AP Statistics – Ch. 5 Notes

Probability

Chance behavior is unpredictable in the short run but has a regular and predictable pattern in the

long run.

The Law of Large Numbers: As we observe more and more repetitions of any chance process, the proportion of times that a specific outcome occurs approaches a single value, called the probability of the outcome. For example, the proportion of heads flipped varies a lot for the first several flips, but as the number of flips increases, the proportion of heads will approach 0.5.

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Probability: The probability of any outcome of a chance process is a number between 0 and 1 that describes the proportion of times the outcome would occur in a very long series of repetitions. Example: In the popular Texas hold ’em variety of poker, players make their best five-card poker hand by combining the two cards they are dealt with three of five cards available to all players. You read in a book on poker that if you hold a pair (two cards of the same rank in your hand), the probability of getting four of a kind is 88/1000.

a) Explain what this probability means. If you play many, many hands of poker in which you hold a pair, the fraction of times you will end up with four of a kind will be close to 88/1000.

b) Why doesn’t this probability say that if you play 1000 such hands, exactly 88 will be four of a kind?

This probability describes what happens, on average, in a very large number of hands like this. It does not mean that if you’ve had 87/999 such hands result in four of a kind, the next hand in which you hold a pair is guaranteed to result in four of a kind.

Myths about Randomness

• The myth of short-run regularity: Most people think that runs and streaks are extremely unlikely, but they are actually quite common in the short term. It is only in the long run that things balance out.

o Roll a die 12 times and record the result of each roll. Which of the following outcomes is more probable: 123456654321 or 154524336126? (In fact, both have the same probability: 1 in 2,176,782,336)

• The myth of the “law of averages”: You hear people say something like, “this batter’s struck out 5 times in a row. He’s due for a hit.” or “You’ve flipped ten heads in a row? The next one is bound to be tails.”

o Coins and dice and other chance processes don’t have a memory. The next outcome is not influenced by the previous outcomes. If you’ve flipped ten heads in a row, the probability of flipping tails on the next flip is still 0.5.

108642

Number of Winning Caps

Simulation: The imitation of chance behavior, based on a model that accurately reflects the situation. Performing a Simulation:

1. Labels: What do the different outcomes of your chance process represent? Make sure the numbers of labels assigned to different events are proportional to the probabilities of the events.

2. Process: How will you actually carry out the simulation? 3. Ignore or don’t ignore: Are there any possible outcomes of your random process that weren’t used as

labels? Should you ignore a label if it appears more than once? (If the labels stand for specific people or specific objects that are being selected, arranged, or divided up, you should ignore repeats. If the labels stand for general characteristics in a large population of unknown size like handedness, gender, or color-blindness, you should not ignore repeats.)

4. Stopping rule: How do you know when you’re done with a trial? 5. What to count: What do you actually count from your simulations?

Use the word “approximately” when reporting a probability based on a simulation! “The probability

of ____ is approximately ____.”

Simulations can tell us whether an outcome is likely or unlikely, but they can’t prove that a claim is

definitely true or that the evidence proves something. In your conclusions, use phrases like, “There is

convincing evidence that _______” or “There isn’t convincing evidence that _______”. (In statistics,

we often consider an outcome that would happen less than 5% of the time by chance to be

“convincing evidence” that something important or fishy is going on, but 5% is a completely arbitrary

choice, so don’t start thinking it’s a magic value.)

Example: As a special promotion for its 20-ounce bottles of soda, a soft drink company printed a message on the inside of each bottle cap. Some of the caps said, “Please try again!” while others said, “You’re a winner!” The company advertised the promotion with the slogan “1 in 6 wins a prize.” The prize is a free 20-ounce bottle of soda. A statistics class wonders if the company’s claim holds true for the bottles at a nearby convenience store. To find out, all 30 students in the class go to the store and each student buys one bottle of the soda. Only two of them get caps that say, “You’re a winner!” Does this result give convincing evidence that the company’s 1-in-6 claim is inaccurate? Assume the company is telling the truth. Use a die to model each bottle of soda. Let 1-5 represent a losing bottle and 6 represent a winning bottle. Roll the die 30 times to represent the 30 bottles of soda bought by the class. Count the number of rolls on which the die lands on 6. Repeat the process many times. In what proportion of trials did the simulated class end up with two or fewer winning caps?

