GEN BUS 306: how to solve

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You enter an online cooking tournament. The entry fee is $15, and the entrants make the recipes at home and vote for their favorite, which cannot be their own recipe. The top 3 finishers win money. First place wins $59, and second place wins $47. Third place wins $3. You give yourself an X chance of finishing in the top 3, and if you finish top 3, then you have an equal chance of finishing first, second, or third. What is X such that entering this tournament is a fair gamble for you. A fair gamble is one where the expected value of entering the tournament is $0.

1 - 59 2 - 47 3 - 3 (1/3)(59)+(1/3)(47)+(1/3)(3) continue late r

You have a trick die. This die is weighted so that rolling a 6 is more likely than rolling any other number. Specifically, the chance of rolling a 6 is 14-times the chance of rolling any other number. And the chance of rolling a {1, 2, 3, 4, or 5} is the same for each number. What is the probability of rolling a 1 with this trick die?

1,2,3,4,5,6 x,x,x,x,x,6x calculate probability over total x's

On any given day, there is a 27% chance that I go running. On any given day, there is a 51% chance that I lift weights, or I run. On days that I lift, there is a 1 in 3 chance that I also go running. On any given day, what is the chance that I lift weights?

A = Run B = Lift P(A) = 0.27 P(A or B) = 0.51 P(A|B) = 1/3 Find P(B) Because we have an "or" statement, we will use the Addition Rule P(A or B) = P(A) + P(B) -P(A and B) Multiplication Rule: P(A and B) = P(B) * P(A|B) Substitute back into the Addition Rule: P(A or B) = P(A) + P(B) -P(B) * P(A|B) 0.51 = 0.27 + P(B) -(1/3) * P(B) 0.24 = (2/3) * P(B) P(B) = 0.36

On any given day, there is a 40% chance it is cloudy. On any given day, there is a 60% chance it is cloudy or rainy. On any given day, there is a 50% chance it is rainy. What is the chance it is cloudy and rainy?

A = cloudy, P(A) = 0.4 B = rainy, P(B) = 0.5 P(A or B) = 0.6 Solve for P(A and B) Addition Rule: P(A or B) = P(A) + P(B) -P(A and B) 0.6 = 0.4 + 0.5 -P(A and B) P(A and B) = 0.3

A sample of 1000 people, a mixture of teens and adults,were surveyed. Among the questions asked was "Do you enjoy reading?" Overall, 750 answered yes. Of those surveyed, 520 were teenagers, and 312 teenagers answered yes. a)What is the probability a random person likes reading and is a teenager? b)What is the probability a random person does not like reading and is an adult? c)What is the probability a random person does not like reading or is a teenager? d)What is the probability a random person likes reading or is an adult?

A = enjoy reading, P(A) = 0.75 and by Complement Rule, P(notA) = 0.25 B = teenager, P(B) = 0.52 and by the Complement Rule, P(adult) = P(notB) = 0.48 P(A|B) = 312520= 0.6 and by the Complement Rule, P(notA | B) = 1 -P(A|B) = 0.4 a. P(A and B) = P(B) * P(A|B) = 0.52 * 0.6 = 0.312 b. P(notA and notB). To solve for this, we need to pull more probability statements out of our data. There are 480 adults and (750 -312 =) 438 like to read, so 42 do not like to read. P(notA | notB) = 42480= 0.0875 P(notA and notB) = P(notB) * P(notA | notB) = 0.48 * 0.0875 = 0.042 c. P(notA or B) = P(notA) + P(B) -P(notA and B) P(notA and B) = P(B) * P(notA | B) = 0.52 * 0.4 = 0.208 P(notA or B) = 0.25 + 0.52 -0.208 = 0.562 d. P(A or notB) = P(A) + P(notB) -P(A and notB) P(A and notB) = P(notB) * P(A|notB) We solved forP(notA | notB) in part b), so using the Complement Rule, P(A|notB) = 1 -P(notA|notB) = 1 -0.0875 = 0.9125 P(A and notB) = 0.48 * 0.9125 = 0.438 P(A or notB) = 0.75 + 0.48 -0.438= 0.792

You enjoy bargain hunting at garage sales. From experience you know that at any random garage sale there is a 30% chance of finding a useful item. From experience, you know that at any random garage sale there is a 40% chance of finding a useful piece of furniture. Experience has also taught you there is an 80% chance of not finding a useful item or not finding a useful piece of furniture. Given that you find a useful piece of furniture at a garage sale, what is the probabilityyou do not find a useful item at that same garage sale?

A = item B = furniture P(A) = 0.3, P(notA) = 0.7P(B) = 0.4, P(notB) = 0.6 P(notA or notB) = 0.8 Find P(notA | B) Start with the Addition Rule since we have data with an "or" statement: P(notA or notB) = P(notA) + P(notB) -P(notA and notB) 0.8 = 0.7 + 0.6 -P(notA and notB) P(notA and notB) = 0.5 Multiplication Rule: P(notA and notB) = P(notA) * P(notB | notA) 0.5 = 0.7 * P(notB | notA) P(notB | notA) = 5/7 Complement Rule: P(B | notA) = 1 -5/7 = 2/7 Bayes' P(notA | B) = P(B | notA) * P(notA) / P(B) P(notA | B) = (2/7) * 0.7 / 0.4P(notA | B) = 0.5 There is a faster way to solve thisif yourecognize the complement of P(notA or notB). P(notA or notB) means A does not happen, or B does not happen, or both do not happen. And the Complement is: A and B both happen, or A&B P(A&B) = 1 -0.8 = 0.2 P(A&B) = P(B) * P(A|B) 0.2 = 0.4 * P(A|B) 0.5 = P(A|B) Complement Rule: P(notA|B) = 1 -P(A|B) = 1 -0.5 = 0.5

The probability I buy lunch on any given day is 40%. The probability I buy lunch and dinner on any given day is 20%. What is the probability I buy dinner? Your answer should be a range of probabilities.

