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 in
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
Trial Number
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
Rank
6
7
6
Queen
4
Queen
5
9
8
9
8
7
Queen
8
5
10
9
6
Jack
Jack
3
Queen
6
Queen
6
2
6
6
3
7
4
9
2
6
7
10
2
9
9
2
Ace
7
9
Suit
Diamonds
Spades
Clubs
Spades
Hearts
Hearts
Clubs
Clubs
Hearts
Spades
Clubs
Clubs
Clubs
Hearts
Clubs
Clubs
Hearts
Hearts
Clubs
Diamonds
Spades
Spades
Spades
Diamonds
Clubs
Spades
Clubs
Spades
Hearts
Spades
Hearts
Hearts
Clubs
Clubs
Clubs
Hearts
Spades
Hearts
Diamonds
Diamonds
Clubs
Clubs
Diamonds
Card 2
Rank
7
8
Queen
Jack
King
5
2
Queen
7
King
4
2
King
Ace
6
9
8
4
Ace
Queen
9
6
10
Ace
2
Queen
Jack
King
4
Queen
7
6
4
2
6
Jack
10
Ace
Ace
6
5
Queen
7
Suit
Spades
Clubs
Spades
Diamonds
Hearts
Spades
Diamonds
Hearts
Clubs
Hearts
Hearts
Hearts
Clubs
Diamonds
Diamonds
Hearts
Clubs
Spades
Diamonds
Spades
Diamonds
Clubs
Spades
Hearts
Clubs
Clubs
Hearts
Diamonds
Hearts
Spades
Spades
Hearts
Hearts
Diamonds
Diamonds
Spades
Clubs
Clubs
Spades
Hearts
Diamonds
Clubs
Spades
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
7
2
King
3
King
9
7
4
9
4
10
10
Ace
7
5
5
4
4
5
3
4
4
King
10
Jack
7
7
Jack
9
4
Ace
6
2
7
6
3
King
Queen
Ace
9
Ace
4
7
Ace
8
6
6
Spades
Diamonds
Hearts
Clubs
Diamonds
Diamonds
Diamonds
Clubs
Spades
Spades
Spades
Clubs
Diamonds
Diamonds
Clubs
Diamonds
Diamonds
Diamonds
Hearts
Spades
Diamonds
Diamonds
Clubs
Diamonds
Diamonds
Clubs
Spades
Diamonds
Diamonds
Spades
Diamonds
Diamonds
Diamonds
Diamonds
Spades
Clubs
Hearts
Hearts
Clubs
Spades
Diamonds
Diamonds
Spades
Hearts
Clubs
Spades
Spades
9
3
9
10
5
7
8
King
Ace
Queen
9
5
2
6
Jack
6
3
Ace
7
Ace
Jack
King
6
King
10
Queen
2
8
3
Queen
Ace
8
3
10
Queen
6
Queen
7
7
7
Ace
King
King
Jack
3
10
Queen
Spades
Diamonds
Clubs
Spades
Clubs
Diamonds
Clubs
Hearts
Spades
Clubs
Diamonds
Diamonds
Spades
Spades
Diamonds
Clubs
Hearts
Hearts
Clubs
Clubs
Diamonds
Hearts
Hearts
Hearts
Spades
Hearts
Spades
Clubs
Clubs
Spades
Clubs
Hearts
Diamonds
Diamonds
Hearts
Diamonds
Hearts
Diamonds
Spades
Hearts
Hearts
Clubs
Spades
Spades
Hearts
Clubs
Spades
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
125
126
127
128
129
130
131
132
133
134
135
136
137
9
3
7
10
King
2
5
3
6
Jack
Ace
3
5
Queen
6
2
2
6
3
10
3
5
10
10
10
2
9
Jack
Ace
King
Ace
3
Queen
3
8
6
4
2
3
7
Jack
7
Jack
King
7
6
7
Diamonds
Spades
Diamonds
Hearts
Clubs
Clubs
Hearts
Clubs
Hearts
Diamonds
Diamonds
Hearts
Diamonds
Hearts
Clubs
Diamonds