The simulation was performed 50 times, and the simulated class ended up with 2 or fewer winning caps 5 times. This means that the probability of ending up with 2 or fewer caps by chance is approximately 5/50 or 10%. This is not particularly unusual. There is not convincing evidence that the company’s claim is inaccurate.

Example: Suppose that a basketball announcer suggests that a certain player is “streaky”. That is, the announcer believes that if the player makes a shot, then he is more likely to make his next shot. As evidence, the announcer points to a recent game where the player took 30 shots and had a streak of 7 made shots in a row. Is this evidence of streakiness, or could it have occurred simply by chance? Assuming that this player makes 50% of his shots and the results of a shot don’t depend on previous shots, how likely is it for the player to have a streak of 7 or more made shots in a row? Using a coin, let heads represent a made shot and tails represent a missed shot. Flip a coin 30 times, writing down the outcome after each flip. Record the length of the longest streak of made shots. In what proportion of the trials did the player have a streak of at least 7 made shots in a row? Is there convincing evidence that the player is “streaky”?

The histogram shows the results of 100 trials. In 9 of the 100 trials, the player made 7 or more shots in a row by chance. Thus, we would estimate that the probability of the player making at least 7 shots in a row, assuming that he is equally likely to make each shot is about 9/100 or 9%. Since it is not unlikely for a streak like this to happen by chance, we don’t have sufficient evidence to conclude that the player is “streaky”.

Example: A teacher wants to choose an SRS of size 6 from a group of 60 seniors and 30 juniors. To do this, she writes each person’s name on an equally sized slip of paper and mixes them up in a hat. Just as the teacher is about to select the first name, a thoughtful student suggests that it would be a good idea to stratify by class. The teacher agrees, and they decide to select a sample of 4 seniors and 2 juniors. However, the names are already in the hat and the teacher doesn’t want to separate them by grade level. Instead, she decides to select names one at a time from the bag until 4 seniors and 2 juniors have been selected. It takes 14 names to satisfy the requirement of getting 4 seniors and 2 juniors. Some of the students suggest that the names must not have been thoroughly mixed in the hat. Design and carry out a simulation using a random digit table to estimate the probability that 14 or more names must be drawn to select 4 seniors and 2 juniors. Is there convincing evidence that the names weren’t thoroughly shuffled? Labels: Let 01 to 60 represent seniors and let 61 to 90 represent juniors. Process: Read the random digit table two digits at a time. What to ignore? Ignore repeats and the labels 91 to 99 and 00. Stopping rule: Stop when four students with labels from 01 to 60 and two students with labels from 61 to 90

have been selected. What to count: Count the number of students who were selected. Repeat several times. Example: 96|74|6 1|21|49| 37|82|3 7|18|68| 18|44|2 96: skip, 74: junior, 61: junior, 21: senior, 49: senior, 37: senior, 82: junior, 37: skip (repeat), 18: senior I selected junior, junior, senior, senior, senior, junior, senior for a total of 7 students.

The histogram shows the results of 100 trials. Fourteen or more names were selected in 2/100 trials. Therefore, the probability that 14 or more names must be drawn to get 4 seniors and 2 juniors is about 2/100 or 2%. This is unlikely to happen by chance, so there is convincing evidence that the names weren’t thoroughly shuffled.

