lesson 15: discrete probability …(lesson 15: discrete probability distributions) 15.02 a discrete...

(Lesson 15: Discrete Probability Distributions) 15.01

LESSON 15: DISCRETE PROBABILITY DISTRIBUTIONS What is Possible? What are the Chances?

PART A: RANDOM VARIABLES and their DISTRIBUTIONS

Let X be a quantitative random variable. Its value is basically a number that is not yet known. However, we may know its probability distribution, which gives:

• The possible values of X and • Their probabilities of occurring.

The distribution may be described by a formula, a graph, or a table.

For now, assume that X is a discrete random variable. Its possible values can be listed in table. (See Lesson 3, Part C.)

Example 1 (Discrete Probability Distributions: Roll One Die)

Roll one standard six-sided die. Let X = the result of the die. X has the following discrete probability distribution:

Value (x)

Probability P (x)

1 1/6 2 1/6 3 1/6 4 1/6 5 1/6 6 1/6

We call this a uniform distribution, because all the probabilities for all possible values are equal.

§


A discrete probability distribution must have these two characteristics:

• All the probabilities must be real numbers between 0 and 1.

• All the probabilities must add up to 1.

Example 2 (Discrete Probability Distributions: Computer Repairs)

An office has four computers. We will randomly select a five-day future work week for the office. Let X = the number of computers that need a repair during the week. The IT manager for the office gives you the following incomplete probability distribution for X.

Value (x)

in computers

Probability P (x)

0 a 1 0.200 2 0.100 3 0.040 4 0.010

Find a, the probability that none of the computers need a repair that week. That is, find P (0).

§ Solution

The key idea here is that the sum of the probabilities must be 1.

a = 1− 0.200− 0.100− 0.040− 0.010 = 0.650

Therefore, P (0) = 0.650.

§


Example 3 (Wording of Inequalities: Computer Repairs; Revisiting Example 2)

In Example 2, we found P (0), or P X = 0( ) . We then had the complete distribution for X:

Value (x)

in computers

Probability P (x)

0 0.650 1 0.200 2 0.100 3 0.040 4 0.010

For each of the following, rewrite using an inequality, rewrite as a sum of probabilities, and calculate the indicated probability.

• a) P (at least 1); is there a quick way of calculating this?

• b) P (more than 1)

• c) P (at most 2)

• d) P (fewer than 2)

§ Solution

• a) P (at least 1) =

P X ≥1( ) = P 1( ) + P 2( ) + P 3( ) + P 4( )= 0.200+ 0.100+ 0.040+ 0.010= 0.350

“At least 1” includes 1, and we go higher. Keep in mind:

• Whether we include or exclude the endpoint, or boundary number (1 here), and

• Whether we pick up higher numbers or lower numbers. A faster approach here would be to subtract the complementary probability, P (0), from 1:

P X ≥1( ) = 1− P 0( )= 1− 0.650= 0.350

(Lesson 15: Discrete Probability Distributions) 15.04.

• b) P (more than 1) =

P X >1( ) = P 2( ) + P 3( ) + P 4( )= 0.100+ 0.040+ 0.010= 0.150

“More than 1” excludes 1, and we go higher. • c) P (at most 2) =

P X ≤ 2( ) = P 0( ) + P 1( ) + P 2( )= 0.650+ 0.200+ 0.100= 0.950

“At most 2” includes 2, and we go lower.

It may be easier to subtract the complementary probability,

P 3( ) + P 4( ) , from 1:

P X ≤ 2( ) = 1− P 3( ) + P 4( )⎡⎣ ⎤⎦= 1− 0.040+ 0.010⎡⎣ ⎤⎦= 1− 0.050⎡⎣ ⎤⎦= 0.950

• d) P (fewer than 2) =

P X < 2( ) = P 0( ) + P 1( )= 0.650+ 0.200= 0.850

“Fewer than (or less than) 2” excludes 2, and we go lower.

