Discrete Probability Distributions

If you have any test reviews, homeworks, guides, anything school related that you think can be posted on this website, reach out to me at makingschooleasier@gmail.com  

 Discrete Probability Distributions

This chapter is essential for gaining an understanding of how probability can be used to analyze business activities or processes that generate random data.  Probability models assign a probability to each outcome in the sample space defined by a random process. To correctly use them, it is important that you understand their development.

Definitions

Random variable – a function or rule that assigns a numerical value to each outcome in the sample spac of a random experiment. (Ch. 5 definition)  A capital X refers to a random variable in general while specific values of x are shown in lower case

Discrete random variable – a random variable that has a countable number of distinct values.

Discrete probability distribution – one that assigns a probability to each value ofa discrete random variable X.

Function - is a relation between a set of inputs and a set of potential outputs with the property that each input is related to exactly one output.  In probability, the input, or potential outcome, cannot have more than one output, or likelihood. 

Expected Value – a measure of central tendency. In a discrete probability distribution, it is the sum of all X-values weighted by their respective probabilities.   It is an average, so we call it the mean and use the symbol for a population mean, µ.

Bernoulli Experiment – A random experiment that has only two outcomes.

Binomial Distribution – Arises when a Bernoulli Experiment is repeated n times.

6.1 Discrete Distributions

A discrete random variable is a value that has a countable number of distinct values.   Some have a clear upper limit while others do not.

Examples of decision problems with discrete random values:

Decision Problem	Discrete Random Variable
A university hqas space for 65 new MBA students. In previous years 75% of those accepted actually enrolled.  The decision is made to accept 80 students.  What is the probability that more than 65 admitted students will actually enroll? [Has an upper limit]	X = number of admitted MBA students who actually enroll (X = 0, 1, 2,…80)
On the late morning work shift, LL Bean’s order processing center staff can handle up to 5 orders per minute.  The mean arrival rate is 3.5 orders per minute.  What is the probability that more than 5 orders will arrive in a given minute? [Does not have an upper limit]	X = number of phone calls that arrive in a given minute at the LL Bean order processing center.

Probability Distributions

A discrete probability distribution is one that assigns a probability to each value ofa discrete random variable X.   The distribution must follow the rules of probability defined in Chapter 5.  If there are ndistinct values of X (x1, x2, … xn):

0 ≤ P(xi) ≤ 1              (the probability for any given value of X)

∑n P(xi) = 1              (the sum of overall values of X)

Discrete probability functions follow the rules of functions:  

-                   More than one sample space outcome can be assigned to the same number, but you cannot assign one outcome to two different numbers.  Likewise, more than one random variable value can be assigned to the same probability, but one random variable cannot have two different probabilities.

Ex. An outcome of the experiment of rolling a die will have a single probability, in this case 1/6, for each individual face of the die (X = 1, 2,…6).  A single outcome of such an experiment cannot be 1/6 and, say, 2/3.

-                   The probabilities must sum to 1.  The values of X need not be equally likely, but the probabilities must sum to 1 in any probability distribution.

Expected Value

A discrete probability function is defined only at specific points on the X-axis.  The expected value E(X) of a discrete random variable is the sum of all X-values weighted by their respective probabilities. It is a measure of central tendency.

              E(X) = µ = ∑n xi P (xi)

[This is the sum of the values after they’ve been multiplied by their respective probabilities.]

The expected value is a weighted average because outcomes can have different probabilities. (Different from each other.)  Because it is an average, we usually call E(X) the mean and use the symbol for the mean, µ.

Refer back to the concepts of relative frequency and cumulative frequency:

Ex. Flipping a coin three times.

Sample space will be: {HHH, HTT, TTT, TTH, THH, HTH, THT}: 8 possible outcomes

Number of Heads	Outcomes	Relative Frequency (also known as probability)	Cumulative Relative Frequency
0	1	1/8 (or .125)	1/8
1	3	3/8 (or .375)	4/8
2	3	3/8 (or .375)	7/8
3	1	1/8 (or .125)	8/8 (or 1, as all probabilities must sum)

Application: Life Insurance

Expected Value is the basis for life insurance.  For ex. Based on US mortality statistics, the probability that a 30 year old white female will die within the next year is .00059.  So, the probability of her living another year is 1 - .00059 or .99941. What premium should a company charge in order to break even on a $500,000 1 year term insurance policy (that is, to achieve zero expected payout)?