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Example: Statistics students at Bingham High School are trying to raise money for True Blue by selling cards, each with one of the letters B, H, or S. To win, a student must collect one of each letter and cannot share or trade cards with others. 68% of the cards are labeled B, 25% are labeled H, and 7% are labeled S. Each card costs $1. A complete set of cards can be exchanged for an $8 prize. Design and carry out a simulation using technology to estimate how many cards, on average, a student should expect to buy in order to win. Also estimate the probability that a student will win with fewer than 8 cards.

Labels: We need 68% of the labels to stand for B, 25% to stand for H, and 7% to stand for S. Let 1-68 represent a card with a B, 69-93 represent a card with an H, and 94-100 represent a card with a S.

Process: Use technology to generate random integers from 1 to 100. Ignore or don’t ignore: Don’t ignore repeats. (We are assuming the probability is fixed for each card. The

labels don’t stand for individual cards, just outcomes. We don’t know how many total cards were printed, so we can’t think of the probability as changing each time a card is drawn.)

Stop rule: Stop when at least one label from each category (1-68, 69-93, and 94-100) have appeared. What to count: Count how many total numbers were generated in each trial, representing the total number of

cards the student had to buy to collect one with each label. Example: 45: B, 46: B, 77: H, 17: B, 9: B, 77: H, 55: B, 80: H, 1: B, 95: S The student had to buy 10 cards to “win”. (He spent $10 and won $8.)

The dotplot shows the results of 100 trials. In 32 trials, the student won with fewer than 8 cards. The probability that a student will win with fewer than 8 cards is about 32/100 or about 32%.

For the 100 trials, the mean number of cards required to “win” was 14.9 and the median was 10. Example: A recent study reported that fewer than half of young adults turn off the water while brushing their teeth. Is the same true for teenagers? A group of statistics students asked an SRS of 60 students at their school “Do you usually brush your teeth with the water off?” The dotplot below shows the results of taking 200 SRSs of 60 students from a hypothetical population in which 50% of students brush with the water off.

a) Suppose 27 students in the sample say “Yes”. Explain why this result does not give convincing evidence that fewer than half of the school’s students brush their teeth with the water off.

b) Suppose 18 students in the sample say “Yes”. Explain why this result gives strong evidence that fewer than half of the school’s students brush their teeth with the water off.

a) Based on the simulation, if the true proportion of

students in the school who brush their teeth with the water off is 50%, we would expect that an SRS of size 60 would have 27 or fewer students who brush with the water off about 48/200 = 24% of the time purely by chance. Since this is not an unlikely sample, we don’t have convincing evidence that the true proportion of students who brush their teeth with the water off is less than 50%.

b) None of the 200 SRSs of size 60 from the hypothetical

population had fewer than 18 students who brush with the water off. This means that it would be extremely unlikely for a random sample to have this few students who brush with the water off purely by chance if the true population proportion is 50%. It is much more likely that the actual proportion of students in the school who brush with the water off is less than 50%.

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Number who brush with water off

Probability Rules The sample space S of a chance process is the set of all possible outcomes. A probability model is a description of some chance process that consists of two parts: a sample space S and a probability for each outcome. Examples: Give probability models for the following: a) Flipping a coin three times b) Rolling two dice

{ }HHH, HHT, HTH, HTT, THH, THT, TTH, TTTS =

The probability of each outcome is 1/8. The probability of each outcome is 1/36. Event: Any outcome or collection of outcomes from some chance process is called an event. An event is a subset of the sample space. Events are usually designated by capital letters. We write the probability of event A

as ( ).P A

Complement: If A is any event, we refer to the set of outcomes not in A as the complement of A. This event is

abbreviated .CA (read “not A”).

Basic Rules of Probability

• The probability of any event is a number between 0 and 1.

o For any event A, ( )0 1.P A≤ ≤

o The probability of an event is the long-run proportion of repetitions on which that event occurs. An event with probability 0 never occurs, and an event with probability 1 occurs on every trial.

Tip: If you choose to express probability as a fraction, it’s okay if you don’t reduce it, but never express probability as a decimal or percent without showing the fraction it came from!