A = lunch, P(A) = 40% B = dinner, P(B) = ? P(A and B) = 20% Here we are given the overlap is 20%, and we are trying to determine what possibilities for P(B) will match this overlap, given P(A) = 40%. If A and B overlap as much as possible, then P(B) = 20%. In this case, every time B happens, A also happens. Notice that B cannot be more than 20% in this case, as then the overlap would be greater than 20% and that violates the information we are given. If A and B are separate as much as possible, then P(B) = 80%. This is because we must have an overlap of20% based on our data. Since A = 40%, this also means that 20% of the time (40% -20% overlap) = 20% only A happens. And this leaves 60% of the time that A does not happen, in which case B could theoretically happen without conflicting with our information. Thus P(B) = 20% (given overlap) + 60% (when A does not happen, so B can happen) = 80% Answer: P(B) = [0.2, 0.8]

A medical studyon a new drugcollected data from sick volunteers, finding that 70% of the volunteerswere better at the end of the study. The volunteers were randomly assigned to one of two groups. One group received the new drugunder study, and theother group received a placebo(no drug). The study found thatof those in the placebo group, 40% were still sick at the end of the study. And of those in the new drug group, only 15% were still sick at the end of the study. What percentof the volunteers were in the placebo group?

A = placebo group, then notA = new drug group B = better at theend of the study, and thus notB = still sick at the end of the study We are given: P(B) = 0.7 This question can be challenging to write the probabilities accurately. "those in placebo group, 40% still sick at the end"is a conditional statementwhere the condition is in the placebo group (or Ain our notation) and "still sick at the end"is "notB"in our notation P(not B | A) = 0.4, and thecomplement is P(B|A) = 0.6 Similarly, P(notB | notA) = 0.15, complement P(B|notA) = 0.85 Find P(A). Use total probability:P(B) = P(A) * P(B|A) + P(notA) * P(B|notA) 0.7 = P(A) * 0.6 + (1-P(A)) * 0.85 0.7 = 0.6 P(A) + 0.85 -0.85P(A) -0.15 = -0.25 P(A) P(A) = 0.6

7 out of 10 days, Taylorgoes runningalong the Lakeshore Path. It rains with probability 0.3 in Madison at this time of the year. When it rains, Taylorrunsonly 10% of the time. If we know that Taylorrunson a given day, what is the probability that it rained on that day?

A = rain, P(A) = 0.3 B = run, P(B) = 0.7 P(B|A) = 0.1 Want P(A|B) = P(B|A) * P(A) / P(B) = 0.1 * 0.3 / 0.7 = 0.043

Data analytics shows that customers have a 40% chance of buying a shirt, and customers have a 70% chance of buying a belt. If a customer does not buy a shirt, that customer has a 75% chance of buying a belt. What is the probability that a customer who buys a belt does not buy a shirt?

A = shirt, P(A) = 0.4and by Complement Rule, P(notA) = 0.6B = belt, P(B) = 0.7 P(B | notA) = 0.75 Solve for P(notA | B) P(B and notA) = P(notA) * P(B | notA) = 0.6 * 0.75 = 0.45 P(B and notA) = P(B) P(notA | B) because we can write the Multiplication Rule either way0.45 = 0.7 * P(notA | B) P(notA | B) = 0.643

You are building your own personal email spam filter. You calculate that 20% of all your emails are spam. Analyzing your spam emails, you find the word "Prince" shows up in 50% of the time. Analyzing your non-spam email, you find the word "Prince" shows up3% of the time (it would have been 0% except you have one friend that loves Purple Rain). When your personal spam filter finds the word "Prince" in an incoming email, what is the probability that email is spam?

Movie studios are trying to determine the impact of streaming on theater revenue. They know thatfor any given movie, streaming and going to the movie theater are mutually exclusive. Recent data finds that 81% of people stream movies. If someone does not stream, there is a36% chance they go to the movie theater to see the movie. Given someone does not go to the theater to see the movie, what is the probability that person streams the movie?

A = stream movie B = watch movie in a theater Given: P(A) = 0.81, complement P(notA) = 0.19 P(B | notA) = 0.36, complement P(notB | notA) = 0.64 Find P(A | notB) Mutual exclusion is the key to this problem. It might appear there is not enough information to answer this question, but there is because we have mutual exclusion. Mutual exclusion tells us 2 events cannot happen together, and thus: P(A|B) = 0 and P(B|A) = 0. What are the complements? If P(A|B) = 0, then P(notA|B)= 1. That makes sense, ifevent B happened, then event A cannot happen. We can use these complements thanks to mutual exclusion. P(A | notB) = P(notB |A) * P(A) / P(notB)by Bayes Use total probability to find P(notB) = P(A) * P(notB| A) + P(notA) * P(notB | notA) = 0.81 * 1 + 0.19 * 0.64 = 0.9316 Note that we use mutual exclusion to find P(notB |A) =1 P(A | notB) = P(notB |A) * P(A) / P(notB)= 1 * 0.81/ 0.9316 = 0.869

60% of the time when it is sunny, Bucky buys an ice cream cone. It is sunny on 80% of all days.30% of the time when it is not sunny, Bucky buys an ice cream cone. Bucky is forgetful, and he only remembers his student ID 10% of the time, where the student ID gives Bucky a 10% discount on ice cream.On a random day, what is the probability that Bucky buys an ice cream cone and gets the student discount?

