Unit 4

Another way of thinking of this is that there are 2 choices for the first birth. Then, either of those will have 2 choices for the second birth, for a total of 2 • 2 = 22 = 4 possible orders. Then, either of those will have 2 choices for the third birth, for a total of 2 • 2 • 2 = 23 = 8 possible birth orders.

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List all outcomes in the sample space for the experiment "possible birth orders for 3 kids" (G=girl, B=boy).

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P(Ace or King)

--A deck of 52 cards has 4 kings and 4 aces and no overlap to worry about in this probability. --P(Ace or King) = P(Ace) + P(King) = 4/52 + 4/52 = 8/52 = 0.154. --These are mutually exclusive events.

Suppose 20 cyclists (13 of which are male) and 30 runners (18 of which are male) are surveyed at a local park on Saturday morning. Calculate P(cyclist or female).

--First construct a table. --Since these two categories overlap (ie. they are not mutually exclusive), we do not want to count the female cyclists twice. --P(cyclist or female) = P(cyclist) + P(female) - P(cyclist and female) --P(cyclist or female) = 20/50 + 19/50 - 7/50 = 32/50 = 0.640 --Or just count all the cyclists and add only the additional females that have not yet been counted. (7 + 13 + 12) / 50 = 32/50 = 0.640.

Suppose 200 college students are enrolled in the nursing program and 150 students are in business. Of the 200 nursing students 155 are female (the rest are male), and 64 of the business majors are female. Calculate the probability that a student is either a nursing student or female.

--First construct a table. --Since these two categories overlap (ie. they are not mutually exclusive), we do not want to count the female nursing majors twice. --P(nursing or female) = P(nursing) + P(female) - P(nursing and female). --P(nursing or female) = (155+45)/350 + (155+64)/350 - 155/350 = 264/350 = 0.754. --Or just count all the nursing majors and add only the additional females that have not yet been counted. (155 + 45 + 64) / 350 = 264/350 = 0.754.

The Addition Rule

--In probabilities, the word "or" implies addition... just be careful not to double count. --If events are mutually exclusive then there is no overlap to worry about, so no chance of double counting.

Suppose 60% of landscapers are certified pesticide applicators. If 5 landscapers are randomly selected, what is the probability that at least one is certified to apply pesticides.

--P(at least one certified) in 5 picks, is the opposite (complement) of picking all not certified in 5 picks. --P(certified) = 60% so P(not certified) = 40% --Since landscapers are independent, P(all 5 not certified) = 0.40^5 = 0.010. --P(at least one certified) = 1 - P(all 5 not certified) = 1 - 0.010 = 0.990.

As chair of a committee, you must randomly select 5 people from your department of 13 men and 16 women. What is the probability that you will select at least one male?

--P(at least one male) in 5 picks, is the opposite (complement) of picking no males (all females) in 5 picks. --Since members of a committee are not independent, P(all 5 female): 16/29 • 15/28 • 14/27 • 13/26 • 12/25 = 0.037. --P(at least one male) = 1 - P(all 5 female) = 1 - 0.037 = 0.963.

Suppose 9% of elementary age children are obese. If 10 elementary age children are randomly selected, what is the probability that at least one is obese?

--P(at least one obese) in 10 picks, is the opposite (complement) of picking all not obese children in 10 picks. --P(obese) = 9% so P(not obese) = 91% --Since weights are independent, P(all 10 not obese) = 0.91^10 = 0.389 --P(at least one obese) = 1 - P(all 10 not obese) = 1 - 0.389 = 0.611.

Calculate P(King, given Heart)

--Since there are only 13 Hearts then my denominator is limited to 13. --Only 1 of these 13 Hearts is a King. --P(King, given Heart) = 1/13 = 0.077.

P(Heart or Ace)

--Since these two categories overlap (ie. they are not mutually exclusive), we need to be careful not to count the overlap twice (ie. do not count the Ace of Hearts twice). --P(Heart or Ace) = P(Heart) + P(Ace) - P(Ace of Hearts). --P(Heart or Ace) = 13/52 + 4/52 - 1/52 = 16/52 = 0.308. --Or just count all the Hearts and the additional Aces that haven't yet been counted. --(13 + 3)/52 = 16/ 52 = 0.308.