• All possible outcomes of a chance process must have probabilities whose sum is 1.

o If S is the sample space in a probability model, ( ) 1.P S =

• If (and only if) all outcomes in the sample space are equally likely, the probability that event A occurs

can be found using the formula

( )# of outcomes corresponding to event

total # of outcomes in the sample space

AP A =

• The probability that an event does not occur is 1 minus the probability that the event does occur. (Since it is certain that an event either will occur or will not occur, the probability that it occurs and the probability that it doesn’t occur always add to 100% or 1.

o ( ) ( )1 .CP A P A= −

Chance Process Sample Space Event Complement

Flip a coin { }heads, tailsS = { }headsB = { }tailsCB =

Roll a die { }1, 2, 3, 4, 5, 6S = Even numbers

{ }2, 4, 6E = { }1, 3, 5C

E =

Pick a letter in the word

“probability” { }P, R, O, B, A, I, L, T, YS =

Vowels

{ }O, A, I, YV = { }P, R, B, L, TC

V =

Mutually Exclusive or Disjoint: Two events are mutually exclusive or disjoint if they can never occur at the same time (they have no outcomes in common). The probability that they both happen is 0! Examples of Mutually Exclusive Events: being male and being pregnant, choosing an odd number and choosing a multiple of 4, etc.

• Addition Rule for Mutually Exclusive Events: If two events can’t happen at the same time, the

probability that one or the other occurs is the sum of their individual probabilities.

o If A and B are mutually exclusive, ( ) ( ) ( ) or .P A B P A P B= +

Example: Buffalo Wild Wings ran a promotion called the Blazin’ Bonus in which every $25 gift card purchased also received a “Bonus” gift card for $5, $15, $25, or $100. According to the company, here are the probabilities for each bonus gift card:

a) Explain why this is a legitimate probability model. All of the probabilities are between 0 and 1, and 0.890 + 0.098 + 0.010 + 0.002 = 1

b) Find the probability that the bonus gift card isn’t worth $5.

( ) ( )not $5 1 $5 1 0.890 0.110P P= − = − =

c) What’s the probability the bonus card is worth $25 or $100?

( )$25 or $100 0.010 0.002 0.012P = + =

Union (((( )))) :A B∪∪∪∪ “Or” (either, at least one)

• All of the outcomes that are in A or B or both.

Intersection (((( )))) :A B∩∩∩∩ “And” (both, joint, overlap, in common)

• All of the outcomes that are in both A and B.

• If the two events are mutually exclusive (disjoint), then ( ) 0.P A B∩ =

The General Addition Rule

( )P A B∪ is the probability that at least one of the events occurs.

( ) ( ) ( ) ( )P A B P A P B P A B∪ = + − ∩

If two events are not mutually exclusive (disjoint), then adding ( ) ( )P A P B+ to find ( )P A B∪ results in

counting the intersection twice. Therefore, it is necessary to subtract ( ).P A B∩

Blazin’ Bonus: $5 $15 $25 $100

Probability: 0.890 0.098 0.010 0.002

Example: A survey of students at a large high school revealed that, in the last month, 38% of them dined at Taco Bell, 16% dined at Chipotle, and 9% dined at both. Suppose we select a student at random. What’s the probability that the student dined at Taco Bell or Chipotle in the last month?

( ) ( ) ( ) ( )Taco Bell or Chipotle Taco Bell Chipotle both

0.38 0.16 0.09 0.45

P P P P= + −

= + − =

Example: A group of AP statistics students surveyed the seniors in their high school about whether or not they use the music streaming apps Pandora and Spotify. The results are shown in the table below. Use this information to complete the Venn diagram, then compute the requested probabilities.