A= Buys Ice Cream B= Sunny P(B) = 0.8 P(A|B) = 0.6 P(A| notB) = 0.3 First,we find the probability of buying ice cream. To do this we use total probability. P(A) = P(B) * P(A|B) + P(notB) * P(A| notB) = 0.8 * 0.6 + 0.2 * 0.3 = 0.54 And we know there is a 10% chance we have our student discount., so the chance of getting the student discount = probability of buying ice cream * probability of remembering student ID = 0.54 * 0.10 = 0.054

Bucky likes to go to the Farmer's Market but will only go if it does not rain. Where he lives, it is rainy on 30% of all days. Bucky checks the weather forecast, but the forecast is not perfect. On days it rains, the forecast correctly predicts rain 80% of the time. On days when it does not rain, the forecast incorrectly predicts rain 10% of the time. The forecast predicts a rainy day tomorrow. What is the probability it rains tomorrow?

You have developed ananalytical model to rain. When it does not rain, you predictedno rain 91% of the time. When it rains, you predicted no rain 8% of the time.In your area, it does not rain 83% of the time. Given your model predicts raintoday, what is the probability it actually rains?

Taylordoes weight training at UW Natatorium regularly and eatshigh protein food. On a given day, Taylorweight trains with probability 0.4. If Taylortrains, Tayloreats high protein food 90% of the time. If Taylordoes not train, Tayloreats high protein food 50% of the time. Taylorate highprotein food today. What is the probability that Taylor trained today?

Bucky flies a lot and likes to upgrade his seat to first class. He knows that if he checks in for his flight at least two hours early, the chance that he will get the upgrade is 0.75; otherwise, the chance of an upgrade is 0.35. With his busy schedule, he checks in at least two hours before his flight only two out of five times. a. What is the probability that, for a randomly selectedtrip, Bucky will be able to upgrade to first class? b. Bucky did not receive an upgrade on his most recent attempt. What is the probability that he arrived late?

Human Resources at your company recently started using an analytical test to screen applicants. The normal distribution is a good approximation to the test results. The test returns a number between 0 and 100, with higher being better suited for the job. The HR group wants to accept approximately 2.5% of applicants with no review and reject approximately 16% of applicants with no review. The remaining 81.5% of applicants will be reviewed. They set their cutoff to reject at 52, and their cutoff to accept at 76. What are the mean and standard deviation of the applicant test results?

Accepting 2.5% with no review means they are scoring at a certain number or higher. The associated z-score is 2. Rejecting 16% means scoring at a certain number or lower. The associated z-score is -1. Set up the 2 equations using z-score: 2 = (76− 𝜇)/𝜎 -1 = (52− 𝜇)/𝜎 2 𝜎 = 76− 𝜇 and -𝜎 = 52− 𝜇 Solve by whichever method you prefer for 2 equations and 2 unknowns to find that 𝜎 = 8 and 𝜇 = 60

You roll a fair die two times. What is the probability you see exactly one 6 in your result?

Addition Rule: P(A or B) = P(A) + P(B) -P(A and B) = 1/6 + 1/6 -1/36 = 11/36. This means A happens, or B happens, or both happen. But since we want exactly 1, we remove both happen. We take our answer 11/36 -1/36 (both happen) = 10/36 which matches our counting work above.

You work in analytics for a hotel chain. You are asked to predict how many guests will not show up for their reservation tonight. There are 250 reservations tonight, and past data suggests that 12% of guests do not show up for their reservations. Assume each guest is independent of all others, and they all have the same chance of not showing up. What is your best prediction?

Because there are a set number of trials (250), with independence between trials and the same probability for each trial, this is a binomial distribution. The mean is our best guess for the number of people who will not show up. The mean of the binomial is np = 250 * 0.12 = 30 guests.

Basketball Free Throw Problem

Draw out the map and different ways of winning or not and use that to calculate the answer.

A quiz has 5 multiple choice questions, with each question having 4 possible answers, of which only one is correct. If a student randomly guesses on every question, what is the probability they get all five questions wrong?Assume each question is independent of all the others.

Each question has 4 choices, only 1 is correct. A random guess has a 25% (1 in 4) chance to be correct and an 75% chance to be incorrect. There are 5 questions, all independent. The probability of getting them all wrong = P(Q1 wrong) * P(Q2 wrong) * P(Q3 wrong) * P(Q4 wrong) * P(Q5 wrong) = 0.75^5 = 0.237

Assume you are rolling two fair, six-sided dice. After the roll, you add the results of each die to find your roll total, a number between 2 and 12.You roll a 9. Your friend now rolls the dice and stops the first time they get either a 9 or a 7 (either roll ends the turn). You win if your friend rolls a 9 first. Your friend wins if they roll a 7 first. What is the probability of you winning this game?

List the ways that two dice can total 9 (listed as first die result, second die result): 3,6 4,5 5,4 6,3 Thus, there are 4 ways out of the 36 total combinations that can be rolled (6*6 = 36) Similarly, the ways to roll a total of 7 are:1,6 2,5 3,4 4,3 5,2 6,1 Thus, there are 6 ways out of the 36 total combinations that can be rolled. Since all we care about is rolling a 7 or a 9, we can ignore the other results and compare the ways to roll 7 and 9 directly. There is a total of 4 ways to roll a 9, and 6 ways to roll a 7. So out of the 10 total ways to roll a 9 or a 7, you win on 4 of those ways. Thus P(you winning by your friend rolling a 9 first) = 4/10 = 40%

Every day to concludebasketball practice, Riley shoots 6 free throws. If Riley makes all 6, Riley marks the day a success. If Riley does not make all 6, Riley marks the day as a failure. Assume that each free throw is independentof the others, and that for any given free throw, Riley has an 84% chance to make it. After 5 practices(days), what is the chance that Riley had at least 1 day marked successfulin those last 5?

