joan did not study and does not learn from each question, so observations are independent • TThere are two possible outcomes, • Answers correctly…….success • Answers incorrectly…..failure • SSince 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) = Note: We do not calculate a standard deviation for a geometric distribution.

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 Binomial vs Geometric

Additional Resources • The Practice of Statistics—YMM • Pg 415 - 452 • The Practice of Statistics—YMS • Pg 438 - 483

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