Exploring Discrete Probability Distributions- Analyzing Outcomes of a Six-Sided Die

Exploring Discrete Probability Distributions- Analyzing Outcomes of a Six-Sided Die

Question 1

a) The sample space of all achievable results after a single six-sided die is rolled onto (S = {1, 2, 3, 4, 5, 6}). Yes, the results are equally possible because the dice are fair, meaning that each face can equally land upwards.
b) All outcomes are equally likely, so we are only dealing with the probability of each outcome equal to 1/6. The sum of the probabilities is equal to 1, and it should be like that because the sum of the probabilities of all mutually exclusive outcomes in a sample space must also be equal to 1 (Grami, 2020). This is valid as a consequence of the stochastic algebra axioms.
c) Finding the probability of a number that is less than 4 on a single throw is equivalent to the combination of the probabilities that the number can be 1, 2, or 3. That is, 1/6 + 1/6 + 1/6 = 3/6 = 1/2.
d) This means there is a 1/3 chance for a 3 or 4 to appear on a single throw, which is equal to the probability of the 3 being presented together with the probability of the 4 being presented, which is 1/6 + 1/6 = 2/6 = 1/3.

Question 2

a. The outcomes on the two cards are not independent because the outcome of the first card affects the probability of drawing the second card. Since the first card is not replaced, the deck is reduced to 51 cards after the first draw, changing the probabilities of the second draw.
b. The probability P (ace on the first card and king on the second) can be calculated as follows:
Probability of drawing an ace on the first card:
= 4/52
Probability of drawing a king on the second card after drawing an ace:
= 4/51 (since there are 4 kings in the remaining 51 cards)
Combined probability:
= 4/52 * 4/51
= 16/2652
= 4/663
c. The probability P (king on the first card and ace on the second) is calculated similarly:
Probability of drawing a king on the first card:
= 4/52
Probability of drawing an ace on the second card after drawing a king:
= 4/51
Combined probability:
= 4/52 * 4/51
= 16/2652
= 4/663
d. The probability of drawing an ace and a king in either order is the sum of the probabilities from parts (b) and (c) above:
= P (ace first, king second) + P(king first, ace second)
= 4/663 + 4/663
= 8/663

Question 3

We have three pancakes:
1. Golden/Golden
2. Brown/Brown
3. Golden/Brown
You pick a pancake at random and see one side is brown. The possible pancakes you could have picked are either the Brown/Brown or the Golden/Brown since the Golden/Golden is out of the equation. Now, two sides are brown (one side from the Brown/Brown pancake and one side from the Golden/Brown pancake) and only one of those sides has brown on the other side. So, the probability that the other side is also brown is 1/2. Therefore, the probability that the other side of the pancake is brown is 1/2.

Question 4

In a lottery where numbers are drawn and then replaced (allowing for repetition), the draws are independent events. This means the outcome of one draw does not affect the others. The probability of drawing the number 8, then 2, then 5 in that order, with replacement, is:
P(8, then 2, then 5)
= P(8) * P(2) * P(5)
Since there are 10 possible digits (0-9) for each draw, the probability for each number is 1/10. Therefore:
P (8, then 2, then 5)
= 1/10 *1/10 * 1/10
= 1/1000
If repetition is not allowed, the draws are no longer independent because each draw affects the outcome of the subsequent draws. The probability would change because after drawing one number, it cannot be drawn again. For example, if you draw an 8 first, there are now only 9 numbers left to draw from for the second number. If you draw a 2 next, there would be 8 numbers left for the third draw. The probability calculation would be:
P (8, then 2, then 5, no repetition)
= 1/10 * 1/9 * 1/8
= 1/720
So, the probability of drawing 8, 2, and 5 in that order without repetition is 1/720.

Question 5

a) The probability of all 12 people being available is the probability of success in all 12 trials.
P = (0.75) ^12
= 0.0317
b) The probability of 6 or more not being available is the sum of the probabilities of having 6, 7, 8, 9, 10, 11, or 12 people not available.
Probability = 1 – P (0 not available) – P(1 not available) – P(2 not available) – P(3 not available) – P(4 not available) – P(5 not available)
Probability = 1 – (C(12, 0) × (0.75)^12 × (0.25)^0 + C(12, 1) × (0.75)^11 × (0.25)^1 + C(12, 2) × (0.75)^10 × (0.25)^2 + C(12, 3) × (0.75)^9 × (0.25)^3 + C(12, 4) × (0.75)^8 × (0.25)^4 + C(12, 5) × (0.75)^7 × (0.25)^5)
P = 0.0544
c) Find the expected number of those available to serve on the jury. What is the standard deviation?
The expected number of successes (people available) = n × p = 12 × 0.75 = 9
Standard deviation
= sqrt (n × p × (1 – p))
= sqrt (12 × 0.75 × 0.25)
= 1.5

