1. Applications of Probability

8. Practice (make a 2-way & tree also)

10. Aladdin is having a rough day
Aladdin is having a rough day. Thrown into the cave of wonders by Jafar he isn’t liking his odds of getting out of here alive.
If he only lives by finding the lamp, what is the chance that he finds the lamp the first time?
If he finds gold the first time, he still has another chance to not die, yeah! What is the chance he finds gold or the lamp?
Let’s say he does die.  Like Indiana Jones he really doesn’t like snakes. What is the probability that he dies by snake?

11. VENN DIAGRAMS

17. Conditional Probability and Independence
Define, use, and generate the rule.

18. Conditional Probability
Define in your own words conditional probability Write the probability of soccer given that they play tennis as a ratio of words Write the probability of tennis given that they play soccer as a ratio of words.

19. Conditional Probability Rule

20. Use the data in the table below, which shows the employment status of individuals in a particular town by age group.
Age Group Full-time Part-time Unemployed
Determine the amount of people (in total) that are full time, part time, or unemployed. Then determine the percent of people (in total) that fall into each category.
If a person in this town is selected at random, find the probability that the individual is employed part-time, given that he or she is between the ages of 35 and 49.
If a person in the town is randomly selected, what is the probability that the individual is unemployed, given that he or she is over 50 years old?
A person from the town is randomly selected; what is the probability that the individual is employed full-time, given that he or she is between 18 and 49 years of age?
A person from the town is randomly selected; what is the probability that the individual is employed part-time, given that he or she is at least 35 years old?
Are the probabilities of the age groups differently distributed or are they all the same? What does that mean about age and employment?

21. Independence and Conditional Probability
An example of number (2): Imagine a scenario where you are drawing cards from a deck. You are looking to pull an ace followed by a king. There is one major qualification on how this plays out: did you replace the first card you drew from the the deck or not?
P(A) = P(K) =
P(A and K) with replacement =
P(A and K) without replacement =

22. On school days, Janelle sometimes eats breakfast and sometimes does not. After studying probability for a few days, Janelle says, “The events ‘I eat breakfast’ and ‘I am late for school’ are independent.” Explain what this means in terms of the relationship between Janelle eating breakfast and her probability of being late for school in language that someone who hasn’t taken statistics would understand.

23. Rain and Lightening
Today there is a 55% chance of rain, a 20% chance of lightning, and a 15% chance of lightning and rain together. Are the two events “rain today” and ”lightning today” independent events? Justify your answer.
Now suppose that today there is a 60% chance of rain, a 15% chance of lightning, and a 20% chance of lightning if it’s raining. What is the chance of both rain and lightning today?

24. Solutions Given P(rain) =. 55, P(lightning) =
Solutions Given P(rain) = .55, P(lightning) = .2, and P(lightning and rain) = .15. Two events are independent if P(lightning and rain) = P(lightning) ⋅ P(rain.) Since P(lightning) ⋅ P(rain) = (.55) ⋅ (.2) = .11 ≠ .15 = P( lightning and rain) the two events are not independent. Given P(rain) = .6, P(lightning) = .15, and P(lightning | rain) = .2. We need to find P(rain and lightning). We use the formula P(lightning | rain) = P(lightning and rain)/ P(rain) Since we have two of the three pieces of information, we have to solve for the third one. P(lightning and rain) = P(lightning | rain) ⋅ P(rain) = (.2) ⋅ (.6) = .12 Answer: There is a 12% chance of both rain and lightning today.

26. Paper Prompt: Given the information below choose a positon regarding the survival rates on the Titanic. Did gender or class or some other factor play the biggest role in survival? You are welcome to do additional research. Your position must be supported with mathematical evidence.

27. The Addition Rule

28. Addition Rule
Be wary of the double count!

29. And vs Or

30. Write a statement that describes how many students play soccer or tennis. (think total tennis, total soccer and the overlap)
Generate a rule that describes “or” scenarios.

31. In a group of 45 boys; 17 own laptop, 19 own desktop computers, and 7 own both. Find the probability that a boy picked from this group at random owns either a laptop or a desktop computer?
Create a venn diagram and solve
The probability that a student owns a car is 0.65, and the probability that a student owns a computer is The probability that a student owns both is 0.55.
Create and venn diagram of this information and determine what the probability is that a randomly selected student owns a car or computer?
What is the probability that a randomly selected student does not own a car or computer?
What is the probability that a student owns a computer given that they own a car?

32. Rain and Lightening Part 2
Now suppose that today there is a 55% chance of rain, a 20% chance of lightning, and a 15% chance of lightning and rain. What is the chance that we will have rain or lightning today?
Now suppose that today there is a 50% chance of rain, a 60% chance of rain or lightning, and a 15% chance of rain and lightning. What is the chance that we will have lightning today?

33. Solution
c. Given P(rain) = .55, P(lightning) = .2, and P(lightning and rain) = .15. We need to find P(rain or lightning,)which is the same as P(lightning or rain.) Using the Addition Rule we P(lightning or rain) = P(lightning) + P(rain) − P(lightning and rain) = − .15 = .6 Answer: There is a 60% chance of rain or lightning today. d. Given P(rain) = .5, P(lightning) = .6, and P(lightning and rain) = .15. We need to find P(lightning). We use the Addition Rule: P(rain or lightning) = P(rain) + P(lightning) − P(rain and lightning). Since we have three of the four pieces of information, we have to solve for the fourth one, the probability of lightning. Subtracting P(rain)and adding P(rain and lightning)to both sides of the equation, we obtain P(lightning) = P(rain or lightning) − P(rain) + P(rain and lightning) = .6 − = .25 Answer: There is a 25% chance of lightning today.