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- Ap Stats Casino Lab Answer Key Pdf
- Ap Statistics Casino Lab
- Ap Stats Casino Lab Answer Key Free
- Ap Answer Key
AP S TATISTICS C ASINO L AB: I NSTRUCTIONS The purpose of this lab is to allow you to explore the rules of probability in the setting of real-life games. You will get a chance to simulate playing several casino-type and other popular games of chance. Your task is to collect information/data about each game and answer the corresponding questions. Ap statistics test 8a answer key pdf - amazon s3 guide is also related with ap statistics test 8a answer key PDF, include: Answers To Stopping On A Dime Lab, Answers To Test Form 2a Algebra 1, and many other ebooks. Dec 31, 2013 AP S TATISTICS C ASINO L AB: I NSTRUCTIONS The purpose of this lab is to allow you to explore the rules of probability in the setting of real-life games. You will get a chance to simulate playing several casino-type and other popular games of chance. Your task is to collect information/data about each game and answer the corresponding questions.
1. P(green) = 0.70, P(red) = 0.45, P(redor green) = 0.8 Are red and green disjoint? Show your work and explain youranswer.
If red and green are disjoint, thenP(red and green) = 0. If this isn’t equal to 0, then they are not disjoint.
So you need to calculate P(red andgreen).Use the addition rule:
P(red or green) =P(red) + P(green) – P(red and green)
0.8 = 0.7 + 0.45 – P(red and green)
When you solve this equation, youget P(red and green) = 0.35.
Because it’s not equal to 0, red andgreen are not disjoint.
2.40 people are surveyed. Half of them likecats, half like dogs. Of the people who like cats, 10 live in houses. Of thepeople who like dogs, 15 live in houses.
Ap Stats Casino Lab Answer Key Pdf
lives in a house | lives in an apartment | ||
Download poker game for pc. likes cats | 10 | 10 | 20 |
likes dogs | 15 | 5 | 20 |
25 | 15 | 40 |
a.P(likes cats or lives in a house) = 20/40 + 25/40 – 10/40 = 35/40
Ap Statistics Casino Lab
b.P(likes cats and lives in a house) = 10/40
c.P(likes dogs or lives in an apartment) = 20/40 + 15 / 40 – 5/40 = 30/40
d.P(likes dogs | lives in an apartment) = 5/15
3.Suppose that among all U.S. students, 80% have gone to an amusementpark and 45% have gone to a beach. Only 15% have done neither.
a. If you select a student at random, what is theprobability that the student has gone to the beach but not to an amusementpark?beach yes | beach no | ||
park yes | 40 | 80 | |
park no | 5 | 15 | 20 |
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5% have been to the beach but not to the amusement park.
b. What is the probability that a randomlyselected student who has been to the beach has also been to an amusement park?This is a conditional probability, because weknow they have been to the beach before we even consider whether they've been to the park.
P(amusement park | beach) = 40/45 = .89
Out of the 45% who have been to the beach,40/45 = 89% have been to the amusement park.
4.If there is a high pressure front, there is a 60% chance of snow. If there isno high pressure front, there is a 5% chance of snow. There is a 25% chance ofa high pressure front. What is the probability that it will snow?
There are two ways it can snow: when there isa high pressure front, and when there is no high pressure front.
P(high pressure front and snow) = (0.25)(0.6)= 0.15
P(no high pressure front and snow) = (0.75)(0.05)= 0.0375
Ap Stats Casino Lab Answer Key Free
P(snow) = 0.15 + 0.0375 = 0.1875
Ap Answer Key
There is an 18.75% chance of snow.
You could also make a tree diagram with HighPressure Front vs. No High Pressure Front as the first branch, and Snow vs. NoSnow as the second. You could then multiply through the tree to get P(highpressure front and snow) and P(no high pressure front and snow), and thenadd them to get P(snow).