Question 6

To create a probability distribution for a coin-flipping game, I will toss a coin at least 25 times and record the number of heads and tails. Here are the results of my 25-coin tosses:
a) The probability distribution from the simulation is as follows:
Outcome Frequency Probability
Head 14 0.56
Tails 11 0.44

This distribution meets the properties for a probability distribution because the probabilities add up to 1.
b) A bar graph to represent this distribution visually.

c) For this data, the random variable will be ‘Heads’ and ‘Tails’, respectively referring to the outcomes of a coin landing either on heads or tails. For each random variable, the simulation looks into the frequency of the event and bases the probability on that.
d) The values more or less have shown what might be expected with small sample sizes. A fair coin flip is expected to have a 50% chance for each outcome (Douneva et al., 2019); however, because randomness includes variations, the actual experiment figures showed it had 56% for heads and 44% for tails. Over more significant sample relationship numbers, we should normally observe the distribution to be close to 50% and a bit more balanced due to the law of large numbers (The Investopedia Team, 2019).

Question 7

To calculate the probability of getting heads 10 times or less in 25 coin flips, we use the binomial distribution formula:
P (X = k) = nkp^k 1-p^n-k
Where:
n is the number of trials (coin flips), which is 25.
k is the number of successful outcomes (getting heads), which ranges from 0 to 10 in this case.
p is the probability of success on a single trial, which is 0.5 for a fair coin.
We need to calculate this probability for each value of k from 0 to 10 and then sum them up to get the total probability of getting heads 10 times or less.
P (X ≤ 10) = Σ P (X = k) from k=0 to k=10
= P (X = 0) + P(X = 1) + … + P(X = 10)
= (C (25, 0) * (0.5)^0 * (0.5)^25) + (C(25, 1) * (0.5)^1 * (0.5)^24) + … + (C(25, 10) * (0.5)^10 * (0.5)^15)
= 0.2122
After performing the calculations, the sum of these probabilities is approximately 0.2122, confirming the probability of getting heads 10 times or less in 25-coin flips is around 21.22%. The calculation assumes that the coin is fair, meaning the probability of getting heads is equal to the probability of getting tails, and each flip is independent of the others.

Question 8

To solve these problems, we need to find the total number of employees and then use it to calculate the probabilities.
Total number of employees = 8 + 21 + 33 + 18 + 7 + 2 = 89
a. Probability of having a Ph.D.
= Number of employees with Ph.D. / Total number of employees
= 8 / 89
a. Probability of having an Associate degree
= Number of employees with an Associate degree / Total number of employees
= 18 / 89
b. Probability of not having a Ph.D.
= 1 – Probability of having a Ph.D.
= 1 – 0.0899
= 0.9101

References

Douneva, M., Jaffé, M. E., & Greifeneder, R. (2019). Toss and turn or toss and stop? A coin flip reduces the need for information in decision-making. Journal of Experimental Social Psychology, 83, 132–141. https://doi.org/10.1016/j.jesp.2019.04.003
Grami, A. (2020). Probability, random variables, statistics, and random processes: fundamentals & applications. John Wiley & Sons.
The Investopedia Team. (2019). Law of large numbers. Investopedia. https://www.investopedia.com/terms/l/

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Week 2 assignment:
The following assignment will allow you to master the concepts you have learned on discrete probability distributions.

1. Answer the following questions based on rolling a single six-sided die. (2 points for each part)

a) If you roll a single die and count the number of dots on top, what is the sample space of all possible outcomes? Are the outcomes equally likely?

b) Assign probabilities to the outcomes of the sample space of part (a). Do the probabilities add up to 1? Should they add up to 1? Explain.
c) What is the probability of getting a number less than 4 on a single throw?
d) What is the probability of getting a 3 or 4 on a single throw?

2. You draw two cards while playing Blackjack from a standard deck of 52 cards without replacing the first one before drawing the second. (2 points for each part)

a) Are the outcomes on the two cards independent? Why?
b) Find P(ace on first card and king on second)
c) Find P(king on first card and ace on second)
d) Find the probability of drawing an ace and a king in either order

3. You have three pancakes. One is golden on both sides, one is brown on both sides and one is golden on one side and brown on the other. You choose one pancake at random and see that one side is brown. Find the probability that the other side of the pancake is brown. (10 points)

4. You wish to play a three-number lottery with your favorite data 8, 2 and 5. After a number is drawn, the number is placed back in the container for the next draw. What is the probability that the three winning numbers will be 8, 2, and 5? Are the numbers independent of each other? How would the probability change if no repetition is allowed? (10 points)

5. Have you ever tried to get out of jury duty? About 25% of those called will find an excuse to avoid it. If 12 people are called for jury duty, (2 points for each part)

a) What is the probability that all 12 will be available to serve on the jury
b) What is the probability that 6 or more will not be available to serve on the jury
c) Find the expected number of those available to serve on the jury. What is the standard deviation

6. Create a probability distribution for a coin-flipping game. That is, toss a coin at least 25 times and keep up with the number of heads and the number of tails. (8 points for each part)

a. Compile your data into a probability distribution. Be sure to show that your distribution meets the properties for a probability distribution.
b. Use a bar graph to graph the distribution.
c. Explain the random variables for your data.
d. Did the values come out as you would expect? Explain what you would expect to happen with the outcome of the game versus the actual expected value of your distribution.

