STA 2023 Florida International Univ Probability and The Law of Large Numbers Mathab Please help me solve my lab 2 questions. Excel Lab attached with pdf instructions. Will need answers to questions in a word doc. LAB 2: Probability and The Law of Large Numbers

Learning Objectives

Construct a finite sample space and define a probability function satisfying the Axioms of Probability.

Compute the probability of an event using the Classical Method.

Invoke the Law of Large Numbers to estimate the probability of an event based on

simulated data.

Apply the Combinations Rule to the calculation of card hand probabilities.

Key Concepts

Probability

We discuss probability often. What is the probability that it will rain? What is the probability that The New York Mets will win the World Series? Probability has many useful

applications but is also a very abstract mathematical concept! The following example

illustrates the mathematical concept of probability.

A three coin toss is a random experiment. We cannot predict with certainty which outcome will occur at a specific instance. We can outline all possible outcomes, and these

outcomes make up a sample space. A probability map, or function, assigns numbers to

the outcomes in a sample space, ideally reflecting the frequency of their occurrence. A

probability map for a three coin toss is depicted in the graphic below:

Probability Map for a Three Coin Toss

Note: H indicates heads and T indicates tails.

Included in the sample space are the eight possible outcomes of a three coin toss.

The probability map is constructed so that each outcome is assigned a probability of 18 .

Any valid probability map must satisfy the Axioms of Probability:

1. The probability of each outcome must be greater than or equal to zero.

2. The probabilities of all outcomes in the sample space must sum to 1.

3. Suppose there are two collections of outcomes (A and B) that do not overlap.

Then: Probability(A OR B) = Probability(A) + Probability(B).

There are many ways to assign probabilities that satisfy these Axioms. Ideally, we assign meaningful numbers to the random outcomes reflecting their chance (or frequency)

of occurrence if the random experiment is repeated many times.

Note: A collection of outcomes in a sample space is an event. Each possible outcome

in a sample space is a sample point.

Classical Method for Computing an Event Probability

A three coin toss has a finite sample space. There are just eight possible outcomes.

For finite sample spaces, we can use the Classical Method to compute the probability

of an event of interest. The Classical Method has five steps:

1. Describe the random experiment and identify several sample points.

2. List (or count) all of the sample points in the finite sample space.

3. Assign a probability to each sample point satisfying the Axioms of Probability.

4. List (or count) the sample points in the event of interest.

5. Sum the sample point probabilities associated with the event of interest.

What is the probability of observing exactly one head in a three coin toss? Lets follow

the steps of the Classical Method to find out:

1. The random experiment consists of tossing a coin three times and recording the

outcome. Some sample points are HHH, HTH, HTT.

2. There are eight sample points in the finite sample space.

3. Each sample point has a probability of 18 .

4. The sample points consistent with exactly one head are: HTT, THT, TTH.

5. The sum of the sample point probabilities for HTT, THT, TTH is 83 .

Therefore, the probability of observing exactly one head equals 83 .

Dice Example

In the dice game of Craps, a player rolls two six-sided dice and sums the two die values

together. If a Craps player hits a seven or an eleven on the first roll, he or she wins

immediately. The sample space consisting of all possible combinations of two six-sided

dice is depicted below:

Sample Space for a Roll of Two Six-Sided Dice

What is the probability that a player hits a seven or an eleven on a single roll? Lets

follow the steps of the Classical Method to find out:

1. The random experiment consists of rolling two six-sided dice. Some sample

points (Die 1/Die 2) are: Six/Five (Sum=Eleven) or Three/Two (Sum=Five).

2. There are thirty-six sample points in the finite sample space.

3. Each sample point has a probability of

1

.

36

4. There are six sample points that produce a sum of seven and two sample points

that produce a sum of eleven.

5. The sum of the sample point probabilities in (4) is

8

.

36

Therefore, the probability of hitting a seven or an eleven on a single toss is

8

.

36

The Law of Large Numbers

Suppose we repeatedly toss two six-sided dice and record a value of one every time

a sum of seven or eleven appears and a zero otherwise. The average of ones and

zeroes equals the relative frequency that a sum of seven or eleven occurs in N trials.

The Law of Large Numbers guarantees that the relative frequency will eventually come

8

very close to 36

! The Law of Large Numbers states that if we repeat a random experiment many times and compute the relative frequency that an event occurs, the relative

frequency will eventually come very close to the true event probability.

Note: It is critical that the occurrence of random outcomes adheres to the probability model specified by the Classical Method or this method will not work!

Simulation Example

To illustrate the Law of Large Numbers, we can use computer simulation to repeatedly

roll two six-sided dice, record the relative frequency of observing a sum of seven or

eleven, and plot this relative frequency as a function of N trials.

Relative Frequency of a Seven or an Eleven over N Rolls of Two Six-Sided Dice Versus

Classical Probability

As the number of trials increases, the relative frequency of a sum of seven or eleven

8

. Note that the Law of LARGE

becomes extremely close to the classical probability of 36

Numbers only guarantees that as the number of trials becomes large, the relative frequency and classical probability will converge. For small numbers of trials, there is no

guarantee that these numbers will even be reasonably close together.

Computing Card Probabilities

Now lets move on to the Blackjack table! In the game of Blackjack, a dealer deals out

cards from a 52-card deck. A blackjack occurs when the sum of the cards equals 21.

What is the probability that the first two cards dealt form a blackjack? Lets follow the

steps of the Classical Method to find out:.

1. The random experiment consists of dealing two cards from a 52-card deck.

Some sample points are: King of Clubs/Five of Hearts (Sum=15) or Four of Diamonds/Nine of Clubs (Sum=13).