Suppose we choose a student from the sample at random. Find the following probabilities, and write the probabilities in symbols using the events P: uses Pandora, and S: uses Spotify.

a) uses Spotify b) uses Spotify and Pandora

( )285

0.38750

P S = = ( )180

0.24750

P S P∩ = =

c) uses Spotify or Pandora d) uses neither Spotify nor Pandora

( )180 105 130 615

0.82750 750

P S P+ +

∪ = = = ( )( ) ( )135

or 0.18750

C C CP S P P S P∪ ∩ = =

Example: A political ad is run on TV and on radio. A simple random sample of 100 voters is taken. The individuals are asked if they have seen the ad on TV and if they have heard the ad on the radio. The results are as follows:

• 33% of people saw it on TV.

• 21% heard it on the radio.

• 10% of people saw it on TV and heard it on the radio. Find the probability that an individual selected at random from the sample

a) only saw it on TV b) only heard it on the radio 23% 11%

c) neither heard it nor saw it d) did not see it on TV 56% 67%

Pandora

Uses Doesn’t Use Total

Spotify Uses 180 105 285

Doesn’t Use 330 135 465

Total 510 240 750

TV

Saw Did not see Total

Radio Heard 10% 11% 21%

Did not hear 23% 56% 79%

Total 33% 67% 100%

33% saw – 10% saw & heard =

23% only saw

21% heard – 10% saw & heard =

11% only heard

Example: Suppose I pick an animal at random from the following list: human, elephant, octopus, whale, Henry (my cat, who does indeed have four legs, and occasionally wakes up long enough to walk on them), crocodile, robin, iguana, worm.

• Event F: Walks on four legs

{ }elephant, Henry, crocodile, iguanaF =

• Event M: Is a mammal

{ }human, elephant, whale, HenryM =

Draw a Venn diagram representing the sample space with subsets F and M. Find the following probabilities and interpret each in context:

( ) ( )4

walks on 4 legs9

PP F = = ( ) ( )4

is a mammal9

PP M = =

{iguana, crocodile, Henry, elephant} {Henry, elephant, human, whale} ( ) ( )walks on 4 legs or is a mammal

6 2

9 3

PP F M =

=

∪

=

( ) ( )walks on 4 legs and is a mammal

2

9

PP F M =∩

=

{iguana, crocodile, Henry, elephant, human, whale} {Henry, elephant}

( ) ( )5

does not walk on 4 legs9

CPP F = = ( )( ) ( )neither walks on 4 legs nor is a mammal

3 1

9 3

CP F M P

=

∪ =

=

{human, whale, worm, octopus, robin} {worm, octopus, robin}

( ) ( )walks on 4 legs but is not a mammal

2

9

CPP F M =∩

=

( )( ) ( )is not both a 4-leg-walker and a mammal

7

9

CPP F M =∩

=

{iguana, crocodile} {iguana, crocodile, human, whale, worm, octopus, robin}

Example: A random sample of 180 people showed that:

• 120 like pizza

• 95 like chicken

• 60 like hamburgers

• 55 like pizza but not chicken

• 45 like hamburgers and pizza

• 10 like hamburgers only

• 30 like all three Determine the probability that an individual chosen at random from the sample:

a) does not like any of the three 20/180

b) likes chicken only 25/180

c) likes at least one of the three 160/180

d) likes hamburger and chicken 35/180

e) likes hamburger and chicken, but not pizza 5/180

f) likes exactly one of the 3 75/180

g) likes pizza or chicken 150/180

h) does not like either hamburger or pizza 45/180

i) likes exactly two of the three 55/180

Filling in the Venn diagram:

1. Fill in 30 like all three and 10 like H only.

2. 45 like H & P.

30 like H, P, & C.

That leaves 15 who like H & P, but not C.

3. 60 like H.

10 (H only) + 15 (H & P, not C) + 30 (H, P, & C) = 55

That leaves 5 who like H & C, but not P.

4. 55 like P, but not C.

15 like P & H, but not C.

That leaves 40 who like P only.

5. 120 like P.

40 (P only) + 15 (P & H, not C) + 30 (P, H, & C) = 85

That leaves 35 who like P & C, but not H.

6. 95 like C.

