module 7 part 4

Getting a hockey stick and a green item, using P(A and B)=P(A)×P(B) .

dependent is Correct × The answer is dependent. Let A= getting a hockey stick. Let B= getting a green item. P(A and B) is the probability that both a green item and a hockey stick are chosen. There is one green hockey stick, so this probability is 1/12. Meanwhile, P(A)=1/12 and P(B)=612=12 since there are 6 green items out of 12 total items. P(A)×P(B)=1/12×1/2=1/24. So, P(A and B) ≠ P(A) × P(B). Therefore, the events are dependent.

Getting a toy lawnmower and a brown item, using P(B|A)=P(B) .

dependent is Correct × The answer is dependent. Let A= getting a toy lawnmower and B= getting a brown item. There are two lawnmowers, and one of them is brown. Therefore, P(B∣∣A)=12. Meanwhile, the probability of choosing a brown item is 4/12=1/3. Since P(B|A)≠P(B), the events are dependent.

Getting a truck and an orange item, using P(A and B)=P(A)×P(B) .

dependent is Correct × The answer is dependent. Let A= getting a toy lawnmower and B= getting a brown item. There are two lawnmowers, and one of them is brown. Therefore, P(B∣∣A)=12. Meanwhile, the probability of choosing a brown item is 412=13. Since P(B|A)≠P(B), the events are dependent.

A and B are independent events. P(A)=0 and P(B)=1. Calculate P(A and B). a) 0 b) 0.1 c) 0.5 d) 1

he correct answer is a. P(A and B)=P(A)⋅P(B)=0⋅1=0 .

Getting a beach ball and a green item, using P(A and B)=P(A)×P(B) .

independent is Correct × The answer is independent. Let A= getting a beach ball. Let B= getting a green item. P(A and B)=112 since there are 1/2 items and only 1 green beach ball. P(A)=2/12=16. P(B)=6/12=1/2. P(A)×P(B)=1/6×1/2=1/12. Since P(A and B)=P(A)×P(B), the events are independent according to this formula.

Getting a hockey stick and a red item, using P(B|A)=P(B|not A) .

independent is Correct × The answer is independent. Let A= getting a hockey stick and let B= getting a red item. P(B|A)=0 since there is only one hockey stick and it isn't red. P(B|not A)=0, as there are 11 non-hockey stick items, and none of them are red. Since 0=0, the events are independent.

Getting a toy car and a green item, using P(B|A)=P(B|not A) .

independent is Correct × The answer is independent. Let A= getting a toy car. Let B= getting a green item. P(B∣∣A)=1/2 since there are 2 toy cars and one of them is green. P(B∣∣not A)=12 since there are 10 items that are not toy cars, and 5 of them are green. Since P(B|A)=P(B|not A), the events are independent according to this formula.

Getting a toy truck and a brown item, using P(A|B)=P(A) .

independent is Correct × The answer is independent. Let A= getting a toy truck. Let B= getting a brown item. P(A∣∣B)=1/4 since there are 4 brown items and 1 of them is a toy truck. The P(A)=3/12=14, since 3 of the 12 items are toy trucks. Since P(A|B)=P(A), the events are independent, according to this formula.

The definition of independent events* is

that the occurrence of one does not affect the probability of the other. So for formula 2 above [P(A|B)=P(A)] , we are saying that it does not matter if B has occurred or not; A still has the same chance of happening either way. Similarly, in formula 3 above, [P(B|A)=P(B)] , we are saying that the probability of B does not change even if A has already happened. The third formula says this a different way, saying that the odds of B happeneing are the same if A has happened or if A has not happened. The first formula on the list P(A and B) is defined as P(A)⋅P(B) for independent events. If A and B are dependent events*, remember the formula you should use is P(A and B)=P(A)⋅P(B|A) .