Note the free throws are all independent. This means we do not have to worry about conditional probabilities. The chance to make 6 out of 6 shots = P(making 1st) * P(2nd) * P(3rd) * P(4th) * P(5th)* P(6th)= (0.84)6= 0.3513 On any given day, Riley has a 35.13% chance of that day being a success.Or, to be a pessimist and use the complement, on any given day, Riley has a 64.87% chance of failing. Next note the wording of the question "at least 1". When we see words like "at least"or "at most", that is often a signal that we should consider whetherit is easier to use the complement. We wantP(at least 1 success out of 5 days) = P(1 success) + P(2 successes) + .. + P(5 successes), or we can use the complement = 1-P(0 successes) It is much faster to calculate P(0 successes) = P(1stday fail) * P(2ndday fail) * P(3rdday fail) * P(4thday fail) * P(5thday fail), and since we have independence, this is= P(fail)5= (0.6487)5= 0.11488 P(at least 1 day out of5 is a success)= 1 -P(0 successes out of 5 days) = 1 -0.11488 = 0.88512

A survey finds that 0.66 of all students have an iPhone. And the survey finds that 0.37 of all students have a MacBook and an iPhone. What is the probability that a student has a MacBook given that student has an iPhone?

P(A)=MacBook=0.66 P(B)=iPhone P(A&B)=0.37 0.37=0.66x x=0.56

You have taken a new analytics role for a bookstore. The previous person left you a mess of unorganized data. In that mess you find a few probabilities. The probability of a customer buying a book is 0.54. The probability of a customer buying a magazine and book is 0.15. The probability of a customer buying a magazine or book is 0.51. What is the probability a customer buys a magazine?

P(A)=books=0.54 P(B)=magazine=? P(A&B)=0.15 P(A or B)=0.51 P(A or B)=P(A)+P(B)-P(A&B) 0.51=0.54+x-0.15

You work for a company that administers drug tests for athletes. Based on your data analytics and collected data, you find that 20% of athletes use illegal drugs. From your lab results, you know that 24% of drug-free athletes mistakenly test positive. And you know that 16% of drug-taking athletes mistakenly test negative. If an athlete tests positive with your test, what is the probability that athlete uses illegal drugs?

P(A)=drugs=0.2 P(B)=positive test P(BIA)=0.84 P(BI/A)=0.24 P(/BIA)=0.16 P(B)=(0.2)(0.84)+(0.8)(0.24)=0.36 P(AIB)=((0.84)(0.2)/(0.36))

You have an unfair coin. When flipped, it has a 88% chance of landing on heads, and a 12% chance of landing on tails. You flip the coin 4 times. What is the probability you see at most 3 heads in the results? Please answer as a decimal, rounded to 2 digits. For instance, if your answer is 3%, you would enter 0.03.

P(A)=heads=0.88 P(B)=tails=0.12 P(at most 3 heads)? Complement Rule: (.88/100)^4 1-^^

Where you live there is rain 19% of the time. On days it rains, there is an 62% chance of traffic during rush hour. On days it does not rain, there is a 25% chance of traffic during rush hour. On any given day, what is the probability there is traffic during rush hour?

P(A)=rain=0.19 P(/A)=0.81 P(BIA)=0.62 P(BI/A)=0.25 P(B)=(0.19)(0.62)+(0.81)(0.25)

I roll a fair die 4 times. What is the probability I see at least one 6 in my result?

P(A)=roll a 6=1/6 P(/A)=5/6 complement is P(zero 6's) = P(/A)^4 P(A)=1-0.4823=0.5177

You are the manager of a clothing store. From collecting data, you know a random customer has a 0.27 chance of buying a shirt. And you know that a random customer has a 0.26 chance of buying pants. Finally, your data tells you that a random customer has a 0.03 chance of buying a shirt and pants. What is the probability a random customer buys a shirt or pants?

P(A)=shirt=0.27 P(B)=shorts=0.26 P(A&B)=0.03 P(A or B)=P(A)+P(B)-P(A&B) 0.5=0.27+0.26-0.03

You run analytics for an electronics store. You find that a random customer has a 0.67 chance of buying a video game. And your data analytics finds that a random customer who buys a video game has a 0.76 chance of buying a gift card. If a random customer comes into your store, what is the chance that customer buys a video game and a gift card?

P(A)=video game=0.67 P(B)=gift card P(BIA)=0.76 P(A&B)=0.67*0.76=0.51

A survey determines that in a particular town, 32% of the residents swim, 35% bike, and 12% do both activities. What is the probability that a randomly selected person in this town does neither activity?

P(S) = 0.32 P(B) = 0.35 P(S&B) = 0.12 The chance that someone does neither activity is equal to 1 minus the chance a person does either activity. P(notS & notB) = 1 -P(S or B) P(S or B) = P(S) + P(B) -P(S&B) = 0.32 + 0.35 -0.12 = 0.55 P(notS & notB) = 1 -0.55 = 0.45

Data analysis at the Apple store finds that any random customer has a 41% chance of buying a phone. And the analysis finds that there is a 78% chance that I random customer buys a phone or a watch. Finally, the analysis finds that if a random customer buys a watch, there is a 23% chance they buy a phone. For a random customer, what is the chance that customer buys a watch?

