module 7 part 3

P(B|A)=P(A and B) . True or False?

This is a false statement. P(B∣∣∣A)=P(A and B)/P(A) .

P(B|A) indicates

conditional probability. If A and B are independent events, P(B|A)=P(B) .

There are 8 blue, 8 brown, and 8 black marbles in a bag. If the first one drawn from the bag is blue and not returned, then what is the probability the next one drawn is brown?

8/23 is Correct × Correct. The answer is 8 23 . There are 23 marbles left in the bag and 8 of them are brown. The probability is 8 23 .

A frequency table is read by the row and column entries. True or False?

This is a true statement. Each entry is read by the row and column entry.

One deck of cards is numbered from 1−10 and the second deck is numbered 1−5 . A card is drawn from each deck. What is the probability of a 5 and a 5 ?

1/50 is Correct × Correct. The answer is 1/50. The probability is 1/10 for 5 in the first deck. The probability is 1/5 for 5 in the second deck. The probability of 5 and 5 is (1/10)(1/5) = 1/50.

For the situation above, what is the probability that a student is female or already has a degree in a non-nursing major?

17/20 is Correct × Correct. The answer is 1720. This requires the general addition rule. P(female)+P(non-nursing degree)−P(female and non-nursing degree)= 4/5+1/4 −(45×14) =1720

A puppet-making booth has the choice of 3 different color bodies and 3 types of head. What is the probability of making one particular type of puppet?

19 Correct. The answer is a. There are 3 body types and 3 types of head, there are 9 different kinds of puppets. The probabilities are 13×13=19

There are 52 playing cards in a complete set, also known as a deck. Within each deck, there are four categories, known as suits, with 13 cards in each suit. There are two black suits (clubs and spades) and two red suits (diamonds and hearts). Each suit has a card numbered two through ten, as well as a "jack", a "queen", a "king", and an "ace". The 13 cards of each suit are usually thought of in this ascending order. If the first spot taken is a handicapped spot, what is the probability the next spot taken is a handicapped spot?

19/99 is Correct × Correct. The answer is 19 99 . .

Sally has 3 coats - one is blue, one is red, and one is green. She also has red and green gloves. Everyday she wears one coat and one set of gloves. Which of the following uses the General Addition Rule to calculate the probability that Sally wears green on any given day? (The fractions are unsimplified to make calculation easier.) a. 1/6+3/6−2/6 b. 2/6+3/6−1/6 c. 2/6+4/6−2/6 d. 3/6+3/6−2/6

2/6+3/6−1/6 Correct. The answer is b. The General Addition Rule gives us a way of calculating the probability of two independent events. You add the probability of each event and subtract their intersection. Construct a table like the one below:

There are 20 handicapped and 80 regular parking spaces. If the first spot taken is a regular spot, what is the probability the next spot taken is a handicapped spot?

20/99 is Correct × Correct. The answer is 20 99 . There are 19 + 80 = 99 parking spaces remaining. The probability is 20/ 99 .

One deck of cards is numbered from 1−10 and the second deck is numbered 1−5 . A card is drawn from each deck. What is the probability of an even from the first deck and odd from the second deck?

3/10 is Correct × Correct. The answer is 3/10. The probability is 5/10 = 12 for even in the first deck. The probability is 3/5 for an odd in the second deck. The probability of an even and an odd is (1/2)(3/5) = 310.

There is a jar with 11 marbles: seven red marbles and four green ones. Your sister then takes a red marble and still has it. What is now the probability of choosing a green marble, if you select a marble at random? a. 4/10 b. 4/11 c. 3/11 d. 3/10

4/10 Correct. The answer is a. There are now four green marbles and 10 total marbles in the jar.

There are 10 cards in a bag numbered 1−10 . If a 3 is drawn first, what is the probability of an odd card being drawn next if there is no replacement?

4/9 is Correct × Correct. The answer is 4 9 . There are 9 cards remaining after the first draw with 4 odd cards. The probability is 4/9 .

At a university, 45 of the nursing majors are female. At the same university 25 are over 40, and 14 already have a degree in a non-nursing major. If these are independent events, what is the probability of a student being female and over 40 ?

8/25 is Correct × Correct. The answer is 8/25. This is an intersection problem: the intersection of the probability of a female and the probability of over age 40. P(female)×P(age over 40)= 4/5×2/5=8/25

There are 10 cards in a bag numbered 1−10 . If a 2 is drawn first, what is the probability for a card greater than or equal to 3 being drawn next if there is no replacement?

8/9 is Correct × Correct. The answer is 8 9 . There are 9 cards remaining after the first draw with 8 cards greater than or equal to 3. The probability is 8/9 .

Examples of Dependent Events

Being a woman and living past the age of 80 Exercising and living past the age of 80 Being dealt an ace and drawing a second ace Living in a nursing home and developing a urinary tract infection Being read to as a toddler and knowing more than 800 words by age 3 Consuming 400 mg of caffeine daily and having normal blood pressure Spending more than 2 hrs each day in a car and having been in an car accident during the last year

Conditional Probability

Conditional probabilities represent a prevalent category of probabilities. From card games to medical conditions, people want to know how one event or condition affects the probability of another event or condition. There is a special notation to indicate conditional probability, the probability that a second event B will occur, given the occurrence of event A . We write P(B|A) , which means "the probability of event B , given event A has occurred." The probability of event B occurring is dependent on whether event A has occurred. Please note,

How can we use the table to determine whether drinking tea and being obese are dependent or independent events? If drinking tea and being obese are independent events, the occurrence of one will not affect the probability of the other. How do we estimate the probability of "drinking 2 or more cups of tea a day"?