Hearts
Hearts
Clubs
Diamonds
Diamonds
Hearts
Spades
Diamonds
Spades
Hearts
Spades
Spades
Clubs
Diamonds
Diamonds
Diamonds
Spades
Diamonds
Diamonds
Clubs
Spades
Diamonds
Hearts
Clubs
Spades
Diamonds
Hearts
Hearts
Clubs
Spades
Clubs
2
Queen
4
King
8
2
4
2
2
King
Jack
9
3
7
King
Queen
Jack
10
2
10
10
3
7
4
7
7
3
5
9
5
2
Jack
6
King
9
10
7
5
9
Ace
8
9
King
Ace
Jack
6
7
Spades
Hearts
Spades
Clubs
Clubs
Spades
Clubs
Diamonds
Diamonds
Clubs
Clubs
Clubs
Spades
Diamonds
Hearts
Hearts
Spades
Hearts
Hearts
Clubs
Hearts
Hearts
Clubs
Hearts
Spades
Clubs
Clubs
Diamonds
Diamonds
Hearts
Spades
Spades
Diamonds
Clubs
Hearts
Spades
Spades
Spades
Hearts
Spades
Spades
Hearts
Diamonds
Clubs
Spades
Clubs
Hearts
138
139
140
141
142
143
144
145
146
147
148
149
150
151
152
153
154
155
156
157
158
159
160
161
162
163
164
165
166
167
168
169
170
171
172
173
174
175
176
177
178
179
180
181
182
183
184
7
3
2
4
8
5
9
10
8
3
5
8
8
10
2
5
10
10
3
Queen
9
10
9
Ace
4
2
2
6
Ace
5
3
7
King
Jack
8
9
10
5
5
2
9
King
4
9
5
8
3
Clubs
Spades
Diamonds
Clubs
Spades
Clubs
Clubs
Clubs
Hearts
Diamonds
Spades
Clubs
Hearts
Spades
Diamonds
Spades
Spades
Diamonds
Clubs
Hearts
Hearts
Clubs
Clubs
Hearts
Diamonds
Diamonds
Hearts
Clubs
Spades
Hearts
Diamonds
Diamonds
Spades
Diamonds
Spades
Hearts
Diamonds
Spades
Clubs
Hearts
Spades
Clubs
Hearts
Diamonds
Spades
Hearts
Clubs
9
Jack
3
Queen
7
King
4
9
3
9
Jack
Jack
8
Jack
Jack
7
3
8
9
Ace
King
Queen
7
7
9
3
6
3
2
4
10
Queen
6
Queen
2
2
Jack
9
5
6
3
5
Ace
2
8
3
King
Clubs
Diamonds
Spades
Diamonds
Hearts
Clubs
Diamonds
Hearts
Spades
Hearts
Clubs
Clubs
Diamonds
Spades
Diamonds
Diamonds
Clubs
Spades
Clubs
Diamonds
Hearts
Spades
Hearts
Spades
Hearts
Clubs
Diamonds
Clubs
Hearts
Hearts
Hearts
Clubs
Spades
Clubs
Hearts
Clubs
Hearts
Hearts
Diamonds
Spades
Clubs
Clubs
Clubs
Diamonds
Diamonds
Clubs
Diamonds
185
186
187
188
189
190
191
192
193
194
195
196
197
198
199
200
201
202
203
204
205
206
207
208
209
210
211
212
213
214
215
216
217
218
219
220
221
222
223
224
225
226
227
228
229
230
231
Ace
9
2
5
6
4
Jack
5
Jack
6
Queen
7
7
10
Ace
10
2
8
4
6
9
5
3
7
Ace
4
Jack
Jack
9
8
10
7
3
8
King
6
10
2
3
5
4
5
6
3
6
6
6
Spades
Hearts
Hearts
Hearts
Hearts
Diamonds
Diamonds
Clubs
Spades
Diamonds
Hearts
Spades
Spades
Clubs
Spades
Clubs
Clubs
Diamonds
Clubs
Spades
Spades
Spades
Clubs
Diamonds
Spades
Diamonds
Clubs
Diamonds
Spades
Clubs
Hearts
Clubs
Hearts
Clubs
Spades
Hearts
Hearts
Spades
Clubs
Spades
Spades
Hearts
Diamonds
Diamonds
Hearts
Clubs
Diamonds
3
Queen