Set A = buy phone, and B = buy watch We are given: P(A) = 0.41 P(A or B) = 0.78 P(A | B) = 0.23 Want to find P(B) Addition rule states P(A or B) = P(A) + P(B) -P(A and B) Multiple rule states P(A and B) = P(B) * P(A | B) Substitute multiplication rule into the addition rule: P(A or B) = P(A) + P(B) -P(B) * P(A | B) 0.78 = 0.41 + P(B) -P(B) * 0.23 and solve for P(B) 0.78 -0.41 = P(B) -0.23 * P(B) 0.37 = P(B)* (1 -0.23) 0.37 = P(B) * (0.77) P(B) = 0.37 / 0.77 P(B) = 0.481

A nationwide survey finds that 50% of people like to read books. Of those people that like to read, 40% also like going to movies. Of the surveyed people that like going to movies, whatis the minimum percent that like to read?

Set A = read, and B = movies We are given: P(A) = 0.5 P(B|A) = 0.4 We want minimum of P(A | B). We ask for the minimum as we do not have enough information to give an exact number, all we do is provide a range. Use what we are given to find P(A & B) = 0.5 * 0.4 = 0.2. This tells us 20% of people like to go to movies and read books What do we know about the people who don't like to read? Nothing! When we work on probability range, we look at the endpoints: •Case 1: Of the people who do not like to read, 0% like to go to movies. Then the people who go to movies are the 20% from the readers + 0% from the non-readers = 20% total •Case 2: Of the people who do not like to read, 100% like to go to movies. Then the people who go to movies are the 20% from the readers + 50% from the non-readers (this is 100% of the 50% non-readers) = 70% total Solve for P(A | B) = P(A & B) / P(B) in each case: •Case 1: P(A | B) = 0.2 / 0.2 = 100% •Case 2: P(A | B) = 0.2 / 0/7 = 2/7 The minimum is thus 2/7

When more than 6 pitchers are used in a baseball game, 75% of those baseball games go longer than 3 hours. When 6 or fewer pitchers are used in a baseball game, 80% of those baseball games go less than 3 hours. 60% of baseball games use more than 6 pitchers. You saw a game that went less than 3 hours. What isthe probability that game used more than 6 pitchers?

Some questions may appear confusing because they have many "not"statements in them. Do not let this throw youoff. Work carefully and be sure you correctly define and write your probabilities. A = more than 6 pitchers B = longer than 3 hours P(B|A) = 0.75 P(notB | notA) = 0.8 P(A) = 0.6 Find P(A | notB) P(A|notB) = P(notB|A) *P(A) / P(notB) P(notB) = P(A) P(notB|A) + P(notA) P(notB|notA) = 0.6 * 0.25 + 0.4 * 0.8 = 0.47 Note P(notB|A) = 0.25 from the completement rule -completement of P(B|A) Thus P(A|notB) = 0.25 * 0.6 / 0.47 = 0.319

A casino offers a gamble where you bet $100 and draw two cards from a standard deck of cards. If the cards match suits, you win $300, and if they do not match suits, you lose your $100. a) What is the expected value of this gamble? b) What is the variance of this gamble?

The chance of drawing 2 cards from a standard deck and having the suits match is 12/51. There are 52 cards in the deck. View this sequentially. We draw the first card, and it has a suit. Since there are 13 cards in each suit, if we draw 1 of that suit, thereare 12 left. And there are 51 cards total since we drew one. Thus, the chance of matching the suit is 12/51. a) E(gamble) = $300 * (12/51) + (-$100) * (39/51) = -$5.88 b) Var(gamble) = (12/51) * (300 - -5.88)^2 + (39/51) * (-100 - -5.88)^2 = 28,788.93 $2

National results for the SAT test show the average combined SAT Writing, Math and Verbal score is 1490 and the standard deviation is 250. National results for the ACT test show that the average composite ACT score is 21 and the standard deviation is 4.5. You took the ACT and your friend took the SAT. You scored a 25 on the ACT, and your friend had a combined SAT of 1700. Who scored higher relative to the other takers of the same test? Assume that a normal distribution is a good fit for both the composite ACT and the combined SAT score.

The z-score = (X -mean) / standard deviation SAT z-score = (1700 -1490) / 250 = 0.84 ACT z-score = (25 -21) / 4.5 = 0.89 Recall the z-score tells us how many standard deviations you are away from the mean, and that the higher the z-score, the further from the mean you are. Since your z-score is higher than your friend's, you scored higher in terms of having more of the population of test takers scoring below your score. If you wanted the percentile of the scores, you could use the Excel function NORM.S.DIST to find that: NORM.S.DIST(0.89,true) = 0.81 NORM.S.DIST(0.84,true) = 0.80

You flip 4 fair coins. What is the probability you see exactly 3 heads? As above, for now try to answer this by counting outcomes.

There are 16 total possibilities, and 4 of them have exactly 3 heads. Thus P(exactly 3 heads in 4 flips) = 4/16 = ¼ = 0.25

A basketball player has a 75% chance of making a free throw. Assuming each free throw is independent of the others, what is the mean and variance of 100 free throws attempted by this player?

This is a binomial distribution, where n = 100 and p = 0.75 Mean of a binomial = np = 100 * 0.75 = 75 Variance of a binomial = np(1-p) = 100 * 0.75 * 0.25 = 18.75

A local carnival has a game using a modified roulette wheel with numbers 1 to 40 on it. Assume the wheel is fair, with each number being equally likely to be the result of a spin. The carnival charges $20 to spin the wheel and pays out a dollar amount equal to the result of the spin. Is this a fair game? And if not, who does the game favor?

