1) A casino in Las Vegas classifies its hotel guests as tourists, gamblers or convention-goers. Furthermore, they look to see if the guests use room service or not. Here are the results for one weekend:

Number of Guests: / Used Room Service / Did Not Use Room Service
Tourist / 437 / 1598
Gambler / 775 / 1392
Convention-goer / 1142 / 810

a)  If a guest is selected at random, what is the probability the guest is a Convention-goer or used room service?

437+775+1142+810 / 6154 = .514 = 51.4%

b)  If a guest is selected at random, what is the probability the guest is a tourist and used room service?

437 / 6154 = .0710 = 7.1%

c)  If a guest is selected at random, what is the probability the guest is a gambler given that they used room service?

775 / (437+775+1142) = .329 = 32.9%

d)  If a guest is selected at random, what is the probability the guest used room service given they were a convention-goer?

1142 / (1142+810) = .585 = 58.5%

e)  Are being a gambler and using room service independent?

P(gambler) = 2167 / 6154 = 35.2%

P(gambler | room service) = 775 / (437+775+1142) = 32.9%.

Since 35.2% ≠ 32.9% -- they are not independent.

2) You are part of a group of 10 people who take a cruise to Mexico and back. Upon return, the probability that a person will be selected to go through a time-consuming high-level scrutiny from customs is .05.

a)  What is the probability of no one in your group receiving additional scrutiny?

binompdf(10,.05,0) =.599 = 59.9%

b)  What is the probability of one person in your group getting additional scrutiny?

Binompdf(10,.05,1) = .315 =31.5%

c)  What is the probability of one or more people in your group getting additional scrutiny?

Use complement rule of x=0. 1 - binompdf(10,.05,0) = .401 =40.1%

d)  If two people or more in your group receive additional scrutiny, is this an unusual event, and why?

2 or more getting scrutiny is the complement of 1 or less.

1 – binomcdf(10,.05,1) = .0861 = 8.6%

So this not a rare event if a rare event probability is 5% (or 1% or even less than that)

e)  What is the mean and standard deviation of how many in your group will get additional scrutiny?

μ= np = 10*.05 = .5

σ = sqrt(np(1-p)) = sqrt (10*.05*.95) = sqrt(.475) = .689

3) Shark attacks. The rate of shark attacks on swimmers at California beaches is 2.3 per year. (note: shark attacks usually do not result in a swimmer’s death or even serious injury)

a)  What is the probability of no one being attacked in a year?

Poissonpdf(2.3,0) = .100 = 10%

b)  What is the probability of more than 2 people being attacked in a year?

Use complement rule. Complement of more than 2 people is a cdf of x=2

1 – poissoncdf(2.3,2) = .404 = 40.4%

c)  What’s the probability of at least 1 person being attacked in 3 months?

Use complement rule. At least 1 is 1 – pdf (or cdf) of x=0

3 months makes t=3/12 = ¼ or .25

1-poissonpdf(2.3*.25,0) = .437 = 43.7%

d)  What is the mean and standard deviation of shark attacks in one decade?

A decade is 10 years, so rate * t = 23.

μ= λt = 2.3*10 = 23

σ = sqrt(λt) = sqrt (23) = 4.80

4) Jury Selection

4 additional people need to be selected for a jury. The four people will be randomly selected from a pool of 7 students, 12 retirees, and 11 employed individuals.

a)  What’s the probability of the four people being made of 2 students and 2 retirees?

Chose 2 students from 7, and 2 retirees of 12: 7C2 * 12C2 / 30C4 = .0506 = 5.1%

Using probability rules: = 5.1%

The reason for the 6 on the end of that line is because there are six ways to arrange (4 combination 2): RRSS, RSRS, RSSR, SRRS, SRSR, SSRR

b)  What’s the probability of the four people being made of 1 student, 1 employed, and 2 retirees?

Chose 1 students from 7, 1 employed from 11, and and 2 retirees of 12: 7C1 * 11C1 * 12C2 / 30C4 = .185 = 18.5%

Using probability rules:

c)  What’s the probability of the group being made up of only retirees?

12C4 / 30C4 = 1.8%

Using probability rules:

5) The Oakland A’s baseball team is planning a promotional “ice cream social” as a fund raiser for community services. The amount of ice cream that will be served (in gallons) is a random variable whose value depends on who and how many come and that day’s weather. The ice cream consumption is normally distributed with a mean of 31.8 gallons and a standard deviation of 5.5 gallons.

a)  What’s the probability that people will consume more than 35 gallons?

Normalcdf( 35, 1e99, 31.8, 5.5) = .280 = 28.0%

b)  What’s the probability that people will consume between 30 and 33 gallons?

Normalcdf( 30, 33, 31.8, 5.5) = .215 = 21.5%

c)  How many gallons need to be ordered so that there’s a 95% chance that they will NOT run out of ice cream?

invNorm( .95, 31.8, 5.5) = 40.85 gallons. You can go ahead and say 41 gallons

6) What is the concept of “regression to mean” when applied to a random series? A local sun-worshipper hates cloudy weather and loves sunny weather. When it’s cloudy, this person forms a ritual to chase away the clouds. When it’s sunny, they perform a ritual to entice the sun to stay. They believe one of the rituals works, and one does not. Which ritual do they believe, and why?

(You’re on your own on this one).