Find the number of heavy tea drinkers and divide it by the total number of people. This will give us the relative frequency of being a heavy tea drinker: 201/1000=0.201 or 20.1% .

examplt of multiplication rule

For example, the probability of rolling a one with a six-sided die is 16 . When rolling two dice with six sides each there are 16×16=136 probability of rolling double ones. You can verify this probability on the table below. Notice there is only one cell with double ones and there are 36 ( 6×6 ) cells in total. Therefore, the probability of double ones is 136 . Likewise, the probability of rolling triple ones when rolling 3 dice is 1/6×1/6×1/6=1/216 .

What is the probability of drinking 2 or more cups of tea if a subject is obese?

If the events are independent, being obese should not affect the probability of drinking tea. Here we look at the numbers in the "obese" column: 285 and 72 . They total to 357 . Of the 357 obese people, 72 drink 2 or more cups of tea a day. The relative frequency of drinking a lot of tea while being obese is 72/357≈0.20167 (rounded to the nearest hundred thousandth), or 20.2% . The relative frequencies are very similar, suggesting that the probabilities are the same and these two events, therefore, are independent.

Independent Events and The Multiplication Principle

Is there a way to determine the probability of a complex series of options, such as the having-three-children situation discussed on the earlier page, without taking the time to draw out a tree or make up a table? The answer is yes, but before you learn the process, you need to know another definition.

Disjoint A and B (intersection)

It is relatively easy to calculate the probability of two or more disjoint events. If there is no overlap, there is no possibility of belonging to both sets. The probability that someone is born in both March and January, for example, is 0 . This is true of the intersection of all disjoint events: If A and B are disjoint, P(A and B)=0

In another game, you win if you roll a number less than 3 or if you roll an even number. What is your probability of winning this game?

P(A or B)=P(A)+P(B)−P(A and B). There are 2 numbers less than 3, so the probability of event A is 26 or 13. There are 3 numbers that are even, so P(B)=12. But the number 2 is both less than 3 AND even, so P(A and B)=16 . The formula gives us 1/3+1/2−1/6=2/3 67% chance of winning

You roll two six-sided dice. What is the probability of rolling a sum less than 5 or greater than 10 ?

The answer is 14. There are 6 results less than 5 (rolling 1, 1; 1, 2; 1, 3; 2, 1; 2, 2; 3, 1) and 3 results greater than 10 (rolling 5, 6; 6, 5; 6, 6). The probability is 6/36+3/36=9/36=1/4.

The frequency table does not need labels for rows or columns. True or False?

This is a false statement. A frequency table needs labels for rows and columns to display the data accurately.

Conditional probability does not rely on another event happening. True or False? a. True b. False

This is a false statement. Conditional probability is the probability of an event occurring after one event has happened.

A frequency table consists of two categorical variables. True or False?

This is a true statement. A frequency (or two-way) table contains two different categorical variables.

A sample space S is comprised of 13 equally likely outcomes. Suppose event E contains 5 of those outcomes. The probability of E is 513 . True or False?

This is a true statement. Each event has a 1/13 probability of occurring, so one of 5 occurring is 1/13+1/13+1/13+1/13+1/13=5/13 .

Events that do not influence each other are independent. True or False?

This is a true statement. Independent events are events that do not affect each other.

Relative frequency helps determine if an event is independent. True or False?

This is a true statement. The relative frequency is related to the independence of events.

If two events are disjoint there is no intersection between the events. True or False?

This is a true statement. There are no common elements between 2 disjoint events.

The addition rule for disjoint events would yield the same result as the general formula for theoretical probability, counting the ways of getting a certain result. True or False?

This is a true statement. We have done problems both ways in this module. We add the individual probabilities for the disjoint events of, say, being born in February OR being born in April. Or we count up all the "ways" of being born in either February or April and put that number over the total size of the sample space. Either method for this problem would yield 212=16

Probability of Dependent Events

Thus far we have calculated the probability of a second event occurring, given a first event. Often we want to determine the probability of both events occurring or some other situation. Here we give some general rules for calculating conditional probabilities

What are dependent events? a. Not independent b. The occurrence of one affects the probability of the other c. All of the above

What are dependent events? a. Not independent b. The occurrence of one affects the probability of the other c. All of the above

Frequency Tables and Independent Events

remember the definition of independent events*, that the occurrence of one does not affect the probability of another? In practice many times researchers use the relative frequency* of events to determine their independence. A researcher might wonder if gender affects the probability of having high cholesterol. They use study results to determine independence. If men and women have high cholesterol equally frequently, we would determine that gender is independent of having high cholesterol.

P(B|A) is read

"The probability of B , given A ". For example, if B is "a person dying from lung cancer" and A is "that person is a life-long smoker," P(B|A) is the probability of "a person dying from lung cancer" given "that person is a life-long smoker." B is the event we are interested in; A is the event that might affect it.