This is a discrete uniform distribution, where a = 1 and b = 40 Mean of a discrete uniform distribution = (a+b)/2 = (1+40)/2 = 20.5 Since you expect to win $20.50, and you pay $20 to play, you expect to win $0.50 per spin. This game favors you.

Badger's Bakery sells blueberry muffins and chocolate cookies. On a typical weekday, the demand for cookies is normally distributed with a mean of 450 and standard deviation of 80. How many cookies must the restaurant stock to limit the probability of running out to 2.5%?

To limit the probability of running out to 2.5%, Badger's needs to make enough cookies such that the z-score is 2. Thus 2=(X-450)/80. ThusX=610. Badger's must make 610 cookies a day.

Three friends are at a carnival and playing the ball toss. In this carnival game, you have 1 throw of a baseball to knock down 3 stacked bottles. The probability that the first friend wins is 34%. The second friend has a 28% chance to win. And the third friend, a pitcher on the baseball team, has an 82% chance to win. Assume each friend plays the game once, and their results are independent of the others. What is the probability that they win exactly 1 prize?

We are given independence, so the chance of winning exactly 1 prize is the sum of the different ways that the 3 friends collectively win 1 prize. That is, the first friend wins and the next 2 lose, or the first and third lose while the second wins, or the first and second lose while the third wins. P(first wins) = 0.34, and thus by complement rule, P(first loses) = 0.66. P(second wins) = 0.28, and thus by complement rule, P(second loses) = 0.72. P(third wins) = 0.82, and thus by complement rule, P(third loses) = 0.18. P(exactly 1 win) = (0.34 * 0.72 * 0.18) + (0.66 * 0.28 * 0.18) + (0.66 * 0.72 * 0.82) = 0.044 + 0.033 + 0.390 = 0.467

Your company makes VR headsets. Past data analysis finds that 80% of the headsets pass final inspection. For the next 3 headsets: a) What is the chance exactly 2 pass inspection? b) What is the chance at least 1 passes inspection?

Write the different ways exactly 2 out of 3 can pass: PPF, PFP, FPP The probability is then the sum of the probabilities for each of those 3 situations 0.8*0.8*0.2 + 0.8*0.2*0.8 + 0.2*0.8*0.8 = 3*0.2*0.8*0.8 = 0.384 P(at least 1 passes) = 1 - P(0 pass) P(0 pass) = 0.2 * 0.2 * 0.2 = 0.008, thus P(at least 1 passes) = 1-0.008 = 0.992

You have an unfair die that is 4 times as likely to roll a 6 as it is to roll any other number. And the probability of rolling any other number is the same. What is the probability of rolling a 6 on this die?

X = probability of rolling any number other than 6 4X = probability of rolling 6 Total all possibilities = P(roll 1) + P(roll 2) + P(roll 3) + P(roll 4) + P(roll 5) + P(roll 6) = X + X + X + X + X + 4X = 9X = 1 since probabilities sum to 1 as one of these numbers must be rolled. X = 1/9 P(roll 6) = 4/9

I make 65% of my free throws. I take 5 free throws(assume the probability is the same for each free throw) a.What is the probability I make at most 4? b.What is the probability I make at least 1? c.What is the probability I make at most 3?

a. A = free throw, P(A) = 0.65 and P(𝐴̅) = 0.35 P( at most 4) = 1 -P(make 5), using the complement rule as P(make 5) is much easier to calculate. P(make 5) = 0.65^5= 0.116, thus P(at most 4) = 1 -0.116 = 0.884 b. P(at least 1) = 1 -P(make 0), again with complement rule P(make 0) = 0.355= 0.005, thus P(at least 1) = 1 -0.005 = 0.995 c. P(at most 3)? Complement rule will still be easier as P(at most 3) = P(make 0) + P(make 1) + P(make 2) + P(make 3), whereas the complement, P(4 or more) = P(make 4) + P(make 5). P(make 5) = 0.116 as we calculated above.P(make 4) we count out as there are 5 ways to make exactly 4 shorts. S = shot made and M = shot missed, here are the sequences that are exactly 4 out 5: SSSSM, SSSMS, SSMSS, SMSSS, MSSSS Notice that the probabilities S = 0.65 and M = 0.35 donot change for each shot, so we have S4* M1= probability for each sequence, and then we multiply by 5 since there are 5 sequences listed for our probability. 5 * 0.654* 0.351=0.312 Answer: P(at most 3) = 1 -P(4 or more) = 1 -0.312 -0.116 = 0.572

You flip a fair coin 4 times, counting each head as 1 and each tail as 0. Assume each flip is independent of the others. a. What is the mean of the 4 coin flips? b. What is the probability of scoring at least 3? c. What is the probability of scoring at least 1? d. What is the probability of scoring exactly 2?

a. For each flip, there is a 0.5 chance of heads (1) and a 0.5 chance of tails (0). So the mean of each flip = 0.5 * 1 + 0.5 * 0 = 0.5. Since there are 4 flips, the mean of 4 flips = 4 * mean of a single flip due to independence = 4 * 0.5 = 2 b. P(at least 3) = P(3 heads) + P(4 heads) There are 4 ways of getting exactly 3 heads: HHHT, HHTH, HTHH, THHH Thus P(3 heads) = 4 * (1/2) * (1/2) * (1/2) * (1/2) = 0.25 There is only 1 way to get 4 heads, thus P(4 heads) = (1/2)4= 0.0625 P(at least 3 heads) = 0.25 + 0.0625 = 0.3125 c. P(at least 1) = 1 -P(0 heads) P(0 heads) = (1/2)4= 0.0625 And P(at least 1 head) = 1 -0.0625 = 0.9375 d. There are 6 ways to flip exactly 2 heads out of 4 flips (try to list out yourself)P(2 heads) = 6 * (1/2)4= 0.375