7. Use your example from question 6 to set up a binomial distribution. Find the probability based on your data that if the coin is flipped 25 times, what is the probability that a heads would appear 10 times or less. (10 points)

8. The following information has been given that shows the highest level of education received by employees of a company. (4 points for each part)

# Employees
Ph.D. 8
Master’s 21
Bachelor’s 33
Associate 18
High School Diploma 7
Other 2
Answer the following questions based on the data in the table:
a. Find the probability an employee chosen at random has a Ph.D.
b. Find the probability an employee chosen at random has an Associate degree.
c. Find the probability an employee chosen at random does not have a Ph.D.
d. Are the events independent? Explain.

Resource:
Brase, C. H., Brase, C. P., Dolor, J. M., & Seibert, J. A. (2024). Understandable Statistics (13th ed.). Cengage Learning US. https://online.vitalsource.com/books/9798214119830

Chapters: 4-5

02 Weblinks
Probability Games
This is a FUN site that allows you to play various games such as trying to win a car in order to help with your understanding of probability. The site has many games to choose from and gives a complete analysis of how the probability games work.

Dartmouth Chance Site
Are you a visual learner? Then this Chance site is the place for you. This site goes through video lectures to help you understand how probability works. From Bad Science to Quantifying Nature, this is sure to be a favorite for the visual learner.

The Probability Web
This probability site was actually put together for teachers and researchers. However, it offers many much of information on newsgroups and specific probability pages by subject area. Take a look to obtain to see a greater understanding of the application of probabilities.

Probability Activities
This is a great site to show you how probability works and then allow you to be able to construct distributions based on the probability data. There are some great activities such as chips in a bowl experiment and playing the lotto (which is always fun!). Be sure and check out the Yahtzee example for constructing two-way tables of probabilities.

HyperStat
Give it to me simple! This site starts out with great simple probability examples. It then grows to show how conditional probabilities work. There are even exercises that you allow you to can practice your probability skills.

02 Lecture Notes
Slide 1: Welcome!

Welcome to the second lecture of this course! You have already learned what statistics is all about and how to organize data. Next, we will discuss the uncertainty part of statistics, known as probability.

Slide 2: What is Probability?

In the world we live in there are many things that are uncertain. Everyday examples include the uncertainty of the weather, traffic, job promotions, etc. This uncertainty can be thought of as the likelihood of observing a particular occurrence. This likelihood is known as probability.

Slide 3: Events and Sample Spaces

When an experiment is conducted the outcome of the experiment is known as an event. The collection of all the events of an experiment is called the sample space. The sample space becomes important when listing all the possibilities for a particular experiment. For example, when a coin is flipped the sample space consists of the events the coin lands on heads or tails. The events are listed as a sample space for these events: S = {heads, tails}.

The event would be that we flipped the coin and got heads (or tails).

Slide 4: Sample Space Examples

Let’s now see if you get the hang of how to identify sample spaces. If a fair six-sided die is rolled, what would be the sample space? Did you get {1,2,3,4,5,6}? What would be the sample space of letter grades for this class? Did you come up with {A,B, C,D,F}?

Slide 5: Properties of Probabilities

Probabilities are expressed in numeric measures. The probability of any event must be between 0 and 1 inclusive. That is, 0 ≤ P(x) ≤ 1, where P(x) is the probability of the particular event x happening. The sum of all the events’ probabilities must also add to 1. That is, .

Slide 6: Classical versus Empirical Probability

Probabilities can be broken down into two parts: classical or empirical. The classical probability requires equally likely outcomes, that is, the probability of each event is the same; whereas empirical probabilities are based on actual data from a probability experiment.

Slide 7: Classical Probabilities

Classical probabilities of an event E are written as

For example, if a coin is flipped the number of possible outcomes is two (heads or tails). Notice that the number of possible outcomes is simply the number of values in the sample space. Thus, if we wanted to find the probability of flipping a fair coin and getting heads would be

because there is one way to get a head and two possible outcomes. Notice that the probability of getting tails would be the same. That is, all events have the same probability and this makes the probability categorized as classical. Also, it is important to recognize that classical probabilities are computed without performing an experiment.

Slide 8: Empirical Probabilities

Empirical probabilities are also referred to as relative frequencies. Empirical probabilities rely on the outcomes of a probability experiment. Empirical probabilities give approximate probabilities because each time an experiment is run the probabilities will lead to different outcomes. Empirical probabilities of an event E are written as