2. There are 52 ways to select the first card and 51 ways to choose the second card.

Initially, it appears there are 52 × 51 (or 2,652) ways to select two cards. However, the order in which the cards are dealt does not matter. King of Clubs/Five

of Hearts is the same hand as Five of Hearts/King of Clubs. There are two orderings for every unique hand of two cards. Therefore, the number of unique card

51

(or 1,326).

combinations is 52×

2

3. All possible two-card hands are equally likely. Each sample point has a probability

1

of 1326

.

4. A blackjack must consist of an Ace (4 possibilities) and a card with a value of

Ten (16 possibilities). Tens, jacks, queens and kings have a card value of Ten.

There are 4 × 16 (or 64) ways to achieve a blackjack with two cards.

5. The sum of the sample point probabilities in (4) is

64

.

1326

Therefore, the probability that a randomly selected two cards form a blackjack is

64

.

1326

Combinations Rule

In the blackjack example, we adjusted for duplicate hands of two cards arising from the

ordering of the dealt cards. Since no two cards in the deck are identical, every hand

has exactly two orderings.

The Combinations Rule is a general formula that can be applied when:

You wish to determine the number of unique combinations that arise from sampling

n items from a group of N items.

The sampling order is not important.

The same item cannot be selected more than once (i.e. items are sampled without

replacement).

The Combinations Rule states that the number of unique combinations

N!

.

(N ?n)!n!

N

n

equals:

In the blackjack example, we sampled n=2 cards without replacement

from the N=52

52

52!

=

cards in the deck. The number of unique two card combinations 2 equals: (52?

2)!2!

52×51

2

(or 1,326). We obtain the same result as before, after applying the Combinations

Rule.

Assignment

Question 1: Computing Event Probabilities Using the Classical Method

Watch the Lab 2 video tutorial entitled Computing Event Probabilities Using the

Classical Method.

Watch the Lab 2 video tutorial entitled Computing Dice Probabilities Using the Classical Method.

Two five-sided dice are rolled. Consider the following events:

Event A: The sum of the two dice equals five.

Event B: The maximum value of the two dice equals five.

Note: If both Die1 and Die 2 equal five, then the maximum value equals five.

(a) How many possible outcomes (sample points) for a roll of two five-sided dice are

possible?

(b) What is the probability associated with each possible sample point?

(c) How many sample points are consistent with Event A?

(d) Compute P(A) using the Classical Method.

(e) How many sample points are consistent with Event B?

(f) Compute P(B) using the Classical Method.

(g) Why cant we use the Combinations Rule to compute the number of sample points?

Question 2: The Law of Large Numbers

The data-set labeled Dice Simulation reports the outcome of a computer simulation in

which two five-sided dice (numbered from 1-5) are repeatedly tossed (N=5,000 trials).

Watch the Lab 2 video tutorial entitled Estimating Event Probabilities from a Dice

Simulation in Microsoft® Excel®.

(a) Using Microsoft® Excel®, complete the following steps i-iii for Events A and B defined in Question 1. Produce two separate graphs.

i: Create an indicator variable (0 or 1) to reflect whether the event occurred at each

trial of the dice simulation.

ii: Compute the relative frequency that the event occurred at each trial of the dice

simulation.

iii: Plot the relative frequency of the event as a function of N trials. On the same graph,

include a constant line reflecting P(Event) computed using the Classical Method.

(b) For Events A and B, compute |f50 ? P(Event)|, where f50 is the relative frequency of

the event on the 50th trial.

(c) For Events A and B, compute |f5000 ? P(Event)|, where f5000 is the relative frequency

of the event on the 5,000th trial.

(d) Is 5,000 iterations a sufficiently large number of simulations to achieve a good approximation of P(A)? P(B)?

Question 3: Computing Card Hand Probabilities

Watch the Lab 2 video tutorial entitled The Combinations Rule.

Watch the Lab 2 video tutorial entitled Computing Card Probabilities Using the

Classical Method.

A dealer deals out two cards from a 52-card deck and the outcome is observed. Consider the following events:

Event A: The two cards form a blackjack.

Event B: The two cards are consecutive (i.e. Five of Spades, Six of Spades).

64

The probability of blackjack (Event A) has already been computed in this text. P(Blackjack)= 1326

.

(a) How many possible outcomes (sample points) are consistent with Event B?

(b) Compute P(B) using the Classical Method.

The data-set labeled Card Simulation reports the outcome of a computer simulation

in which two cards are repeatedly dealt from a deck of 52 cards (N=5,000 trials).

Watch the Lab 2 video tutorial entitled Estimating Event Probabilities from a Card

Simulation in Microsoft® Excel®.

(c) Using Microsoft® Excel®, complete the following steps i-iii for Events A and B. Produce two separate graphs.

i: Create an indicator variable (0 or 1) to reflect whether the event occurred at each

trial of the card simulation (N=5,000).

ii: Compute the relative frequency that the event occurred at each trial of the card

simulation (N=5,000).

iii: Plot the relative frequency of the event as a function of N trials. On the same graph,

include a constant line reflecting P(Event) computed using the Classical Method.

(d) For Events A and B, compute |f50 ? P(Event)|, where f50 is the relative frequency of

the event on the 50th trial.

(e) For Events A and B, compute |f5000 ? P(Event)|, where f5000 is the relative frequency of the event on the 5,000th trial.

(f) Is 5,000 iterations is a sufficiently large number of simulations to achieve a good

approximation of P(A)? P(B)?

Card 1

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