The probability my store sells a t-shirt on any given day is 40%. The probability my store sells a sweater on any given day is 50%. a.What is the probability I sell a t-shirt or sweater on any given day? Your answer should be a range of probabilities. b.What is the probability I sell a t-shirt and a sweater on any given day? Your answer should be a range of probabilities.

a. P(A)=t-shirt=0.4 P(B)=sweater=0.5 P(A or B) = P(A) + P(B) -P(A and B) P(A or B) = 0.4 + 0.5 -P(A and B) Check the endpoints. If A and B do not overlap as much as possible, we can theoretically have only days where we sell a t-shirt (40%), days where we only sell a sweater (50%), and thus 10% of days where we sell nothing. In this case, P(A or B) =90%, while P(A and B) =0%. If A and B overlap as much as possible, then we can have days where if we sell a t-shirt (40%), we also sell a sweater (50%). Thus, the maximum overlap is 40% -every day where we sell a sweater. In this case, P(A and B) = 40% and P(A or B) = 50%. Answer: P(A or B) = [0.5 , 0.9] P(A and B) = [0, 0.4]

You are a business analyst working for VisaStar. Your job is to monitor customers'credit card activity for signs of possible theft. From previous experience, you know the following facts: There is a 0.5% chance that any credit card used is a stolen one. A stolen credit card has a 90% chance of being used for multiple large online purchases in a single day. By contrast, a non-stolen credit card has only a 1% chance of being used for multiple large online purchases in a single day. The two events in this problem are: S = Credit card is a stolen one L = Multiple, large, online purchases in one day a. What are P(S), P(L|S), and P(L| not S)? b. What is P(L), or theprobability of seeing multiple large online purchases in a single day from any credit card? c. A customer you are monitoring purchases Darryl Sims' 1982winning jersey($1,500)and fifty $20 Qdobagift cards($1,000)within minutes of each other. What is the probability that the credit card used for thesetransactionsis a stolen one? Translate this question to a probability in terms of S and L and show your steps using Bayes' Rule.

At the Middle Earth Securities Exchange, stock prices of listed firms will either go up or go down each day. Each day: •The stock price of Mordor Foundries Inc will go up with probability P(MF) = 0.30 •The stock price of Rivendell Textiles Inc will go up with probability P(RT) =0.50 •The stock price of Erebor Constructions Inc will go up with probability P(EC)=0.70 a.True or False: based on the information given above, Rivendell Textiles (RT) and Erebor Constructions (EC) cannot be mutually exclusive events. b.Middle Earth Investments, the leading investment bank operated by Sauron & Sons, offers a financial product that has the following payoff: +$100 (i.e. you gain $100) if prices of both Mordor Foundries (MF) and Rivendell Textiles (RT) go up together over a day, and -$40 (i.e. you lose $40) otherwise. Frodois interested in playing the market and wants to know the probability range for this financial product paying off.

a. True. Mutually Exclusive means that P(A or B) = P(A) + P(B), because P(A&B) = 0.P(RT) + P(EC) = 0.5 + 0.7 = 1.2 > 1 so RT and EC cannot be Mutually Exclusive. b. We are interested in knowing P(MF & RT) = P(MF) + P(RT) -P(MF or RT) = 0.3 + 0.5 -P(MF or RT) = 0.8 -P(MF or RT). P(MF or RT) could range between 0.8 (if MF and RT do not overlap at all) and 0.5 (if MF and RT overlap completely). Thus P(MF & RT) ranges between 0 and 0.3.

You bought 3 loot boxes in your favorite online game. Each box has a 1% of a legendary item, and a 99% chance of getting junk. Your game has a rule that when you buy 3 loot boxes at once and open them immediately, for each junk item you receive, the chance of getting a legendary item increases by 10 percentage points (i.e., after 1 junk item the chance of legendary is 11%, and after 2 junk items, the chance of legendary is 21%). The chance resets after each legendary received. a. What is the probability of finding at least 1 legendary item from the 3 loot boxes? b. What is the probability of getting exactly 1 legendary item from the 3 loot boxes? c. If the probabilities did not change (as is true for most loot boxes), what is the chance of getting at least 1 legendary item from the 3 loot boxes?

a. Use the complement rule: P(at least 1 legendary from 3 boxes) = 1-P(0 legendary from 3 boxes) P(first box has 0 legendary) = 0.99 P(second box has 0 legendary given first has 0 legendary) = 0.89 P(third box has 0 legendary given first and second have 0 legendary) = 0.79 P(0 legendary from 3 boxes) = 0.99 * 0.89 * 0.79 =0.696 P(at least 1 legendary) = 1 -0.696= 0.304 b. P(exactly 1 legendary) = P(1 legendary in first box, 0 in second and third) + P(1 legendary in second box and 0 in firstand third) + P(1 legendary in third box and 0 in first and second) P(1 legendary in first box, 0 in second and third) = 0.01 * 0.99 * 0.89 = 0.009 P(1 legendary in second box and 0 in first and third) = 0.99 * 0.11 * 0.99 = 0.108 P(1 legendary in third box and 0 in first and second) = 0.99 * 0.89 * 0.21 = 0.185 P(exactly 1 legendary) = 0.009 + 0.108 + 0.185 = 0.302 c. P(at least 1 legendary from 3 boxes) = 1-P(0 legendary from 3 boxes) P(0 legendary from 3 boxes) = 0.99 * 0.99 * 0.99 = 0.97 P(at least 1 legendary from 3 boxes) = 0.03