Event	x	P(x)	xP(x)
Live	0	.99941	.00
Die	500,000	.00059	295.00
Total		1.0000	295.00

The insurance company should charge an annual premium of 295.00 (or 24.59/mo) to break even.

Application: Raffles and Lotteries

Expected Value can be applied to raffles and lotteries. If it costs $2 to buy a ticket in a raffle to win a new luxury automobile worth $55,000 and 29,346 raffle tickets are sold, the expected value  of a lottery ticket is:

E(X) = (value if you win)  * P(win) / (value if you lose) * P(lose)

        = ($55,000) * 1/29,346 + (0) (29,345/29,346)

         = (55,000)(.000034076) + (0)(.999965924) = $1.87

The cost of the raffle ticket actually exceeds its value.  Why would you buy it then?  Some would buy it only because the potential win far exceeds the cost of the ticket. Since the idea of a raffle is to raise money (they’re often done for charity), the sponsor tries to sell enough of them to push the E(X) of the ticket to below its price.

Variance and Standard Deviation

The variance V(X) of a discrete random variable is the sum of the squared deviations about its expected value, weighted by the probability of each X-value.  If there are n distinct values of X, the variance is:

V(X) = σ2 = ∑n [xi - µ]2 * P(xi)

Just like the E(X) is used interchangeably with the mean, V(X) denotes variance of a probability distribution.  The standard deviation is the square root of the variance and is denoted σ.

                                          σ = √V(X)

What is a PDF or a CDF?

A known distribution can be described by either a probability distribution function (PDF) or a cumulative distribution function (CDF).    The discrete PDF will show the probability of each X-value, while the CDF shows the cumulative sum of probabilities.

Random variables and their distributions are described by their parameters.  The equations for the PDF, CDF and the characteristics of the distribution (such as the standard deviation or mean) will depend on the parameters of the process.

6.2 Uniform Distribution

The uniform distribution is one of the simplest discrete models. It describes a random variable with a finite number of consecutive integer values from a to b.  This means that the entire distribution depends only on the two parameters at a and b. Each value is equally likely.

This summarizes the characteristics of the uniform distribution:

Uniform Distribution

Parameters	a = lower limit b = upper limit
PDF	P(X = x) = 1\b – a + 1
CDF	P(X ≤ x) = x – a + 1/b – a + 1
Domain	x= a, a+1, a+2,…b
Mean	a+b/2
Standard Deviation	√ [(b-a) + 1]2 – 1/12

Ex. Die Roll.  Equally likely outcome for each side: 1/6

When you roll one die, the number of dots forms a uniform discrete random variable with six equally likely integer values 1, 2, 3, 4, 5, 6.   For this example, the mean and standard deviation are:

PDF:                             P(X = x) = 1/ b- a + 1 = 1/ 6 – 1 + 1 = 1/6  for x = 1, 2,…6

Mean:                             a+b/2 = 1+6/2 = 3.5

Std. Dev.:               √ [(b-a) + 1]2 – 1/12  = √ [(6-1) + 1]2 – 1/12 = √36-1/12 = √35/12 = √2.917 = 1.708

You can see that the mean, 3.5, must be halfway between 1 and 6, but there is no way you can get the standard deviation without using a formula.

6.3 Bernoulli Experiments

A random experiment that has only two outcomes is called a Bernoulli Experiment. The probability of success is π and the probability of failure is 1-π, so that the probabilities will sum to 1. The probability of success remains the same for each trial.

It can be any value between 0 and 1.  Ex. In flipping a coin, π = .50.  But in other applications, π could be close to 1 (e.g. the probability that a customer’s VISA will be approved) or close to 0 (probability that an adult male is HIV positive).  The definitions of success and failure are arbitrary and can be switched, although we usually define the less likely outcome so that π is less than .50.

