Statistics 17 Probability Models

Bernoulli Trials The basis for the probability models we will examine in this chapter is the Bernoulli trial. We have Bernoulli trials if: –there are two possible outcomes (success and failure). –the probability of success, p, is constant. –the trials are independent.

Bernoulli Trials

Binomial Variables

The Binomial Model A Binomial model tells us the probability for a random variable that counts the number of successes in a fixed number of Bernoulli trials. Two parameters define the Binomial model: n, the number of trials; and, p, the probability of success. We denote this Binom(n, p).

The Binomial Model In n trials, there are ways to have k successes. –Read n C k as “n choose k.” Note: n! = n  (n – 1)  …  2  1, and n! is read as “n factorial.”

The Binomial Model Binomial probability model for Bernoulli trials: Binom(n,p) n = number of trials p = probability of success q = 1 – p = probability of failure X = # of successes in n trials P(X = x) = n C x p x q n–x

Example

Binomial Coefficient

Binomial Probability

Binomial Distribution

Example

Calculators

The Geometric Model A single Bernoulli trial is usually not all that interesting. A Geometric probability model tells us the probability for a random variable that counts the number of Bernoulli trials until the first success. Geometric models are completely specified by one parameter, p, the probability of success, and are denoted Geom(p).

The Geometric Model Geometric probability model for Bernoulli trials: Geom(p) p = probability of success q = 1 – p = probability of failure X = number of trials until the first success occurs P(X = x) = q x-1 p

Independence One of the important requirements for Bernoulli trials is that the trials be independent. When we don’t have an infinite population, the trials are not independent. But, there is a rule that allows us to pretend we have independent trials: –The 10% condition: Bernoulli trials must be independent. If that assumption is violated, it is still okay to proceed as long as the sample is smaller than 10% of the population.

Practice A new sales gimmick has 30% of the M&M’s covered with speckles. These “groovy” candies are mixed randomly with the normal candies as they are put into the bags for distribution and sale. You buy a bag and remove candies one at a time looking for the speckles.

Practice A new sales gimmick has 30% of the M&M’s covered with speckles. These “groovy” candies are mixed randomly with the normal candies as they are put into the bags for distribution and sale. You buy a bag and remove candies one at a time looking for the speckles. Does this situation involve Bernoulli trials?

Practice Does this situation involve Bernoulli trials? There are two outcomes: success = speckles, failure = ordinary. The probability of success, based on the information from the candy company, is 30%. Trials can be assumed independent – there is no reason to believe that finding a speckled candy reveals anything about whether the next one out of the bag will have speckles. This situation involves Bernoulli trials

Practice What’s the probability that the first speckled one we see is the fourth candy we get? P(First speckled is fourth candy) = (0.7) 3 (0.3) =

Practice What’s the probability that the first speckled one is the tenth one? P(First speckled is tenth candy) = (0.7) 10 (0.3) ≈ General Formula: P(X = x) = q x−1 p

Practice What’s the probability that the first speckled candy is one of the first three we look at? P(First speckled is among first three) = (0.3) + (0.7)(0.3) + (0.7) 2 (0.3) ≈ 0.657

Practice How many do we expect to check, on average to find a speckled one?

Practice What’s the probability that we’ll find two speckled ones in a handful of five candies? Students need to show the set- up on the AP Exam, even though they will use technology to get the answer.

Mean and Standard Deviation

Example

Geometric Probability Model

Binomial or Geometric Number of successes in a fixed number of trials Number of trials required for one success binomPdf( ) binomCdf( ) geometPdf( ) geometCdf( )

Binomial and Geometric

Example

The Normal Model to the Rescue! When dealing with a large number of trials in a Binomial situation, making direct calculations of the probabilities becomes tedious (or outright impossible). Fortunately, the Normal model comes to the rescue…

The Normal Model to the Rescue As long as the Success/Failure Condition holds, we can use the Normal model to approximate Binomial probabilities. –Success/failure condition: A Binomial model is approximately Normal if we expect at least 10 successes and 10 failures: np ≥ 10 and nq ≥ 10

Continuous Random Variables When we use the Normal model to approximate the Binomial model, we are using a continuous random variable to approximate a discrete random variable. So, when we use the Normal model, we no longer calculate the probability that the random variable equals a particular value, but only that it lies between two values.

What Can Go Wrong? Be sure you have Bernoulli trials. –You need two outcomes per trial, a constant probability of success, and independence. –Remember that the 10% Condition provides a reasonable substitute for independence. Don’t confuse Geometric and Binomial models. Don’t use the Normal approximation with small n. –You need at least 10 successes and 10 failures to use the Normal approximation.

What have we learned? Bernoulli trials show up in lots of places. Depending on the random variable of interest, we might be dealing with a –Geometric model –Binomial model –Normal model

What have we learned? –Geometric model When we’re interested in the number of Bernoulli trials until the next success. –Binomial model When we’re interested in the number of successes in a certain number of Bernoulli trials. –Normal model To approximate a Binomial model when we expect at least 10 successes and 10 failures.

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