Probability Distributions Including Binomial Distributions

# Probability Distributions Including Binomial Distributions

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## Probability Distributions Including Binomial Distributions

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1. Probability DistributionsIncluding Binomial Distributions Created by: K Wannan Edited by: K Stewart

2. Probability • P(x) is the probability that a single outcome, x, will occur. • P(A) is the probability that a single event, A, will occur. The event can include several different outcomes (e.g. rolling an even number)

3. Random Variable, X An experiment can have many outcomes. We use a variable, X, to represent all of them. We use lower case x to represent a specific outcome. Random Variable X represents the set of all possible outcomes. Each of these outcomes can be referred to individually as xn.

4. Notation: P(X), P(X=x) and P(x) • X is the set of all possible outcomes. • P(X) shows the probability of each and every one of the possible outcomes P(xn) usually in a table or graph (can also be modelled by an equation). • The probability of outcome x can be written P(X = x) or just P(x).

5. 1 2 1 3 Probability Distributions • For example, the probability distribution of P(X) where the random variable, X, is the result from spinning this spinner is shown:

6. Random Variable, X Random Variables can be made up of outcomes that are either discrete or continuous. DiscreteContinuous Values that are Values that are separate from each infinite in number other in an interval e.g. shoe sizes e.g. time

7. Ex Classify each of the following random variables as discrete or continuous • The number of phone calls made by a sales person • The length of time the salesperson spent on the telephone • A company’s annual sales • The number of widgets sold by the company • The distance from Earth to the sun D C D D C

8. Uniform Probability Distribution • Examples: rolling a dice, tossing a coin • Uniform probability means each outcome is equally likely to happen in any single trial of an experiment i.e. where n is the number of outcomes.

9. A test consists of 5 multiple choice questions each with 4 possible answers, of which only one is correct. A student guesses the answers without reading the questions. What is the probability distribution for the outcomes of the test? • Other Probability Distributions

10. What is the number of trials? • What is the probability of success for each trial? • What is the probability of failure? • How many ways can r of the 5 answers be correct? 5 1/4 3/4 5Cr

11. Draw a tree diagram to help you answer the following questions: What is the probability of getting all of the questions right? What is the probability of getting the first question right and all of the rest wrong?

12. What is the probability of getting only one question right? (Use your tree diagram to help you figure out how many ways this can happen)

13. This calculation is the process you need to create the rest of this probability distribution P(X). Specifically, to calculate each probability:

14. This is the probability distribution for the results of this five question multiple choice test:

15. What is the sum of the probability distribution? Explain why. The sum is 1 because this is 100% of all of the possible results.

16. What is the probability of the student passing the test?

17. If you guessed on all 5 of these questions, how many would you expect to get right? Since P(correct) = 25% for every question, then you can expect to get 25% of them right. i.e. 25% of 5 questions is 1.25 questions Therefore, you can expect to get one or two correct.

18. Expected Value (Binomial Distribution) Solution: Since 1, 2, 3 and 5 are prime numbers, each roll has a 4/6 chance of being prime, so expect to observe 10 (4/6) = 6.7 Therefore you can expect to observe 6 or 7 prime numbers. Example 2: You roll a dice 10 times. How many prime numbers do you expect to observe?

19. For Binomial Distributions… The Expected Value E(X) = npwhere n is the number of trials and p is the probability of success

20. Expected Value Take the average! (5+5+20+40)/4 =70/4 =17.5 Expect to win \$17.50 You spin this wheel. How much can you expect to win?

21. Expected Value = Average When there are equally likely outcomes, you can add up their values and divide by the number of things added. This is the value that you can expect to get. When the outcomes are not equally likely, you need to use a weighted average.

22. Weighted Average Looking at the wheel again: You have a 50% chance of winning \$5 A 25% chance of winning \$20 and A 25% chance of winning \$40 Weighted average: (0.5)(5)+(.25)(20)+(.25)(40) =17.5 Therefore, expect to win \$17.50.

23. Weighted Averages You are randomly assigned to a canoe. Determine the length of canoe that you can expect to get when there are: 7 Canoes that are 4.6m long, 10 that are 5.0m long, 4 that are 5.2m long and 4 that are 6.1m long Hint: create a probability distribution P(X)!

24. 5.096

25. Expected Value If an average is where you “add up everything and divide by the number of things added”, how is this an average? What makes it a weighted average?

26. Ex Consider the following game. You roll a single die. If you roll an even number then you GAIN that amount of points. If you roll an odd number then you LOSE that amount of points. • Generate a Probability Distribution of the points. • How many point can you expect to get on any one roll? • Is this game fair?

27. What is the Probability Distribution of the Points?

28. Is this game fair? • In order for a game of this nature to be fair then points earned and points lost should be equal E(X)=0

29. Homework! Pg 374 # 1-4,7,8 Pg 385 # 1, 3 5,7,10 Yippee hooray