10
Queen
6
5
3
7
6
Queen
3
Jack
7
7
7
10
3
7
9
8
10
10
6
Queen
Queen
King
8
2
Ace
5
7
King
9
Queen
7
Jack
3
10
Ace
3
7
4
King
5
7
7
Ace
Hearts
Spades
Clubs
Diamonds
Diamonds
Hearts
Diamonds
Diamonds
Diamonds
Hearts
Spades
Clubs
Hearts
Clubs
Spades
Hearts
Spades
Clubs
Hearts
Clubs
Hearts
Spades
Clubs
Spades
Hearts
Clubs
Hearts
Diamonds
Diamonds
Spades
Clubs
Clubs
Diamonds
Spades
Clubs
Hearts
Diamonds
Hearts
Clubs
Hearts
Hearts
Spades
Clubs
Diamonds
Clubs
Hearts
Clubs
232
233
234
235
236
237
238
239
240
241
242
243
244
245
246
247
248
249
250
251
252
253
254
255
256
257
258
259
260
261
262
263
264
265
266
267
268
269
270
271
272
273
274
275
276
277
278
5
Queen
9
Queen
King
2
3
6
8
2
King
King
7
Queen
King
Ace
7
7
Jack
Jack
7
King
3
7
6
Ace
5
10
King
Ace
9
2
6
Ace
King
3
6
10
3
4
Ace
10
7
Queen
9
5
5
Hearts
Spades
Hearts
Hearts
Spades
Clubs
Hearts
Diamonds
Diamonds
Spades
Hearts
Clubs
Hearts
Spades
Diamonds
Diamonds
Diamonds
Hearts
Diamonds
Hearts
Spades
Spades
Diamonds
Hearts
Clubs
Spades
Spades
Spades
Spades
Clubs
Clubs
Spades
Clubs
Diamonds
Diamonds
Diamonds
Spades
Diamonds
Clubs
Hearts
Spades
Hearts
Diamonds
Diamonds
Diamonds
Hearts
Clubs
King
4
10
King
Jack
7
Queen
9
9
8
10
3
Queen
3
5
6
4
5
6
Queen
8
King
3
Jack
9
9
5
King
Ace
5
8
Jack
2
3
2
10
Ace
Jack
Jack
2
9
10
8
King
2
6
Ace
Clubs
Clubs
Hearts
Hearts
Spades
Hearts
Diamonds
Spades
Clubs
Spades
Clubs
Diamonds
Clubs
Hearts
Clubs
Hearts
Clubs
Clubs
Diamonds
Spades
Diamonds
Clubs
Hearts
Diamonds
Diamonds
Diamonds
Hearts
Hearts
Clubs
Hearts
Spades
Clubs
Clubs
Spades
Spades
Clubs
Spades
Clubs
Clubs
Hearts
Spades
Clubs
Clubs
Hearts
Diamonds
Clubs
Hearts
279
280
281
282
283
284
285
286
287
288
289
290
291
292
293
294
295
296
297
298
299
300
301
302
303
304
305
306
307
308
309
310
311
312
313
314
315
316
317
318
319
320
321
322
323
324
325
6
2
2
King
5
8
King
6
4
Ace
9
2
King
8
10
4
5
10
5
Queen
Ace
9
Queen
4
10
3
10
Jack
10
4
10
8
7
8
2
4
5
King
6
10
7
3
Jack
Jack
King
3
2
Clubs
Clubs
Hearts
Diamonds
Clubs
Spades
Hearts
Clubs
Spades
Clubs
Diamonds
Spades
Hearts
Clubs
Hearts
Diamonds
Hearts
Hearts
Spades
Hearts
Spades
Spades
Clubs
Spades
Spades
Spades
Clubs
Spades
Spades
Diamonds
Spades
Clubs
Spades
Hearts
Clubs
Diamonds
Clubs
Spades
Diamonds
Diamonds
Hearts
Diamonds
Clubs
Spades
Clubs
Clubs
Hearts
5
9
Ace
9
4
King
2
King
7
King
King
9
King
3
4
Queen
Ace
4
Jack
2
8
7
8
3
6
9
7
8
5
10
10
4
9
8
6
2
2
7
10
King
Ace
10
2
3
5
8
King
Clubs
Diamon…
Purchase answer to see full
attachment