A stockperson checks the coolers for outdated milk cartons. A sample of 4 milk cartons is to be randomly selected from a group consisting of 8 outdated cartons and 20 fresh cartons. What is the probability that all four selected milk cartons will be outdated?

--You are "randomly selecting" 4 cartons, so multiply 4 fractions. --Since you would not want to put outdated cartons back on the shelf, there are less and less to choose from so these events are dependent (the fractions will decrease). --There are 28 cartons total and the chance of selecting an outdated carton is 8/28. --So multiply 8/28 • 7/27 • 6/26 • 5/25 = 0.003.

As chair of a committee, you must randomly select 5 people from your department of 13 men and 16 women. What is the probability that you will select all women?

--You are "randomly selecting" 5 people, so multiply 5 fractions. --Since one person can not be selected for the committee twice, these events are dependent (you have less and less people to choose from). --There are 29 people total and the chance of selecting a female is 16/29. --So multiply 16/29 • 15/28 • 14/27 • 13/26 • 12/25 = 0.037.

During a segment of Nightline, it was reported that there is a 60% success rate for those trying to stop smoking through hypnosis. Find the probability that 8 randomly selected smokers who undergo hypnosis all successfully stop smoking.

--You are "randomly selecting" 8 smokers, so multiply 8 fractions (or their decimal equivalent) --Since the chance that one smoker is successful is independent of whether the other smokers are successful, the chances of success will remain the same for each trial. --Multiply: 0.60 • 0.60 • 0.60 • 0.60 • 0.60 • 0.60 • 0.60 • 0.60 = (0.60)^8 = 0.017.

What is the probability of selecting 3 hearts from a deck of 52 cards if you do not replace the cards after each pick (ie. Without replacement).

--You are "selecting" 3 cards from the deck, so multiply 3 fractions. --The chance of picking a heart is 13/52 (since there are 13 hearts in the deck of 52 cards). --Since the cards are not replaced, both the number of cards and the number of hearts available decrease with each pick. --So multiply 13/52 • 12/51 • 11/50 = 0.013.

A marble is selected from a bag containing 4 red, 5 green, and 6 blue marbles. If the marbles are replaced after each pick, what is the probability that 3 green marbles will be selected in a row?

--You are "selecting" 3 marbles so multiply 3 fractions. --Since the marbles are replaced, the events are independent (the fractions stay the same). --The chance of selecting a green marble from the bag is 5/15. --So multiply 5/15 • 5/15 • 5/15 = (5/15)3 = 0.037.

Create a distribution for the three birth experiment. Calculate P(exactly one boy).

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A complete listing of the sample space together with each outcome's probability is called...

... a distribution.

All probabilities are values between zero and one.

0 ≤ P(x) ≤ 1

An event is ANY subset of the sample space.

An event is ANY collection of simple outcomes.

Classical (theoretical) Probability

Based on a known number of outcomes; each outcome has the same chance of occurring (equally likely events). --the probability that a five will be rolled on a six-sided dice. --the probability of guessing correctly on a multiple choice question. --the probability of picking a King from a deck of cards.

Subjective (intuitive) Probability

Based on personal experience; just educated guesses based on what "you" have experienced. --the probability that you will run out of gas on the way to school. --the probability that you will experience anxiety during your next math test. --the probability that you will get an A on the next test.

Mutually Exclusive Events

Mutually Exclusive Events are events that can't happen at the same time (no overlap) Examples: --Drawing an ace and drawing a king are mutually exclusive events (one card can not be both). --Drawing an ace and drawing a diamond are not mutually exclusive since one card could be an Ace and a Diamond. --Being male and pregnant are mutually exclusive (one person can not be both). --Being female and a student are not mutually exclusive since one person could be both female and a student.