Slide 3 1 out of the 8 outcomes results in boy-girl-boy. Therefore the probability of boy-girl-boy is 18 or 0.125 . We say P(A)=0.125

. Slide 3 1 out of the 8 outcomes results in boy-girl-boy. Therefore the probability of boy-girl-boy is 18 or 0.125 . We say P(A)=0.125 There is also only 1 outcome for girl-boy-girl. So, this probability, P(B) , is also 0.125 . We've seen that if A and B are disjoint, P(A or B)= P(A)+P(B) . Here P(A)=0.125 and P(B)=0.125 , so P(A or B)= 0.125+0.125= 0.25 . End of Last Slidee've seen that if A and B are disjoint, P(A or B)= P(A)+P(B) . Here P(A)=0.125 and P(B)=0.125 , so P(A or B)= 0.125+0.125= 0.25 . End of Last Slide

What is the probability that a coin shows either a heads or a tails?

1 is Correct × The answer is 1. A coin will show either a heads or a tails because that is the complete sample space.

What is the probability that a die shows a 2 or a 3 when rolled?

1/3 is Correct × The answer is 13. There is a 16 probability of rolling a 2 and a 16 probability of rolling a 3. 16 +16 =26 = 13.

Out of class 20 students, what is the probability that you or your 3 friends are picked first?

1/5 is Correct × The answer is 15. The probability is the sum of the probabilities that you or your friends are chosen. The probability is 120+320=420=15.

There are 52 playing cards in a complete set, also known as a deck. Within each deck, there are four categories, known as suits, with 13 cards in each suit. There are two black suits (clubs and spades) and two red suits (diamonds and hearts). Each suit has a card numbered two through ten, as well as a "jack", a "queen", a "king", and an "ace". The 13 cards of each suit are usually thought of in this ascending order.

13/51 is Correct × Correct. The answer is 13 51 . There are 51 cards left in the deck and 13 are spades. The probability is 13/51 .

What is the probability that a traffic light is green or yellow (assuming each light lasts for the same duration)?

2/3 is Correct × The answer is 23 (assuming, erroneously, that the lengths of the lights are equal). There is13 chance of the light being yellow and 13 probability of the light being green.

In 2012 approximately 4.2% of the population delayed needed medical care because of cost and were without health insurance. Approximately 16.9% of the population was without health insurance. You are asked to find out the probability of someone delaying care, given that he or she lacks health insurance.

24.9 is Correct × The answer is 24.9. P(B∣∣A)=P(A and B)P(A)=0.042 0.169=0.2485 or 24.9%.

If there are 6 red scrubs and 7 green scrubs, what is the probability that a red scrub is chosen and then another red scrub? These scrubs are chosen without replacement.

5/26 is Correct × The answer is 526. The probability that one red scrub is chosen is 6/13. The probability that a second red scrub is chosen is 5/12. Canceling the 6 and reducing the 12 leaves a reduced fraction of 5/26

Using the same clothing, what is the probability a green scrub, and then another green scrub is chosen? These scrubs are chosen without replacement.

7/26 is Correct × The answer is 7/26. The probability for 1 green is 7/13. The probability for the second green is 6/12 =1/2. The probability is (7/13)(1/2)=7/

2. In the above situation, what is the probability that a red scrub is chosen and then a green scrub? These scrubs are chosen without replacement

7/26 is Correct × The answer is 7/26. The probability that a red scrub is chosen is 6/13. The probability that a green scrub is then chosen is 7/12. The product is 7/26.

It has been found that 81% of children have regular check-ups (well-child visits) and 59.3% get the flu shot each year. 52% of children have both. What is the probability that a child has had either a well-child visit or a flu shot? (Enter your answer as a percentage without the percentage sign.)

88.3 is Correct × The answer is 88.3%. The probability of a child having a wellness visit or the flu shot is given by the formula P(A or B)=P(A)+P(B)−P(A and B). Substituting in, we get P(A or B)=0.81+0.593−0.52= 0.883 or 88.3%.

To find probabilities:

Create events A and B based on the givens of the problem Establish what kind of probability you will be finding, such as P(A and B) or P(A|B) , depending on what the problem is asking If necessary, determine if A and B are dependent or independent events Write out the appropriate formula Determine the probabilities needed for the formula (or take them from the givens of the problem) Calculate the probability

For dependent events, the probability of B is always equal to the probability of B , given A . True or False?

For dependent events, the probability of B is not equal to the probability of B , given A .

This is called the addition rule of probability*:

If A and B are disjoint, P(A or B)=P(A)+P(B)

Example Suppose 8.6% of all hospital admissions are children admitted to the hospital through the emergency room and 18% of hospital admissions are children. What is the probability that a child who is in the hospital had an emergency room visit?

Let A= a child in the hospital; Let B= coming through the emergency room P(A and B)=0.086P(A)=0.18P(B∣∣A)=0.0860.18P(coming through the emergency room given being a child) =47.8% Interpretation: close to 50% of the children in the hospital initially came through the emergency room

Disjoint and Complementary

Note that events can be disjoint* without being complementary

If the probability of a sunny day is 0.7 and the probability of the date on the calendar being an odd number is 0.5 , what is the probability of the calendar date being odd or it being a sunny day?

P(A or B)=P(A)+P(B)−P(A and B) Since "calendar date being an odd number" and "being a sunny day" are independent events, P(A and B)=0.7×0.5=0.35 . Calendar date being an odd number or being a sunny day =0.7+0.5−0.35=0.85 85% chance of it being sunny or being an odd numbered day on the calendar.

This is the multiplication rule for independent events*.