§

(Lesson 16: Expected Value) 16.01

LESSON 16: EXPECTED VALUE What Happens in the Long Run?

PART A: EXPECTED VALUE, or MEAN, of a DISTRIBUTION

The expected value, or mean, of a probability distribution is denoted by

E X( ) or µ . Its formula resembles the weighted mean “zigzag” formula from Lesson 8,

w ⋅ x∑w∑ , where the weights are probabilities and w∑ = 1.

Expected Value (or Mean) of a Probability Distribution

E X( ) or µ = P x( ) ⋅ x∑ , or x ⋅P x( )∑

E X( ) gives the approximate long-run average per repetition if the random experiment producing the value of X is repeated many times. This is a result of the Law of Large Numbers (see Example 4).

PART B: EXAMPLES

Example 1 (Expected Value: “Deal or No Deal”)

A “Deal or No Deal” contestant has a case. Let X = the amount of money in the case. X has a uniform distribution on the three remaining dollar amounts left in play:

Value (x)

Probability P (x)

$100 1/3 $400,000 1/3

$1,000,000 1/3 Find E X( ) and interpret it.


§ Solution

Think: “Zigzag,” as in Lesson 8.

E X( ) or µ = P x( ) ⋅ x∑= 1

3⎛⎝⎜

⎞⎠⎟

$100( ) + 13

⎛⎝⎜

⎞⎠⎟

$400,000( ) + 13

⎛⎝⎜

⎞⎠⎟

$1,000,000( )= $466,700

If many contestants have the same three dollar amounts in play, then the long-run average value of the case per contestant is about $466,700.

§

If the probability distribution is uniform, as in Example 1, then E X( ) is simply the unweighted average of the possible values. Example 2 (Expected Value for a Uniform Distribution: Roll One Die)

If X is the result of a standard six-sided die that is rolled, then E X( ) is 3.5, the average of the die faces. This is because X has a uniform distribution.

Value (x)

Probability P (x)

1 1/6 2 1/6 3 1/6 4 1/6 5 1/6 6 1/6

§

WARNING 1: Beware of unweighted averages. If X does not have a uniform distribution, then do not simply take the unweighted average of the values.


Example 3 (Expected Value: Computer Repairs; Revisiting Lesson 15)

An office has four computers. We will randomly select a five-day future work week for the office. Let X = the number of computers that need a repair during the week. The probability distribution for X is:

Value (x)

in computers

Probability P (x)

0 0.650 1 0.200 2 0.100 3 0.040 4 0.010

Find E X( ) and interpret it.

§ Solution

E X( ) or µ = P x( ) ⋅ x∑= 0.650( ) 0( ) + 0.200( ) 1( ) + 0.100( ) 2( ) + 0.040( ) 3( ) + 0.010( ) 4( )= 0.560 computers

Over many five-day work weeks, the long-run average number of computers needing a repair per week is about 0.560 computers.

WARNING 2: Making sense. Although 0.560 computers may not make sense for a single week, it makes sense as a long-run average per week.

§


Example 4 (Expected Value: Roulette; Revisiting Lesson 11, Example 4)

In order to play a round of roulette at a particular casino, you first pay $100. If you bet on “red” and a red number comes up, then you win $200. If you bet on “red” and another color comes up, then you win nothing. Let X = your net monetary gain as a result of playing a round in which you bet on “red.”

Find E X( ) and interpret it.

18 red slots P red( ) = 18

38=

919

≈ 47.4%( ) 18 black slots

P black( ) = 18

38=

919

≈ 47.4%( )

2 green slots P green( ) = 2

38=

119

≈ 5.26%( )

38 total

§ Solution

We first figure out the distribution for X. The first column may help.