Mutually Exclusive Events:

P(A or B) = P(A) + P(B)

Overlapping Events:

P(A or B) = P(A) + P(B) - P(A and B) Here the overlap is subtracted out since it has been counted in both events. You could also just add the probabilities without counting the overlap twice in the first place.

P(A) = probability that event A will occur.

P(A') = P(not A) = P(Ac) = P( A ¯ ) = probability that event A will not occur (complement of A).

We might be logical and say that there are two possibilities each time, so a 50-50 split. 0.5^3 = 0.125 every time. Recall the sample space was S = {GGG, GGB, GBG, GBB, BBG, BBB, BGG, BGB} Each of the 8 simple outcomes needs a probability assigned and they all have to add up to 1.

P(GGG) = 1/8, P(GGB) = 1/8, P(GBG) = 1/8, P(GBB) = 1/8, P(BBG) =1/8, P(BBB) = 1/8, P(BGG) = 1/8, P(BGB) = 1/8 P(exactly one boy) = P({GGB, GBG, BGG}) = 1/8 + 1/8 + 1/8 = 3/8 = 0.375

Dependent Events

Prior events do make a difference in subsequent selections. Picking classmates for a group, picking cards without replacement, and selecting broken eggs out of a carton are all dependent events, because each event is affected by the fact that you already picked from the total outcomes, leaving you with less to choose from.

Complements and the Multiplication Rule

Recall that P(not A) is the complement of P(A), or P(not A) = 1 - P(A). When using the Multiplication Rule, complements can really simplify problems that include the phrase "at least one". Rather than determining every possible order in this list of compound events, it is helpful to think in terms of the complement of "none", since P(none) is a simple event. P(at least one) = 1 - P(none).

Relative Frequency (empirical) Probability

Requires collecting data to count outcomes that actually occur. Small samples may not reflect accurate probabilities. --the probability that a car accident will occur at the corner of Broad and High. --the probability that a baby will be born after 5pm at Riverside Hospital. --the probability that a female will live to the age of 90.

The possibilities for the first birth is B or G. If the first birth was G then the second birth could be G or B. If the first birth was B then the second birth could be G or B. That brings our running total to GG, GB, BB, BG. No matter which of these four orders occurs, the possibilities for the third birth are G or B. GG could end with B or G - GGG, GGB GB could end with B or G - GBG, GBB BB could end with B or G - BBG, BBB BG could end with B or G - BGG, BGB

S = {GGG, GGB, GBG, GBB, BBG, BBB, BGG, BGB} for a total of 8 outcomes in the sample space.

S = sample space

Sample space = set of all possible (simple) outcomes of an experiment. Simple means that the outcome cannot be broken up into a collection of other outcomes.

Law of Large Numbers

The more times an experiment is repeated (increased sample size), the closer the relative frequency (empirical) probability gets to the classical (theoretical) probability. For example, if we were to flip a coin 10 times and record the number of tails we may or may not observe 5 of the 10 flips to be tails as we would expect theoretically. But if we continue to flip the coin, the relative frequency probability calculated from say 1000 flips of a coin gets closer to 50%, and 10,000 flips is likely to be even closer to 50%. So with more flips (or increased sample size), the relative frequency probability gets closer to the classical (theoretical) probability. In short, big samples are more reliable than small samples when determining relative frequency probabilities.

The Multiplication Rule

The multiplication rule applies if you are picking "more than one". The number of picks tells you the number of fractions to multiply. Next decide if the events are independent (probability of each pick is the same) or dependent (each pick depends on what you picked previously).

Independent Events

The prior event has no effect on the subsequent selection. Rolling dice, picking cards with replacement, spinning a roulette wheel, and guessing on test questions are all independent since your chances of success don't change for each trial.

Conditional Probabilities

The probability that an event will occur given the outcome of the previous event. The term "given" implies that the denominator is reduced to only what is given, and then the possible successes are limited to that given event as well. P(A, given B) = P(A|B) = probability that event A will occur, given that B has already occurred.

The sum of the probabilities of all outcomes in a sample space is one.

ΣP(x) = 1