The Multiplication Rule for Independent Events When two events are independent, P(A and B)=P(A)×P(B)

What is conditional probability? a. Probability of one event b. Probability of two events occuring, individually. c. Probability of an event occurring given that another event has already occurred. d. All of the above

The answer is c. Conditional probability is the probability of an event occurring given that another event has already occurred.

Dependent Events: P(A and B)

The conditional probability rule gives us the formula we can reformulate to calculate the probability that both events occur. Remember that P(B∣A)=P(A and B)/P(A) We can multiply both sides by P(A) to solve for P(A and B) : The General Multiplication Rule P(A and B)=P(A)×P(B|A) Returning to our earlier example of the cookies, remember that the cookie jar has 4 sugar cookies and 6 chocolate chip cookies. Let A= "choosing a sugar cookie first." Let B= "choosing a second sugar cookie." P(A)=410 or 25 . Remember that once you've chosen a sugar cookie, there is one fewer and one fewer of all the cookies. P(B∣∣A)=39 or 13 . Therefore: P(A and B)=25×13=215 The probability of choosing two sugar cookies from the bowl is 215 .

Independent Events: P(A or B)

The probability of either of the two events occurring is given by a general formula. Consider again the union of two events, A or B. The red line around the space of A, B and their intersection indicates the space of events characterized by "or."

The intersection is always used for conditional probability. True or False?

This statement is true. Conditional probability does use the intersection.

When researching or trying to learn something about a given population, the population and the sample should always be disjoint. True or False?

his is a false statement. While a population and a sample can be disjoint, they shouldn't be if you're trying to use your sample to learn something about your population!

The | between B and A

is a vertical bar indicating a conditional probability. It is a symbol not a division sign.

An example of events that are both disjoint and complementary are

"flipping heads" and "flipping tails." Together these events constitute the universe of possible events, so they are complementary. You also cannot have them at the same time, which means they're disjoint. But the events "being born on a Wednesday" and "being born on a Friday" are disjoint without being complementary. There are other possibilities so even if you are not born on a Wednesday, that does not necessarily mean you are born on a Friday.

A population* and a sample* can be disjoint as well, but they shouldn't be. For example, s

suppose you are studying the prevalence of obesity in pre-school children. You would not survey the local high school in order to gather data about the population of pre-schoolers. That would make your population and your sample disjoint, which is nonsensical if you're trying to use your sample to learn something about your population. On the other hand, taking two disjoint samples can make sense: in your study of obesity in pre-schoolers, you might take disjoint samples—one in California, one in Massachusetts.

When considering whether two events are independent or dependent,

think of one event in isolation, then think of that event again in relation to the second. Let's do this with two randomly selected people and the likelihood that they will die of a heart attack. First, consider the first person in isolation. According to the CDC, 23.5% of deaths in the US are by heart attack. Now, suppose the second person dies of a heart attack, does this change the probability for the first? No! Since the second person has no effect on the first, these events are independent*.

Ricky randomly works 3 out of every 5 days from home. In the area where he lives, it is sunny outside 2 out of every 5 days (on average). What is the probability that he works from home or it is sunny outside?

this is an "or" question, the probability that he works from home (event A) or it is sunny outside (event B). P(A)=35=0.60 P(B)=25=0.40 P(A or B)=P(A)+P(B)−P(A and B)= 0.60+0.40−0.24=0.76 76% of the time, Ricky works from home or it is sunny outside.

Independent events*

those that are not affected by other trials or events. For example, if you were to flip a coin once, that first result (either heads or tails) would not have any impact when the coin is flipped a second time — the first event gives no indication of what could result from the second event. A more real-world example is two people who don't know each other having heart attacks. If someone has a heart attack, that has no effect on the probability the second person will have a heart attack

Example: An Event That is Not Independent

In a candy container, there are five white chocolates and ten dark chocolates. If you pick a chocolate at random out of the jar, the probability of what you will get on the next pick is affected. If you pick a dark chocolate on your first pick, there will be fewer dark chocolates the next time you pick, lowering the probability you get a dark one on the second draw. On the other hand, if you pick a white chocolate on your first pick, there will be fewer white chocolates the next time you pick, lowering the probability you will get a white on the second draw. Unlike the coin toss, what you get in the first trial is not independent from what you get in the second trial. Another, more real world example, is the event "smoking cigarettes" and "getting lung cancer." The occurrence of the first event affects the probability of the second event.

Table Construction

In two-way tables, the cells represent the number of objects (usually people) who fall into the two categories signaled by the cell's row and column. To read a table, match each cell with its row and column. So in the table above, there were 375 people who were given treatment A and who survived. In the next cell in the same row, there were 25 people who were given treatment A and who died. An important aspect of two-way tables is that each person can fall into only one cell. In the example above, someone cannot receive both treatment A and treatment B. Obviously, someone cannot both die and survive. To find the total number of people involved in the study, you can add up the numbers in each cell. You will not double count anyone, because everyone appears only once in the tally.

Dependent Events

Just as events can occur independently, there are also those whose occurrence can affect the probability of one another. If there is one winning ticket in a bowl and 49 losing tickets, each time someone chooses (and then discards) a losing ticket, there are fewer losing tickets. Therefore, the probability of picking the winning ticket increases slightly with each losing ticket picked out of the bowl. Events in which the probability changes depending on other events are categorized as dependent events*.

26% of Americans age 65 or older have diabetes. 17.68% of people over 65 have diabetes and die from heart disease. 25% of people over 65 die from heart disease. What is the probability of being 65 or older with diabetes or dying from heart disease?

