discrete random variables: the binomial distribution

Discrete Random Variables:

The Binomial Distribution

Bernoulli’s trials

J. Bernoulli (1654-1705) analyzed the idea of repeated independent trials for discrete random variables that had two possible outcomes: success or failure

In his notation he wrote that the probability of success is denoted by p and the probability of failure is denoted by q or 1-p

Binomial distribution The binomial distribution is just n

independent individual (Bernoulli) trials added up.

It is the number of “successes” in n trials. The sum of the probabilities of all the

independent trials totals 1. We can define a ‘success’ as a ‘1’, and a

failure as a ‘0’.

Binomial distribution “x” is a binomial distribution if its probability function is:

Examples (note – success/failure could be switched!):

p

px

10

1 probability of success

probability of failure

Situation x=1

(success)

x=0

(failure)

Coin toss to get heads

Turns up heads Turns up tails

Rolling dice to get 1 Lands on 1 Lands on anything but 1

While testing a product, how many are found defective

Product is defective Product is not defective

Binomial distribution The binomial distribution is just n

independent ‘Bernoulli trials’ added upIt requires that the trials be done “with replacement” ex. testing bulbs for defects:

•Let’s say you make many light bulbs•Pick one at random, test for defect, put it back•Repeat several times•If there are many light bulbs, you do not have to replace (it won’t make a significant difference)•The result will be the binomial probability of a defective bulb (#defective total sample).

Binomial distribution - formula Let’s figure out a binomial random

variable’s probability function or formula Suppose we are looking at a binomial with

n=3 (ex. 3 coin flips; ‘heads’ is a ‘success’) We will start with ‘all tails’ P(x=0):

Can happen only one way: 000Which is: (1-p)(1-p)(1-p) Simplified: (1-p)3

Binomial distribution - formula Let’s figure out a binomial probability

function (for n = 3) This time we want 1 success plus 2

failures (ex. 1 heads + 2 tails, or P(x=1)): This can happen three ways: 100, 010, 001 Which is: p(1-p)(1-p)+(1-p)p(1-p)+(1-p)(1-p)p Simplified: 3p(1-p)2


function (for n = 3) We want 2 ‘successes’ P(x=2):

Can happen three ways: 110, 011, 101, or… pp(1-p)+(1-p)pp+p(1-p)p, which simplifies to.. 3p2(1-p)

Binomial distribution - formula

Let’s figure out a binomial probability function (n = 3)

We want all 3 ‘successes’ P(x=3): This can happen only one way: 111 Which we represent as: ppp Which simplifies to: p3


function – in summary, for n = 3, we have: P(x)

(Where x is the number of successes; ex. # of heads)

3

2

2

3

3

132

131

10

p

pp

pp

p

x

The sum of these expressions is the binomial distribution for n=3. The resulting equation is an example of the Binomial Theorem.

Binomial distribution - formula

A quick review of the Binomial Theorem: If we use q for (1 – p), then…

p3 + 3p2q +3pq2 + q3 = (p + q)3

which is an example of the formula:

(a + b)n = ____________________

(if you forget it, check it in your text)

Binomial distribution - formula Let’s figure out a binomial r.v.’s probability function

(the quick way to compute the sum of the terms on the previous slide) - now here’s the formula…In general, for a binomial:

xnx ppxP 1nscombinatio of #

(the # of x successes with probability p in n trials)

Binomial distribution - formulaWhich can be written as:

xnx ppxnx

nxP

1

!!

!

or P(x) = nCx px(1 – p)n-x

This formula is often called the general term of the binomial distribution.

Expected Value

The expected value of a binomial distribu-tion equals the probability of success (p) for n trials:

npXE

E(X) also equals the sum of the probabilities in the binomial distribution.

Binomial distribution - Graph

Typical shape of a binomial distribution:Symmetric, with total P(x) = 1

x

PNote: this is a theoretical graph – how would an experimental one be different?

Binomial distribution - example

A realtor claims that he ‘closes the deal’ on a house sale 40% of the time.

This month, he closed 1 out of 10 deals. How likely is his claim of 40% if he only

completed 1/10 of his deals this month?

Binomial distribution - example By using the binomial distribution function,

it’s possible to check if his assessment of his abilities (i.e. 40% ‘closes’) is likely:

251.00467.00256.0

234

789106.04.0

!6!4

!104

040.0010.04.0106.04.0!9!1

!101

006.0006.0116.04.0!10!0

!100

64

9

100

P

P

PP(0 deals)=

Binomial distribution - example So it seems pretty unlikely that his assess-

ment of his abilities is right:The probability of closing 1 or fewer

deals out of 10 if (as he claims) he closes deals 40% of the time is less than 5% or less than 1/20.

What % of ‘closes’ do you think would have the highest probability in this distribution, if his claim was right?

Binomial distribution - example

Now see if you can determine the expected number of closings if he had 12 deals this month, assuming 40% success.

We need the values of n (= ___) and of p (= ____). E(X) = np = ______ - this means that we would expect him to close about _____ deals, if his claim is correct. [End of first example.]

Binomial Distribution – ex. 2

Alex Rios has a batting average of 0.310 for the season. In last night’s game, he had 4 at bats. What are the chances he had 2 hits?

You try this one! First ask 3 questions…


Is each time at bat an independent event?

Is ‘getting a hit’ a discrete random variable? Is this a Bernoulli trial? How would you

define a “success” and a “failure”?

If you can answer ‘yes’ to the three questions above, then you can use the binomial distribution formula to answer the problem.


First determine the following values:The number of trials (Alex is at bat __ times) –

this is the value of nThe probability of success (Alex’s average is

___) – this is pThe probability of ‘failure’: 1 – p = ___The # of successes asked for (his chances of

getting ___ hits) – this is x

Now you can use the formula:


xnx ppxnx

nxP

1

!!

!

Put in the values from the previous screen, and discuss your answers. [pause here]

Did you get P(2) = 0.275? Is 2 the most likely number of hits for Alex last night? How about 1 or 3?

P(0 or 1 or 2 or 3 or 4 hits) = _____?

Hypergeometric distribution

What happens if you have a situation in which the trials are not independent (this most often happens due to not replacing a selected item).

Each trial must result in success or failure, but the probability of success changes with each trial.

Hypergeometric distribution Consider taking a sample from a

population, and testing each member of the sample for defects.

Do this sampling without replacement.

As long as the sample is small compared to the population, this is close to binomial. But if the sample is large compared to the population, this is a hypergeometric dist.

Hypergeometric dist. - formula

A hypergeometric distribution differs from binomial ones since it has dependent trials.

Probability of x successes in r dependent trials, with number of successes a out of a total of n possible outcomes:

rn

xranx

C

CCxP

a)(

Hypergeometric dist. - formula

Expected Value – the average probability of a success is the ratio of success overall (a/n) times r trials:

n

raXE

!!!

!!!

!!!

rnrn

xranxnan

xaxa

xP

The full version of this formula is:

discrete random variables: the binomial distribution

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