Binomial probability model describes the number of successes in a specified number of trials.

You need:* 2 parameters (success, failure)* Number of trials, n* probability of success, p

The Binomial Probability Model

Example:A cereal manufacturer puts cards of 3 famous athletes in its cereal boxes. 20% of the boxes contain pictures of Derek Jeter, 30% contain a picture of David Beckham, and the rest contain a picture of Serena Williams.

Suppose you want to know what the probability is of getting 2 Derek Jeter cards if you buy 5 boxes. Could you use a binomial probability model?

We can use a binomial probability model because:1 – there are two outcomes (success – getting a Derek Jeter card, failure – not getting a Derek Jeter card.2 – there is a set number of trials (n = 5)3 – we know the probability of success (p = .2)

Using the model:

Let X = the random variableLet n = 5Let p = .2 (the probability of 1 success)Let k = 2 (the number of successes)

2 successes in 5 trials means 2 successes & 3 failures.

The possible order in which the outcomes can occur are disjoint (e.g., if 2 successes came in the first 2 trials, they couldn’t come in the last 2).

Each different order in which we can have k successes in n trials is called a combination. It can be represented by nCk (n choose k) or .

k

n

To figure out the number of combinations in our trial:

10220

1231212345

)!(!!

knk

nk

n

This means there are 10 ways to get 2 Derek Jeter pictures in 5 boxes.

Next step

2048.)8(.)2(.10)2( 32 XPNbr of successes

Nbr of combinations

Probability of success k

Probability of failure

(n- k)

The probability of success in getting 2 Derek Jeter cards in 5 boxes of cereal is .2048.

Another example: The count of X children with type O blood among 5 children whose parents carry genes for both the O and the A blood types is B(5, 0.25). Find P(X=3).

Note: B means binomial setting, n = 5, p = 0.25

knk ppk

nXP

)1()3(

23 )25.1(25.3

5)3(

XP

23 )75(.)25)(.20()3( XP0879.0)3( XP

Note: 0! = 1

One step further:

Suppose the number of electrical switches that fail inspection is B(10, .1). What is the probability that no more than 1 switch fails.

7361.)1(

3487.3874.)1(

)3487)(.1)(1()3874)(.1)(.10()1(

)9(.)1(.0

10)9(.)1(.

1

10)1(

)0()1()1(

10091

XP

XP

XP

XP

XPXPXP

µ = np

σ )1( pnp

These formulas are only good for binomial distributions. They can’t be used for other discrete random variables.

Example:Using the previous cereal example, in 100 boxes of cereal, how many Derek Jeter cards do you expect to find?

Step 1: E(x)= np = (100)(.2) = 20Step 2: Step 3: Summary: In 100 cereal boxes, we expect to find 20 Derek Jeter cards, with a standard deviation of 4 cards.

)8)(.2)(.100()1( pnp 4

Rule of thumb: If np ≥ 10 and n(1-p) ≥ 10, Normal approximation can be used.

Ex: (from your book page 527) A survey asked a nationwide random sample of 2500 adults if they agreed or disagreed that “I like buying clothes but shopping is often frustrating and time consuming.” Suppose that in fact 60% of all adult U.S. residents would say “Agree” if asked the same question. What is the probabiltiy that 1520 or more would agree?

Since both criteria for Normal Approximation are satisfied, we can use Normal distribution calculations.

µ = (2500)(.6) = 1500

σ = = 24.49)4)(.6)(.2500(

2061.7939.01)82.()1520(

49.2415001520

49.241500

)1520(

)49.24,1500(

ZPXP

XPXP

N

Using the calculator1) Probability distribution function – given a discrete random variable X, the probability distribution function assigns a probability to each value of X. See page 520, example 8.7 for binompdf(n,p,x)2) Cumulative distribution function – given a random variable X, the cumulative distribution function of X calculates the sum of probability of obtaining at most X successes in n trials. See page 522 example 8.10 for binomcdf(n,p,x)3) Look at the Technology Tool Box pages 530 - 532