Let A= being 65 or older with diabetes Let B= dying from heart disease after age 65 P(A)=26% P(B)=25% P(A and B)=17.68% What is the P(A or B) ? P(A or B)=P(A)+P(B)−P(A and B)P(A or B)=0.26+0.25−.1768=0.3332 There is a 33.32% chance of getting diabetes or dying of heart disease after age 65

Example 26% of Americans age 65 or older have diabetes. 68% of people over 65 with diabetes die from heart disease. What is the probability of having diabetes and dying from heart disease?

Let A= being 65 or older with diabetes Let B= dying from heart disease after age 65 P(A)=26% P(B|A)=68% What is the P(A and B) ? P(A and B)=P(A)×P(B|A)P(A and B)=0.26×0.68=0.176 There is a 17.68% chance of getting diabetes and dying of heart disease after age 65

The chances of having a child with autism are 1 in 88 . Researchers think there is a genetic component to autism because the probability of giving birth to an autistic child if you already have a child with autism is 7% . Given these statistics, what is the probability of having two autistic children (if you are planning to have two children)?

Let A= having a first child with autism Let B= having a second child with autism P(A)=188=0.01136P(B∣∣A)=7% What is the P(A and B) ? P(A and B)=P(A)×P(B|A)P(A and B)=0.01136×0.07=0.000795 There is a 0.08% chance of having two autistic children

The probability of going fishing on a Saturday is represented as P(F) . The probability of catching a fish on a Saturday is P(C) . How would we write the probability of catching a fish on a Saturday, given that you have gone fishing? a. P(C&F) b. P(F|C) c. P(C|F)

P(C|F) Correct. The answer is c. This is read as "the probability of catching a fish, given that you have gone fishing."

What is the probability that a person was born on a Tuesday, Thursday, or a Friday?

3/7 is Correct × The answer is 37. There is a 17 chance of being born on each day of the week. 17+17+17=37.

If the probability that your favorite pitcher is pitching is 50% , the probability that your baseball team will win is 50% , and the probability that your team will win and your favorite pitcher is pitching is 30% , what is the probability that either your favorite team will win or your favorite pitcher will be pitching? (Enter your answer as a decimal.)

0.7 is Correct × The answer is 0.7. The probability of the team winning or your favorite pitcher pitching is given by P(A or B)=P(A)+P(B)−P(A and B). Substituting in, we have P(A or B)=0.5+0.5−0.3=0.7

What is the probability of selecting a weekend day or Memorial Day (Monday) in May if the first day is a Sunday?

10/31 is Correct × 10/31. There are 9 weekend days and 1 Memorial Day as well as 31 days in May. The probability is 9/31+1/31=10/31.

A coin is flipped three times. What is the probability of flipping three heads in a row (do NOT construct a tree; instead, use the multiplication principle)?

18 Correct. The answer is c. There are 2 possibilities for each coin toss. The probabilities are 12×12×12=18

What is the probability that a 10 -sided fair die shows a multiple of 3 or 10 ?

2/5 is Correct × The answer is 25. There are 3 multiples of 3 (3, 6, 9) and one value of 10. The probability is 310+110=410=25 .

For the situation above, what is the probabilty that a student is female or over age 40 ?

22/25 is Correct × Correct. The answer is 22/25. This requires the general addition rule. P(female)+P(age over 40)−P(female and age over 40)= 4/5 + 2/5 − (4/5×2/5)= 22/25

There are 52 playing cards in a complete set, also known as a deck. Within each deck, there are four categories, known as suits, with 13 cards in each suit. There are two black suits (clubs and spades) and two red suits (diamonds and hearts). Each suit has a card numbered two through ten, as well as a "jack", a "queen", a "king", and an "ace". The 13 cards of each suit are usually thought of in this ascending order. There are 52 cards in a deck of cards. If the first card drawn is red and not put back into the deck, then what is the probability the second card drawn is also red?

25/51 is Correct × Correct. The answer is 25 51 . There are 51 cards left in the deck and 25 are red. The probability is 25/51 .

One deck of cards is numbered from 1−10 and the second deck is numbered 1−5 . A card is drawn from each deck. What is the probability of a multiple of 3 from the first deck and 3 from the second deck?

3/50 is Correct × Correct. The answer is 3/50. The probability is 3/10 for a multiple of 3 in the first deck. The probability is 1/5 for a 3 in the second deck. The probability of a multiple of 3 and a 3 is (3/10)(1/5) = 3/50.

There are 52 playing cards in a complete set, also known as a deck. Within each deck, there are four categories, known as suits, with 13 cards in each suit. There are two black suits (clubs and spades) and two red suits (diamonds and hearts). Each suit has a card numbered two through ten, as well as a "jack", a "queen", a "king", and an "ace". The 13 cards of each suit are usually thought of in this ascending order. If the first card drawn is a heart and not returned to the deck, then what is the probability the second card drawn is also a heart?

4/17 is Correct × Correct. The answer is 4 17 . There are 51 cards left in the deck and 12 are hearts. The probability is 12/51 = 4/17 .

. A container contains 6 blue, 4 black, 5 yellow, and 3 purple marbles. Two marbles are drawn with replacement. What is the probability of drawing a black marble twice?

4/81 is Correct × Correct. The answer is 481. There are 6+4+5+3=18 marbles in the container. The probability of black is 418=29. The probability of two black is (2/9)(2/9) = 4/81 .