When calculating probabilities, it is important to understand

the difference between independent and dependent events. Examine the tables below to familiarize yourself with features of independent and dependent events. Independent Events

A and B are independent events. P(A)=0.8 and P(B)=0.2. Calculate P(B|A). a) 0.16 b) 0.2 c) 0.8 d) 1

) 0.2 Correct. The correct answer is b. P(B|A)=P(B)=0.2.

A and B are independent events. P(A)=1 and P(B)=0.9. Calculate P(A and B). a) 0.09 b) 0.9 c) 1 d) 1.9

) 0.9 Correct. The correct answer is b. P(A and B)=P(A)⋅P(B)=1⋅0.9=0.9

What do we call two events for which the occurrence of the first affects the probability that the second event occurs? a) Independent events b) Dependent events c) Mutually exclusive events d) All of the above

) Dependent events Correct. The correct answer is b. If the occurrence of the first affects the probability that the second event, they are known as dependent events.

You flip a coin twice. Determine if the following two events are independent or dependent: Flipping a heads the first time and flipping a tails the second time. a) Independent b) Dependent

) Independent Correct. The correct answer is a. Flipping a heads the first time does not affect the probability that you will flip a tails the second time. Therefore, they are independent events.

What is the probability of this branch of the probability tree?

.0564 is Correct × The answer is .0564. The probability is 0.12(0.47)=0.0564.

What is the probability of this branch of the probability tree?

.0636 is Correct × The answer is .0636. The probability is

. What is the probability that a patient is retired/not working?

.21 is Correct × The answer is .21.The probability is 1−0.67−0.12=0.21.

What is the probability that a patient makes less than $50,000 per year, given that the patient works from home?

.53 is Correct × The answer is .53. The probability is 1−0.47=0.53

If a patient works from an office, what is the probability that the patient makes less than $50,000 per year?

.59 is Correct × The answer is .59. The probability is 1−0.41=0.59.

What is the probability that a patient makes less than $50,000 per year, given that the patient is retired?

.89 is Correct × The answer is .89. The probability is 1−0.11=0.89.

A and B are independent events. P(A)=0.7 and P(B)=0.5 . Calculate P(A|B) . a) 0.35 b) 0.5 c) 0.7 d) 1.2

0.7 Correct. The correct answer is c. P(A|B)=P(A)=0.7 .

You can use the same tree to calculate a different probability. What is the probability that your boss gives you a chocolate donut followed by a strawberry donut?

3/8×5/7=15/56 or about 28.8%.

For example, what is the probability that your boss gives you a strawberry donut followed by a second strawberry donut? To do this problem, multiply along the branch of the tree that involves selecting two strawberry donuts.

5/8×4/7=20/56=5/14 , or about 35.7% .

The probability of the union of two dependent events is P(A)+P(B|A)

False Correct. This is a false statement. The probability is P(A)+P(B)−P(A and B).

Extrapolating from the table above, if a man is a current heavy smoker, what is the probability for developing lung cancer?

Find the total of the men who were current heavy smokers: 38 Find the men of those current heavy smokers who developed lung cancer: 9 Divide the number of lung cancer victims by the total number of current smokers: 938=23.68%

Of the men who were current smokers, what proportion developed lung cancer?

Find the total of the men who were current smokers: 152 Find the men of those current smokers who developed lung cancer: 25 Divide the number of lung cancer victims by the total number of current smokers: 25/`152=16.45%

Let us use a variation on our simple example to go through these equalities

Let A= "flipping a head on the first flip," let B= "flipping a head on the second flip." Unlike our earlier example, where B= "flipping two heads," this B is independent of A . What you flip on the first coin toss has no effect on the outcome of the second coin toss. Using this A and B , let us consider the equalities:

1.) What is the probability that a teenager does not use drugs and gets a positive drug test result?