The only parameter needed to define a Bernoulli process is π.  It has a mean of π and a variance of 1-π.

6.4 Binomial Distribution

To understand binomial distribution, you’ll need to understand factorials, combinations and permutations from Chapter 5.  I’ll cover those here.

Factorials

The factorial of a non-negative integer n, denoted by n!, is the product of all positive integers less than or equal to n. For example:

              5! = 5 x 4 x 3 x 2 x 1 = 120

These can be performed on your calculator using the ! function.   The value of 0! is 1.

Permutations

Chose r items at random without replacement from a group of n items. In how many ways can the r items be arranged, treating each arrangement as a different event (i.e. treating the three-letter sequence XYZ as different from the three-letter sequence ZYX.)?  A permutation is an arrangement of the r sample items in a particular order. 

The number of possible permutations of n items taken r at a time is denoted nPr.

              nPr = n!/(n-r)!

This is also available on your calculator in the same function area as you will find !.  (On the TI-30XIIs, it can be found by pressing the Probability key, labeled PRB, just right of the LOG key.  You need to first enter the number you wish to be n and then hit PRB, select nPr with the arrow keys and then enter the number for r.  Hit enter afterwards.)

Combinations

A combination is a collection of r items chosen at random without replacement from n items where the order of the selected items is not important.  (i.e. treating the three-letter sequence XYZ as being the same as the three-letter sequence ZYX.) The number of possible combinations or r items chosen from n items is denoted nCr.

              nCr = n!/r!(n-r)!

We use combinations when the only thing that matters is which r items are chosen, regardless of how they are arranged.

Now, onto the Binomial Distribution

Bernoulli experiments lead to an important and more interesting model: the binomial distribution.  This arises when a Bernoulli experiment is repeated n times.

Detailed Characteristics for the Binomial Distribution

Parameters	n = number of trials π = number of successes
PDF	P(X = x) n!/x!(n-x)!(πx)(1-π)n-x for X = 0, 1, 2, 3…n (non-negative integers)
Domain	X = 0, 1, 2, 3…n (non-negative integers)
Mean	nπ
Standard Deviation	√nπ(1-π)
Comments	Skewed right if π < .50, skewed left if π > .50 and symmetrical if π = .50.  Skewness decreases as nincreases.

Ex.

Consider a shop that specializes in quick oil changes. It is important to this type of business to ensure that a car’s service time is not considered “later” by the customer.  Therefore, to study this process, we can define service times as being either “late” or “not late” and define the random variable X to be the number of cars that are late out of the total number of cars serviced.  We further assume that cards are independent of each other and the chance of a car being late stays the same for each car. Based on our knowledge of the process we know that P(car is late) = π = .10.

Think of each car as a Bernoulli experiment and let’s apply the binomial distribution.  Suppose we would like to know the probability that exactly 2 of the next 12 cars serviced are late.  In this case n = 12, and we want to know P(X=2):

P(X=2) = 12!/2!(12-2)! (.10)2 (1-.10)12-2 = .2301

Compound Events

We can add the individual probabilities to obtain any desired event probability.  For example, the probability that the sample of four patients will contain at least two uninsured patients is

P(X ≥ 2) = P(2) + P(3) +P(4) = .1536 + .0256 +.0016 = .1808

Probability that fewer than 2 patients have insurance is

P(X < 2) = P(0) + P(1) = .4096 + .4096 = .8192

When faced with questions asking about “fewer than” or “more than”, “greater than”, etc, it’s most helpful to sketch a diagram of a number line to indicate the range of values being sought.

Recognizing Binomial Applications

The binomial distribution has five defining elements:

-                   A fixed number of trials

-                   There are only two outcomes of a trial: success or failure

-                   The trials are independent

-                   Probability of a success (π) remains constant

-                   The random variable (X) is the number of successes.

Seek out the five characteristics of a binomial distribution in order to determine if you’re being asked to find one (and therefore, use the formula).  Review the question with this list in mind.

If you have any test reviews, homeworks, guides, anything school related that you think can be posted on this website, reach out to me at makingschooleasier@gmail.com