A container contains 6 blue, 4 black, 5 yellow, and 3 purple marbles. Two marbles are drawn with replacement. What is the probability of drawing a blue and yellow?

5/54 is Correct × Correct. The answer is 5/54. There are 6+4+5+3=18 marbles in the container. The probability of blue is 6/18 = 13 and yellow is 5/18. The probability of blue and yellow is (1/3)(5/18) =5/54.

There are 10 cards in a bag numbered 1−10 . If an even card is drawn first, what is the probability of an odd card being drawn next if there is no replacement?

5/9 is Correct × Correct. The answer is 5 9 . There are 9 cards remaining after the first draw with 5 odd cards. The probability is 5/9 9 .

A 12 -sided die is rolled. What is the probability of a multiple of 4 or a number less than or equal to 5 ?

7/12 is Correct × Correct. The answer is 7/12. The probability of a multiple of 4 is 3/12. The probability of a number less than or equal to 5 is 5/12. There is one value that overlaps (4) or 112. The probability is 3/12 +5/12 − 1/12 = 8/12 −1/12 = 7/12./

A 12 -sided die is rolled. What is the probability of an even number or a number greater than or equal to 10 ?

7/12 is Correct × Correct. The answer is 7/12. The probability of an even number is 6/12. The probability of a number greater than or equal to 10 is 3/12. There are two values that overlap (10, 12) or 2/12. The probability is 6/12+3/12−2/12=9/12−2/12=7/12

There are 8 blue, 8 brown, and 8 black marbles in a bag. If the first one drawn from the bag is black and not returned to the bag, then what is the probability the next one drawn is black?

7/23 is Correct × Correct. The answer is 7 23 . There are 23 marbles left in the bag and 7 of them are black. The probability is 7 23 .

Again using the same clothing, what is the probability that a green scrub is chosen first and then a red scrub? These scrubs are chosen without replacement.

7/26 is Correct × The answer is 726. The probability that a green scrub is first chosen is 7/13. The probability that a red scrub is then chosen is 6/12. Notice the numerators of the previous calculation have just been reversed. The reduced fraction is 7/26.

example

A cookie jar has four sugar cookies and six chocolate chip cookies. With conditional probability, we're looking to find the probability of choosing certain cookies in different circumstances. a picture of cookies What is the probability that a sugar cookie is chosen first? There's 4/10 or 2/5 chance a sugar cookie is chosen first, but what is then the probability of choosing another sugar cookie? There is one less sugar cookie in a jar that now has only 9 cookies. Picking a second sugar cookie has a probability of 3/9 or 1/3 . Since it is affected by the event before, the probability of picking another sugar cookie is smaller. The probability of this event is dependent on the event prior. The probability of picking a sugar cookie on the second draw, given you have already picked a sugar cookie on the first draw is 1/3 . If the first cookie chosen is chocolate chip, the cookie jar would then have all four sugar cookies remaining, among the nine cookies left in the jar. Therefore, probability of choosing a sugar cookie on the second draw, given you have already picked a chocolate cookie is 4/9 . Notice how the numbers change and will continue to change as the events occur — because the circumstances are changing and the events are dependent upon one another, the probabilities also change.

Frequency Tables

A frequency table (two-way table) is a table consisting of frequency counts of two categorical variables. One variable is called the row variable, and the other variable is called the column variable. These tables are very helpful when it comes to analyzing data.

Disjoint Events

Before we introduce formulas for probabilities, we will continue to define different categories of events. Disjoint events have special characteristics that make their intersection* and union* unlike those of other events. f two events cannot both occur at the same time, they are called disjoint*. This concept applies to any number of multiple events. For example, a coin cannot be both "heads" and "tails" at the same time. The three statements below are disjoint: Examples of Disjoint Statements I was born in January. I was born in March. I was born in May. No two of those statements can both be true.

Disjoint A or B (union)

Calculating the probability of the union of A and B is also logical. Since there is no overlap, the probability of being in either one or the other is the sum of their individual probabilities. What is the probability that someone is born in January or March, for example? The sample set of "month of birth" is comprised of the 12 months of the year. If you think about it, there are 2 ways to be born in January or March. You can be born in January ( 1 chance out of 12 months) or you can be born in March ( 1 chance out of 12 months). These 2 chances lead to a probability of 212 . P(being born in January)=112P(being born in March)=112 The probability of being born in January OR March is 112+112=212=16 This is called the addition rule of probability*: If A and B are disjoint, P(A or B)=P(A)+P(B)

Conditional Probability Rule

Conditional Probability Rule P(B∣∣∣A)=P(A and B)/P(A) Let us illustrate this rule by using a very simple example. Imagine you are flipping a coin twice. Let A= "getting a head on the first flip" and let B= "getting two heads." Let us look at a number of categories related to probability: The sample space* of this event is HH, HT, TH, TT P(A) is 2 possibilities out of 4 or 12 P(B) is 1 possibility out of 4 or 14 A and B , the intersection, is the set of elements that belong to both categories, which is just the element HH. So the P(A and B)=14 If you have already flipped a head, you have two possibilities for the second flip, heads and tails. P(B∣∣A)=12 So substituting the probabilities above into our general conditional probability formula P(B∣∣∣A)=P(A and B)P(A) gives 12=1/4 divided by1/2 This very useful formula generally cannot be verified by writing out the entire sample space. Instead, we often use it with probabilities that we are given./

Conditional probability*

Conditional probability is used to calculate these kinds of events. Conditional probability* is the probability of an event occurring (such as picking the winning ticket), given that another event has already occurred (such as someone else having already picked one of the losing tickets).