Let A= not using drugs Let B= getting a positive test result P(not using drugs and getting a positive test result)=P(not using drugs)×P(getting a positive result given no use of drugs) P(A and B)=P(A)×P(B|A)=.835×0.07=0.05845 Interpretation: 5−6 teenagers out of 100 will test positive for drugs even though they did not use drugs. c

Brianna likes blue items and boats. Let A= "getting a blue item" and B= "getting a boat" Are A and B independent events? Brianna likes blue items and boats. Let A= "getting a blue item" and B= "getting a boat" Are A and B independent events?

Let us use P(A and B)=P(A)⋅P(B) to test for independence. P(A and B)=18 There is one item that is both blue and a boat. The intersection of "getting a blue item" and "getting a boat" is the blue boat. P(A)=12 There are 4 blue items and 8 items total. P(B)=14 There are 2 boats and 8 items altogether. P(A)⋅P(B)=12×14=18 Therefore P(A and B)=P(A)⋅P(B) and the events are independent

Note that the probabilities of each line of the probability tree sum to 1 . This is because the probability tree includes all possible scenarios, meaning that the sum of all the possibilities must add up to 100% (or 1 ) since all possibilities are accounted for.

Notice that the probabilities of each branch sum to 1 : 20/56+15/56+15/56+6/56=56/56

Which statement does not belong for two independent events?

P(A|B)=P(B) Correct. The answer is b. The correct statement is P(A|B)=P(A).

P(A|B)=P(A)

P(A∣∣B)=12 There are two outcomes, TH and HH, that comprise event B (tossing a H on the second throw). One of those outcomes, HH, is included in event A (tossing a H on the first throw). Therefore the probability that A occurs given B has occurred is 1/2 . (Notice, this fraction is not 24 reduced. We are only considering the half of the sample space in which B has occurred.) P(A)=1/2 There are two outcomes that comprise event A : HT and HH. There are four total outcomes. Therefore P(A)=1/2 P(A|B)=P(A)

P(B|A)=P(B|not A)

P(B∣A)=1/2 P(B∣∣not A)=1/2 P(not A) has two possible outcomes, TH and TT. Of those two outcomes, only one satisfies event B occurring, TH. Therefore the probability is 1/2 P(B|A)=P(B|not A)

P(B|A)=P(B)

P(B∣A)=1/2 The logic here is the same as that used to calculate P(A|B) . There are two outcomes, HT and HH, that are included in event A (tossing a H on the first throw). One of those outcomes, HH, satisfied event B (tossing a H on the second throw). Therefore the probability that B occurs given A has occurred is 12 P(B)=12 There are two outcomes with which B can occur, TH and HH; there are four outcomes total P(B|A)=P(B)

Why does this work?

Recall the formula P(A and B)=P(A)×P(B|A) . To find the probability of one branch of the tree you are solving for P(A and B) —in this case strawberry and strawberry. The first branch gives you P(A) (getting an initial strawberry donut) and the second gives you P(B|A) (getting a second strawberry donut, given you already have one). In a probability tree, the formula can be extended P(C|A and B) etc. for each further branch.

Probability Trees

Remember that earlier we built tree diagrams to help sort out the sample space of an event. For example, we used a tree structure to determine the eight ways of having three children. Probability trees are similar in structure, but they record information along the stems of the tree as well as at the nodes.

Eight toys are being given out to eight children in a hospital ward: a green boat a blue boat a pink sparkly fairy doll a blue sparkly fairy doll a yellow toy telephone a blue whale plush animal a blue light-up star a pink light-up star Ari likes pink items and boats, so let A= "getting a pink item" and B= "getting a boat." Are A and B independent events?

Solution: checking if P(A|B)=P(A) Let us select one of the equalities above to determine if A and B are independent events. Calculate P(A|B) and P(A) . P(A|B)=0 B is given, so Ari has gotten a boat. The boats don't come in pink, so the probability that Ari gets a pink item is 0 . P(A)=14 There are 2 pink items and 8 items total. P(A|B)≠P(A) , therefore the events are not independent.