35 (C & P, not H) + 30 (C, P, & H) + 5 (C & H, not P) = 70

That leaves 25 who like C only.

7. 180 people surveyed.

40 + 35 + 25 + 15 + 30 + 5 + 10 = 160 like at least one

That leaves 20 who don’t like any.

Conditional Probability and Independence

The probability that one event happens given that another event is already known to have happened is called a conditional probability. The probability that event B happens given that event A has happened is denoted by

( ) ,P B A read “the probability of B given A.”

For example, ( )cancer smokerP would mean the probability of a person getting cancer given that the person is

a smoker. ( )smoker cancerP would mean the probability that a person is a smoker if we know that he or she

has cancer. When looking at a table, the condition (the given part – the part after the bar) tells you what group to focus

on. Focus on that group and ignore the rest of the table.

Example:

a) If we know that a student in the sample uses Pandora, what is the probability the student uses Spotify? How would you write this probability in symbols?

( )180

Spotify Pandora 0.353510

P = ≈

b) If we know that a student in the sample uses Spotify, what is the probability the student uses Pandora? How would you write this probability in symbols?

( )180

Pandora Spotify 0.632285

P = ≈

Two events A and B are independent if knowing whether or not one event has occurred does not give you any additional information about the probability that the other event will occur.

Events A and B are independent if and only if ( ) ( ) ( )CP A B P A B A= = and ( ) ( ) ( ).C

P B A P B A B= =

Example of Independent Events: Event A: First flip of a coin comes up heads. Event B: Second flip of a coin comes up heads. These two events are independent, because the likelihood of flipping heads on the second flip

is not affected by the outcome of the first flip. In other words, ( ) ( ) 1 2.P B A P B= =

Example of Dependent Events: Event A: The first card in a deck is a heart. Event B: The second card in the deck is a heart. These two events are dependent, because knowing that the first card is a heart changes the

probability that the second card is a heart. ( ) 13 52,P B = but ( ) 12 51.P B A =

Example: Consider the previous example. Are the events P: uses Pandora and S: uses Spotify independent? Justify your answer.

( )180

Spotify Pandora 0.353510

P = ≈ ( )105

Spotify not Pandora 0.438240

P = ≈ ( )285

Spotify 0.38750

P = ≈

Using Pandora and using Spotify are not independent for students in the sample. Knowing a student uses Pandora decreases the probability that the student uses Spotify.

Pandora

Uses Doesn’t Use Total

Spotify Uses 180 105 285

Doesn’t Use 330 135 465

Total 510 240 750

Example: Is there a relationship between gender and subject preference (math or English)? The two-way table summarizes the relationship between gender and subject preference for a class of 25 AP Statistics students.

Suppose we choose one of the students in the sample at random. Are the events “male” and “prefers math” independent? Justify your answer.

( )8

prefers math male 0.810

P = = ( )12

prefers math female 0.815

P = = ( )20

prefers math 0.825

P = =

The events “male” and “prefers math” are not independent because ( ) ( )prefers math male prefers math .P P=

Knowing that a student is male does not change the probability that the student prefers math.

Example: Australian high school students were asked about their gender and whether they are right- or left-handed. The results are summarized in the table at the left. Calculate the following probabilities.

a) ( )0.07

0.152left-hand.

m0 07 0.3

e le9

d aP = ≈+

About 15.2% of the males in the sample are left-handed.

b) ( )female left-hand 0e0.03

.30.03

d0.07

P = =+

30% of the left-handed students in the sample are female.

c) ( )0.39

0.433male right.

-ha0 39 0.

de51

n dP = ≈+

About 43.3% of the right-handed students in the sample are male.

d) ( )right-handed femal .e0.51

0 9440.51 0.03

P = ≈+

About 94.4% of the females in the sample are right-handed. e) A student from the sample is left-handed. What is the probability that the student is male?