Assume you are faced with the following car-insurance type lottery: with probability 80%, you lose nothing; with probability 15%, you incur a $1,000 loss; with probability 5%, you incur a $10,000 loss. This lottery is a random variable. a. What is the expected value of this random variable? b. What is the variance of this random variable?

a. EV= (.8)(0)+(.15)(-1000)+(.05)(-10,000) EV= -$650 b. Variance=0.8(0 - (-650))2 + 0.15(-1000 - (-650))2 + 0.05(-10,000 - (-650))2=4,727,500 $2

You have 10 extra tickets to a virtual concert. Assume each of your friends has a 74% chance of wanting to attend the virtual concert, and assume each friend decides independently of all the other friends. a) If you invite 13 friends, what is the probability that exactly 10 accept? b) What are the fewest number of friends you should invite if you want to have at least a 90% chance of having all 10 tickets used?

a. This is a binomial since we have independent trials of an event that either happens or does not happen. As we are looking for exactly 10 successes out of 13 attempts, we should use the Binomial Formula. In Excel, use "=BINOM.DIST(10,13,0.74,FALSE) = 0.25 b. We again use the Binomial Formula, only now we want to use the cumulative version as we want to determine what n is to ensure that 10 or more success is 90% or greater. We want to set up a spreadsheet where we can vary the number of friends (n) that we invite until we find a cumulative chance of 10+ to be 90% An example of a spreadsheet that would work is below. The answer is n = 16 The "Exact" column uses BINOM.DIST(s,n,p,FALSE) The "0 to s" column is the cumulative BINOM.DIST(s,n,p,TRUE) The ">s" column is the one of interest for our work. Since we want 10 or more, we are specifically interested in the highlighted row for s=9. And that is because the column is ">s", so the highlighted column is "greater than 9" or in other words, "10 or more" The ">s" column is 1 - "0 to s" column. Since the "0 to s" column is the cumulative up to s, the complement is ">s"

The company you work at finds that the normal distribution is a good representation of daily sales, with mean 20 and variance 9. a) How often should you expect to see 29 sales or fewer? b) How often should you expect to see 17 sales or more? c) How many items should you stock daily if you want the chance of running out on any given day to be less than 2.5%?

a. z-score = (29 - 20) / 3 = 3. We divide by 3 since we are using the standard deviation, which is the square root of the variance. We are looking for 29 or fewer sales, so want to read our normal chart from the left. For 3 standard deviations above the mean, our chart shows 99.85%. Thus 99.85% of the time we should expect to see 29 sales or fewer b. Now we are looking for 17 sale or more, and thus want the area to the right in our chart. z-score = (17 - 20) / 3 = -1. From our chart, 1 standard deviation below the mean is 16%, but since we want the area to the right, we use 100% - 16% = 84%. Thus 84% of the time we expect to see 17 sales or more. c. We are looking for the chance to run out to be 2.5% or less. Or in other words, we want to have inventory 97.5% of the time. From our chart, 97.5% corresponds to a z-score of 2. Solving for x, we have 2 = (x - 20) / 3, or x = 26.

A friend proposes a gamble to you. You roll 2 fair, six-sided dice, and add up the results of each die to find your total number. If your roll totals a 5 or 11, your friend pays you $45. If you roll any other total, you pay your friend $8. What is your expected payoff from playing this game 100 times? Please answer to 2 decimal places, and do not use any unit in your answer. A negative answer means you expect to lose money playing this game.

calculate the chances of rolling a 5 and a 11 calculating 5 = 1/9 calculating 11 = 4/36 together = 6/36 (6/36)(45)-(30/36)(8) = 0.833 * 100

A survey of random sports fans finds that 60% like soccer. The same survey finds that 30% like soccer and basketball. What can you say about how the proportion of fans in that survey that like basketball?

draw out the soccer probability then soccer and basketball on both sides to find the range

What is the interquartile range of the following numbers?

find upper median and lower median then subtract both values to find IQR

Your online store only sells one item, in either a regular or deluxe version. Because of high demand and restricted supply, you restrict each customer to buying 1 item - either buying the regular version of that item, or the deluxe version of that item, but not both. On a typical day you have 194 customers visit your online store. Each customer has a 0.31 chance of purchasing an item. If a customer purchases an item, there is a 6% chance they buy the regular item for $5.16, and if they do not buy the regular item, then they buy the deluxe item for $23.28. Assume each customer makes their purchase decision independently of all other customers, and each customer has the same chance of purchasing the item. What is your expected daily sales?

first calculate the EV given mean and SD find the conditional probability and complement (EV)P(buy regular)+(EV)P(buy deluxe)

Calculate the variance, and round to 2 decimal places.

put numbers in excel and use the variance formula

Calculate the average absolute deviation, and round to 2 decimal places.

put the numbers in excel and use the AAD formula

What is the correlation in the chart below?

you have R^2, unsquare it to find the correlation, make sure negative or positive depending on direction

You make special shoes for dogs. The shoe sizes that you sell follow a normal distribution with a mean of 15 and standard deviation of 3(paw sizes follow a different standard than human shoes). You are planning to make 100 sets of shoes to sell. If you want to match expected demand as closely as possible; how many sets of shoes should you make that are size 12 or larger?

z-score=(12-15)/3=-1 From our normal table, z-score= -1 is associated with 16% of the area to the left of -1 standard deviations below the mean. Since we are interested in shoe sizes 12 and larger, we take 1-0.16=0.84, or 84% of our shoes should be made size 12 or above. Since we are making 100. sets of shoes, this means 84 sets should be size 12 or larger.

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