Ap Stats Casino Lab Answer Key
Chapter 7 2 Casino Lab Casino Lab Solutions STATION 1. The probabilities on the branches of the tree diagram should be the answer to parts (c)1, 2, and 3. Chapter 7 3 Casino Lab STATION 3. P(“blackjack” face-up card is ace) = 51 2 b. Home Teachers Last Name S-Z Selvaag, Jay AP Statistics Chapter 9 - Testing a Claim Chapter 9 - Testing a Claim Here are online resources, notes, classwork and homework assignments for Chapter 9.
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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.
lives in a house | lives in an apartment | ||
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
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 | 40 | 80 |
park no | 5 | 15 | 20 |
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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
P(snow) = 0.15 + 0.0375 = 0.1875
Ap Stats Casino Lab Answer Key Free
There is an 18.75% chance of snow.
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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).