Disjoint Probability

Disjoint Probability Notice that when asking about disjoint events, we naturally employ the word "or." Questions that ask about the probability of one or more disjoint events, always use an "or" and can be calculated using the formula: P(A or B) =P(A)+P(B)

Disjoint Events: Dependent

Disjoint events are dependent. For example, if you know someone is born on a Wednesday, you also know that he or she cannot be born on a Friday. If you know someone is not born on a Wednesday, there is a greater possibility, 16 rather than 17 , that he or she is born on a Friday. Because the occurrence of one affects the probability of the other, these events are dependent.

Dependent Events: P(A or B)

Earlier we learned how to calculate the probability of the union of disjoint* events: P(A or B)=P(A)+P(B) We also learned a "General Addition Rule" for independent* events. The rule for dependent events is the same, and is reproduced here: General Addition Rule P(A or B)= P(A)+P(B)−P(A and B) Let us return to our simple example to see how the general addition rule works for dependent events as well. Recall that we are flipping a coin twice and A= "getting a head on the first flip" and B= "getting two heads." P(A or B) is the probability that either a head is flipped first or two heads are flipped. Clearly, event B includes event A ; if we get two heads, we must have flipped heads on the first flip. There are two cases where we flip heads on the first flip: HH and HT. Therefore P(A or B)=1/2 . Notice how this probability follows the general addition rule: P(A or B)=P(A)+P(B)−P(A and B)P(A or B)=1/2+1/4−1/4=12

If the probability of having blond hair is 5% , then the probability of having blond hair, given that you are Swedish, is 5% . True or False? a. True b. False

False Correct. This is a false statement. Being Swedish and having blond hair are dependent events, as Swedes are more likely to have blond hair.

In the problem above, making a puppet with a red body and making a puppet with a blue body are independent events. True or False?

False Correct. This is a false statement. You can only choose one color for the body of the puppet, so choosing a red-bodied puppet and choosing a blue-bodied puppet are disjoint events. Since the events are disjoint, they are dependent

Notice that this number overstates the union by two cells. Where did we go wrong? We counted the probability of wearing both a black suit and black shoes twice. You can see that in the table above two cells have two orange terms.

General Addition Rule P(A or B)=P(A)+P(B)−P(A and B) By subtracting the intersection, P(Bob wears a black suit and Bob wears black shoes) , we avoid counting that area twice in the sum of P(Bob wears a black suit) and P(Bob wears black shoes) . So the formula is to add P(Bob wears a black suit) and P(Bob wears black shoes) , and then subtract P(Bob wears a black suit and black shoes) . Another way to understand it is to look at the Venn diagram above. To find the probability of both circles, you add the "area" of each circle, but then, notice you've counted the area of the intersection twice (once for the blue circle and once for the yellow circle), so it will need to be subtracted to compensate. For disjoint events, there is no area of intersection, so no need to subtract it. The formula above is the formula for events that are not disjoint*. Remember that for events that are disjoint, the general rule simplifies to P(A or B)=P(A)+P(B) since there is no area of intersection.

If balloons come in equal numbers of red, green, yellow, orange, purple, pink, gray, and peach what is the probability that someone gets a ballon that is NOT red?

The answer is 78. Each of the 8 colors has a 18 chance of happening. The sum of these disjoint events is 78.

There are 20 handicapped and 80 regular parking spaces. If the first spot taken is a handicapped spot, what is the probability the next spot taken is a handicapped spot?

There are 19 + 80 = 99 parking spaces remaining. The probability is 19/99

Two-way Table Analysis

There are many ways to analyze the data in a two-way table. To assist in the analysis, we should add a row and a column to record the totals for each:

In a certain game, you roll a fair, six-sided die once. You win if you roll a 3 or a 4 . What is the probability of winning this game?

These are disjoint events. P(A or B)=1/6+1/6 33% chance of winning

Slide 1Drawing from the example of three children, suppose you want to calculate the probability of having a boy-girl-boy (which we'll call event A ) or having a girl-boy-girl (event B ).End of First Slide

These events are clearly disjoint. There is no way to do both. To calculate the probability of one or the other, you use the addition rule

If 2 events can occur at the same time, they are called disjoint. True or False?

This is a false statement. Disjoint events are 2 events that cannot occur at the same time.

The only information you need to perform analysis in a two-way table can be provided by row entries. True or False?

This is a false statement. You also need the totals in columns and rows to perform analysis.

Ricky randomly works 3 out of every 5 days from home. In the area where he lives, it is sunny outside 2 out of every 5 days (on average). What is the probability that he works from home and it is sunny outside?

This is an "and" question, the probability that he works from home (event A) and it is sunny outside (event B). P(A)=3/5=0.60 P(B)=2/5=0.40 P(A and B)=0.60×0.40=0.24 24% of the time, Ricky works from home and it is sunny outside.

A co-worker jokes that Bob always wears black. Referring again to the table, the probability of Bob wearing a black suit OR a black shoes is 1115 . There are 11 cells that have either a black suit or black shoes (or both) and 15 cells total. 1115 is the probability of either event happening, including the situation in which both events happen. How can we get this answer without creating a whole table?