Outcome of round

Value (x)

Probability P (x)

You win. $200 − $100 = $100 9

19

You lose. $0 − $100 = − $100 9

19+ 1

19= 10

19

Here is a cleaned-up version of the distribution table:

Outcome of round

Value (x)

Probability P (x)

You win. $100 9

19

You lose. − $100 1019


E X( ) or µ = P x( ) ⋅ x∑= 9

19⎛⎝⎜

⎞⎠⎟

$100( ) + 1019

⎛⎝⎜

⎞⎠⎟−$100( )

= −$5.26

If you play many rounds of roulette at this casino and always bet on “red,” your long-run average net loss will be about $5.26 per round. WARNING 3: Interpret negative expected values appropriately.

• Reminder: According to the Law of Large Numbers (LLN), the probabilities built into the game will “crystallize” in the long run. In other words, the observed relative frequencies of red, black, and green results will approach the corresponding theoretical probabilities

919

, 919

, and 119

⎛⎝⎜

⎞⎠⎟

after many bets.

Think About It: How do you maximize your chances of winning a huge amount of money?

§

(Lesson 16: Expected Value) 16.06 PART C: VARIANCE and STANDARD DEVIATION (OPTIONAL)

The idea here is that, the lower the standard deviation is, the more reliable a measure E X( ) is as a long-run average.

(Lesson 16: Expected Value) 16.07.

(Lesson 17: Counting) 17.01

LESSON 17: COUNTING How Many Possible Ways?

PART A: FUNDAMENTAL COUNTING RULE

Let’s say there are two decisions to be made: Decision A and Decision B. If there are m possible choices for Decision A

A1, A2 ,…, Am( ) and

n possible choices for Decision B B1, B2 ,…, Bn( ) , regardless of how A is decided,

then there are mn possible ways to decide both A and B. Envision a possibility tree or a grid:

A \B B1 B2 … Bn

A1

A2

!

Am This multiplication rule extends to more than two decisions.

(Lesson 17: Counting) 17.02 Example 1 (Fundamental Counting Rule: Passwords)

Assume that a password must have the following characteristics:

• It must be (exactly) four characters long. • It must begin with an uppercase letter. • Each of the other characters can be an uppercase letter or a digit from 0 to 9.

How many possible passwords are there?

§ Solution

There are 26 possibilities (the uppercase letters) for the first character. There are 36 possibilities (the uppercase letters and the digits 0-9) for each of the others. The number of possible passwords is:

26 × 36 × 36 × 36 = 1,213,0561st 2nd 3rd 4th

Note: If you write 26 ⋅363 , remember that exponentiation precedes multiplication in the order of operations.

§

Example 2 (Fundamental Counting Rule: Passwords)

If a password is randomly selected, what is the probability that it is “IH8U”?

§ Solution

The 1,213,056 possible passwords are equally likely to be selected, so the

probability of getting a particular one (such as “IH8U”) is

11,213,056

.

§


PART B: PERMUTATIONS (ORDERINGS)



PART C: PARTIAL PERMUTATIONS

Look for n Pr on calculators.

The formula is most useful when r is large.

(Lesson 17: Counting) 17.06 PART D: COMBINATIONS

Look for nCr on calculators.


Method 2 (Helps with Lesson 18.)

Before the race, randomly assign the numbers 1 through 7 to the runners. After the race, three win (W) and four lose (L).

Related Question: Let’s say you are given a list of seven essay questions, and you must choose three of them to write on. How many ways are there to choose the three questions? The answer, again, is 35. Related Question: How many ways are there to get exactly three questions right on a test of seven questions? The answer, again, is 35.


PART E: HOW DO I DECIDE WHICH RULE TO USE?

Two key questions:

1) Are possibilities being reduced for future tasks/decisions? If not, maybe just use the Fundamental Counting Rule (Part A). If we have the same numbers of possibilities (choices) for different decisions, then a power may be involved.

2) If possibilities are being reduced, does order matter among the winners/chosen

items, or do we distinguish among them? If so, consider using permutations (Part C), with factorials corresponding to the special case of “complete” permutations (Part B). If order does not matter, consider using combinations (Part D).