( )0.07

male left-handed 0.70.07 0.03

P = =+

70% of the left-handed students are male, so there’s a 70% chance the student is male. f) Are the events M: is a male and L: is left-handed independent? Justify your answer.

( ) 0.7P M L = (see part e) and ( ) 0.39 0.07 0.46.P M = + =

Being a male and being left-handed are not independent in this sample, since ( ) ( ).P M L P M≠

Knowing that a student is left-handed increases the probability that the student is a male.

Conditional Probability Formula: ( )( )

( )

( )

( )

both events occur

given event occurs

P A B PP B A

P A P

∩= =

Male Female Total

Math 8 12 20

English 2 3 5

Total 10 15 25

Example: According to the National Center for Health Statistics, in December 2017, 43% of U.S. households had a traditional landline telephone, 91% of households had cell phones, and 37% had both. What is the probability that a randomly selected household with a landline also has a cell phone?

( )( )

( )

both 0.37cell phone landline 0.86

landline 0.43

PP

P= = ≈

General Multiplication Rule: The probability that events A and B both occur is given by

( ) ( ) ( )P A B P A P B A∩ = ⋅

Example: About 55% of high school students participate in a school athletic team at some level, and about 5% of these athletes go on to play on a college team in the NCAA. What percent of high school students play a sport in high school and go on to play a sport in the NCAA?

( ) ( ) ( )

( )( )

plays in HS plays in NCAA plays in HS plays in NCAA played in high school

0.55 0.05 0.0275 2.75%

P P P∩ = ⋅

= = =

Example: Three cards are drawn at random from the top of a deck. What is the probability that the first two cards are spades and the third card is a heart?

Note: the number of cards remaining changes after each draw, which changes the denominator. Also, the number of spades remaining changes after the first spade is drawn.

( ) ( ) ( ) ( )spade, then spade, then heart spade spade after spade heart after 2 spades

13 12 130.0153 1.53%

52 51 50

P P P P= ⋅ ⋅

= ⋅ ⋅ = =

Multiplication Rule for Independent Events: If A and B are independent events, then the probability that A and

B both occur is ( ) ( ) ( ).P A B P A P B∩ = ⋅ This formula only works for independent events!

Example: If two dice are rolled together several times, what is the probability of rolling a sum of 7 five times in a row? 36 possible outcomes from rolling 2 die.

6 of them have a sum of seven (1, 6); (2, 5); (3, 4); (4, 3); (5, 2); (6, 1)

( )

( )

6 1sum of 7

36 6

1 1 1 1 1 15 7's in a row 0.0001286

6 6 6 6 6 7776

P

P

= =

= ⋅ ⋅ ⋅ ⋅ = =

Probability of “At least one…”: ( ) ( )at least one 1 noneP P= −

Example: The First Trimester Screen is a noninvasive test given during the first trimester of pregnancy to determine if there are specific chromosomal abnormalities in the fetus. According to a study published in the New England Journal of Medicine in November 2005, approximately 5% of normal pregnancies will receive a positive result. Among 100 women with normal pregnancies, what is the probability that there will be at least one false positive?

( )

( ) ( )100

one woman with a normal pregnancy does not get a false positive 1 0.05 0.95

no false positives for 100 women with normal pregnancies 0.95 0.00592

at least one false positive for 100 women with nor

P

P

P

= − =

= =

( )mal pregnancies 1 0.00592 0.99408= − =

Example: Sandy and South Jordan are two neighboring suburbs in Utah. According to the local newspaper, there is a 50% chance of rain tomorrow in Sandy and a 50% chance of rain in South Jordan. Does this mean that

there is a ( )( )0.5 0.5 0.25= probability that it will rain in both cities tomorrow? Justify your answer.

Since Sandy and South Jordan are neighboring suburbs, they are likely to have similar weather. In other words, the weather in one won’t be independent of the weather in the other. It’s not appropriate to use the multiplication rule for independent events in this situation.