To calculate an "or" problem, it makes sense to add the probability of both events. The probability of one of two (or more) events occurring is just the probability of each event added together. But notice that when you add all the times Bob wears black shoes and all the times Bob wears a black suit, you are counting the cells where both occur twice. P(Bob wears a black suit)=15P(Bob wears black shoes)=23P(Bob wears a black suit or black shoes)=1/5+2/3=3/15+10/15=13/15

Suppose there are two different kinds of medical gloves, A and B used by medical assistants. A costs less but has a defective rate of 4.2% . The more expensive brand, B, costs more, but has a defective rate of 1.7% . If 3% of the procedures performed by medical assistants involve potentially infectious materials, what is the percentage chance of exposure for each brand of glove?

Using the multiplication rule for independent events, you multiply 4.2%×3% to find the probability of exposure for brand A. You perform a similar calculation to find the probability of exposure for brand B. Probability of exposure for brand A: 0.042×0.03=0.00126 Remember to convert probabilities to decimals before multiplying and remember to count the total number of places after the decimal point to obtain the correct answer. Probability of exposure for brand B: 0.017×0.03=0.00051 You can convert both decimal probabilities back to percentages if you feel more comfortable interpreting data in that form. Brand A has a 0.00126 probability of exposing medical assistants to infectious material, which is 0.126% . Brand B has a 0.00051 probability of exposing medical assistants to infectious material which is 0.051% .

Kelly gets a donut on the way to work when she has time and the donut shop has chocolate donuts. If the probability of her having time by getting the earlier train is 75% and the probability of the shop still having chocolate donuts is 50% , what is the probability that she gets her chocolate donut?

We are looking for an intersection here, an "and" problem. P(A and B)=P(A)×P(B)=0.75×0.5 37.5% chance of getting a donut.

Obesity and Drinking Tea Consider the following table showing tea consumption and obesity. We might use the table to ask if tea consumption and obesity are linked or independent*.

Weight: Non-obese Obese Total Tea Consumption None or 1 Cup per Day non obese 514 obese 285 total 799 2 or More Cups per Day non obese 129 obese 72 total 201 Totals 643 357 1000

Example 2 Suppose there are two convenient parking spaces at a company building and the remaining eight are far from the door. There are two elderly employees at the company and eight middle-aged employees. The arrival time of employees in the morning is random.

What is the probability that an elderly employee gets the second space, given that a elderly employee has already gotten the first space? Let A= "An elderly employee gets the first near parking spot" P(B|A)= Because one employee has parked, there are now nine employees looking for spaces. Because one of the elderly employees has already gotten the first spot, there is one left. 1/9 What is the probability that an elderly employee gets the second space, given that a middle-age employee has gotten the first space? Let B= "An elderly employee gets the second near parking spot" P(B|not A)= Because one employee has parked, there are now nine employees looking for spaces. Because neither of the elderly employees has gotten the first spot, there are two left 2/9 Notice in the example above, the probability of B changed if A had occurred. P(B)≠P(B|A) shows that A and B are dependent events.

Important Numerical Phrases

When understanding an event, some phrases can be challenging to interpret until you get used to them. For example, the Center for Disease Control and Prevention holds that a BMI (Body Mass Index) above the 95 percentile is obese and a cause for concern. For a six-year-old boy, a BMI of 18.4 falls above the 95 th percentile. One category for measurement, therefore, might be 18.4 or above. A six-year-old with a BMI of 18 would not fall into the category of 18.4 or above, while a six-year-old with a BMI of 19.2 would. So the event "being a six-year-old and having a body mass of 18" and "being obese (that is having a BMI of 18.4 and above)" are disjoint. Similarly, a category might have a range of values or require one of a set of conditions. In all these cases, pay attention to whether the value you are considering belongs to the category or is disjoint.

Independent Events: P(A and B)

You may have noticed a pattern when we were constructing the tree for having three children. For each new option we had to add an additional set to each of the previous options. This, in essence, is multiplication. At each point in the tree the probability of one particular outcome, say "having a boy," is equal to 1 divided by the total number of possibilities, 2 . As the number of options increase, the new probability is captured by multiplying the previous probability by the probability of the new independent event in the sequence. This is the multiplication rule for independent events*.

In 2012 approximately 4.2% of the population delayed needed medical care because of cost and were without health insurance. Approximately 16.9% of the population was without health insurance. You are asked to find out the probability of someone delaying care, given that he or she lacks health insurance. If the situation described above is represented at P(B|A) , what are the events A and B ? A= being uninsured; B= delaying needed care A= delaying needed care; B= being uninsured A= the probability of being uninsured; B= the probability of delaying care A= the probability of delaying needed care; B= the probability of being uninsured

a is Correct × The answer is a. If you are trying to find the probability that someone delays care given he or she lacks health insurance and that is represented as P(B|A), then B= "someone delays needed care" since this is the probability you are trying to find. Meanwhile, you are given A or "someone lacks health insurance." A= being uninsured; B= delaying needed care.

In 2012 approximately 4.2% of the population delayed needed medical care because of cost and were without health insurance. Approximately 16.9% of the population was without health insurance. You are asked to find out the probability of someone delaying care, given that he or she lacks health insurance. Which of the following formulas should be used to solve for P(B|A) ? P(B|A)=P(A)×P(B) P(B|A)=P(A)+P(B) P(B|A)=P(A and B)×P(A) ) P(B∣∣∣A)=P(A and B)P(A)

d is Correct × The answer is d. The formula for conditional probability is P(B∣∣A)=P(A and B)P(A)