(Lesson 17: Counting) 17.09.

The prior probability is low; remember Lesson 14, Example 3.

(Lesson 18: Binomial Distributions) 18.01

LESSON 18: BINOMIAL DISTRIBUTIONS How Many Successes? What are the Chances?

PART A: WHAT IS A BINOMIAL EXPERIMENT?

Example 1 (Binomial Experiments)

I roll a standard six-sided die five times. Let X = the number of rolls coming up a “4.” This is a binomial experiment, and X is binomial random variable. General characteristics of a binomial experiment In this Example 1) There are a fixed number of A trial is a roll of the die.

independent trials (n of them) n = 5 .

2) Each trial results in a success A success is a “4” being rolled. or a failure. A failure is something else being rolled. • The “successes” are the things we want to count; they could be deaths!

3) For each trial, there is a fixed

probability (p) of success.

• P success( ) = p p = 1

6

• P failure( ) = 1− p, or q q = 5

6

§

Similar examples:

• 1/6 of all Americans have cooties. Five Americans are randomly sampled. Let X = the number in the sample with cooties. • A multiple-choice test has five questions, each with six options (A-F). A student guesses randomly on all five questions. Let X = the number of correct answers.


If X is the number of successes among the n trials, then X has a binomial probability distribution: X ∼ Bin n, p( ) .

• In the previous examples, X ∼ Bin n = 5, p = 1

6⎛⎝⎜

⎞⎠⎟

.

• n and p are the parameters of the distribution. If you change either, you change the distribution!

Here is the distribution table for Bin n = 5, p = 1

6⎛⎝⎜

⎞⎠⎟

; probabilities are rounded off

to three decimal places:

Value (x)

Probability P (x)

0 0.402 1 0.402 2 0.161 3 0.032 4 0.003 5 0+

Note: P 5( ) ≈ 0.000129 ; we don’t want to write “0” in that row, so we write “0+.”

(Lesson 18: Binomial Distributions) 18.03 PART B: HOW DO WE FIND BINOMIAL PROBABILITIES? (OPTIONAL)

1) Software / TI calculators

2) Tables

3) Binomial probability formula




Note: P 5( ) ≈ 0.000129


PART C: PASCAL’S TRIANGLE (OPTIONAL)


PART D: SAMPLING RULE (See Lesson 13, Part D.) We assume independence, even if we are sampling without replacement, if n (the sample size) is no more than 5% of N (the population size), and N is large.

PART E: USING TABLES

Example 2 (Using Tables to Analyze Binomial Distributions)

70% of American adults like Senator Smith. That is, P a random American adult likes Senator Smith( ) = 0.7. Six American adults are randomly sampled to form a focus group. Find the probability that at least four of the people in the focus group like Senator Smith. Hint: The table for Bin n = 6, p = 0.7( ) is below.

Value (x)

Probability P (x)

0 0.001 1 0.010 2 0.060 3 0.185 4 0.324 5 0.303 6 0.118

Note: “At least 4” can be rephrased as “4 or more” or “no fewer than 4.” Think About It: Why are the probabilities skewed towards the high end?

§ Solution

P X ≥ 4( ) = P 4( ) + P 5( ) + P 6( )≈ 0.324+ 0.303+ 0.118≈ 0.745

§

(Lesson 18: Binomial Distributions) 18.09 PART F: WHEN IS AN EVENT “UNUSUAL”?

Rare Event Rule for Inferential Statistics


Note: P 49( ) ≈ 7.80%

(Lesson 18: Binomial Distributions) 18.11. Analysis #2 (one-tailed)

Analysis #3 (two-tailed; we use symmetry)

Analysis #4 (z scores)

Coming soon ….

lesson 15: discrete probability …(lesson 15: discrete probability distributions) 15.02 a discrete...

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