In other words, ( ) ( ) ( )rains in Sandy rains in South Jordan rains in Sandy rains in South Jordan .P P P≠ ⋅«

Instead, we would need to find ( ) ( )rains in Sandy rains in South Jordan rains in Sandy .P P⋅ Since the weather

should be similar in the two suburbs, ( )rains in South Jordan rains in SandyP would be much higher than 0.5.

Mutually Exclusive vs. Independent Choose a U.S. adult at random. Define event A: The person is male, and event B: the person is pregnant.

a) Are these two events mutually exclusive? Yes, it is impossible for a person to be both male and pregnant.

b) If you know that event A has occurred, does it affect the probability of event B occurring?

If we know a person is male, the probability that the person is pregnant is 0. ( )pregnant male 0P =

If we know a person is pregnant, the probability that the person is male is 0. ( )male pregnant 0P =

c) Are the two events independent?

No. Knowing that one event has occurred changes the probability of the other event (to 0).

d) If you know two events are mutually exclusive, what conclusion can you draw about whether they are independent? If two events are mutually exclusive, they will never be independent, because knowing that one event has occurred drops the probability of the other event to 0. Being mutually exclusive and being independent are mutually exclusive!

(Note: The converse is not necessarily true. If two events aren’t mutually exclusive, that doesn’t mean that they are independent. There are plenty of events that are neither mutually exclusive nor independent.)

Joint Probability and Tree Diagrams

A joint probability is the probability that two events both occur or ( ).P A B∩ A tree diagram is a useful

technique for solving problems involving joint probability. 1. Represent each outcome of the first random process with a branch of a tree. Label each branch with the

outcome it represents and the probability that the outcome occurs. 2. For each outcome of the first random process, create new branches for the outcomes of the second

random process, labeling each with the outcome it represents and the conditional probability of the outcome given the first outcome occurred.

3. Repeat for any additional random processes. 4. To calculate each joint probability, multiply the probabilities across each set of branches.

Example: A random sample of high school students is taken. Of the 40 students, 18 have allergies and 22 don’t. Suppose that we choose two students at random from the sample. Draw a tree diagram and use it to find the probability that both students suffer from allergies.

( )18 17

both have allergies 0.19640 39

P = ⋅ =

Example: The Kaiser Family Foundation released a study about the influence of media in the lives of young people aged 8-18. In the study, 17% of the youth were classified as light media users, 62% were classified as moderate media users, and 21% were classified as heavy media users. Of the light users who responded, 74% described their grades as good (A’s and B’s), while only 68% of the moderate users and 52% of the heavy users described their grades as good.

a) Draw a tree diagram to model the situation.

b) What percent of the young people described their grades as good?

0.1258 0.4216 0.1092 0.6566+ + = 65.66% of the students in the sample described their grades as “good”.

c) What percent of the young people with good grades are heavy users of media?

( )( )

( )

heavy good grades 0.1092heavy good grades 0.1663 16.63%

good grades 0.6566

PP

P

∩= = = =

Example: At the beginning of the year, Ms. Linford asked all of her AP Statistics students the question, “Who is the current Prime Minister of Great Britain?” She then asked whether the answer was a guess. Only 17.1% of the students knew the correct answer without guessing. The rest guessed. Of the students who guessed, 11.8% ended up choosing the correct answer. If a randomly selected student answered the question correctly, what is the probability that the student actually knew the answer?

( )( )

( )

knew answer right 0.171knew answer right 0.636 63.6%

right 0.0978 0.171

PP

P

∩= = = =

+

Example: Many employers require prospective employees to take a drug test. A positive result on this test indicates that the prospective employee uses illegal drugs. However, not all people who test positive actually use drugs. Suppose that 4% of prospective employees use drugs, the false positive rate is 5%, and the false negative rate is 10%. What is the probability that a person who tests positive actually uses illegal drugs?

( )( )

( )

drugs + test 0.036drugs + test 0.429 42.9%

+ test 0.036 0.048

PP

P

∩= = = =

+