Leni likes pink items and fairy dolls. Let A= "getting a pink item" and let B= "getting a fairy doll" Are A and B independent events

Solution: checking if P(B|A)=P(B| not A) Let us use P(B|A)=P(B| not A) to test for independence. P(B∣∣A)=12 Leni has gotten a pink item, now what is the probability of getting a fairy doll? There are 2 pink items and one is a fairy doll, so the probability is 0.5 that she will get a fairy doll, given that she has gotten a pink item. P(B∣∣not A)=16 Leni has NOT gotten a pink item. There are 6 items that are not pink. One of those 6 items is a fairy doll. Therefore, the probability of B given not A is 16 Therefore, P(B|not A)≠P(B|A) and the events A and B are not independent

. Consider the following situation

Suppose at a work function, the boss is handing out donuts. You're brave enough to ask for another one after waiting 10 minutes, but you're not brave enough to tell your boss that your favorite flavor is strawberry. (Also, no one else is eating donuts.) If there are five delectable strawberry frosted donuts and three chocolate frosted donuts, what is the probability that you get two strawberry ones?

A and B are independent events. P(A)=1 and P(B)=0.1. Calculate P(A and B). a) 0.01 b) 0.1 c) 1 d) 1.1

The correct answer is b. P(A and B)=P(A)⋅P(B)=1⋅0.1=0.1

For independent events, what does P(A|B) equal? a) P(A)⋅P(B) b) P(A) c) P(B) d) P(B|not A)

The correct answer is b. P(A|B)=P(A) .

A and B are independent events. P(A)=0.3 and P(B)=0.8. Calculate P(A|B). a) 0.24 b) 0.3 c) 0.8 d) 1.1

The correct answer is b. P(A|B)=P(A)=0.3 .

A and B are independent events. P(A)=0.5 and P(B)=1 . Calculate P(B|A) . a) 0.5 b) 1 c) 1.5 d) 15

The correct answer is b. P(B|A)=P(B)=1 .

For independent events, what does P(B|not A) equal? a) P(A)⋅P(B) b) P(A) c) P(B|A) d) P(A and B)

The correct answer is c. P(B|not A)=P(B|A) .

law of total probability*

The fact that if you multiply across each branch and then add up the resulting products, you get 1 is called the law of total probability*. Even though the probabilities of individual sequences of events are not the same—some are far more likely than others—the tree gives the complete set of probabilities for any sequence of events and therefore the sum of the branches is 1 .

P(A and B)=P(A)⋅P(B)

The intersection of A and B is the toss HH. That is one toss out of four possible, so the P(A and B)=14 P(A "tossing a heads on the first flip")=12 P(A "tossing a heads on the second flip")=12 P(A)⋅P(B)=12⋅12=14 Therefore: P(A and B)=P(A)⋅P(B)

Find the probability of developing lung cancer given a man is a non-smoker.

The intuitive way to do this problem is to notice that there are 810 non-smokers in the study and then to realize that only 2 of them developed lung cancer. P(developing lung cancer given someone is a non-smoker)= 2810=1405 Let us verify this by using our formula for conditional probability: Step 1: Let A= "being a non-smoker"; Let B= "developing lung cancer" Step 2: Determine P(B|A) Step 3: P(B∣∣∣A)=P(A and B)P(A) Step 4: Fill in the probabilities you know. To do this you must calculate P(A) and P(A and B) P(A)=8101000P(A and B)=21000 Remember that there are 2 men who are both non-smokers and lung-cancer victims out of 1,000 participants P(B|A)= 21000÷8101000 If you keep the probabilities as fractions you can cancel the numerator and denominator after converting the division to a multiplication problem: 2/1000×1000/810 Notice in the calculation above, the 1000 s will cancel, leaving a numerator of 2 and a denominator of 810 . This will always be the case. 2/1000×1000/810 In the future, use the easier method of using the totals for event A as the denominator of your probability fraction.

A frequency table has only 2 rows and 2 columns. True or False?

This is a false statement. A frequency table can have more than 2 rows and 2 columns.

A probability tree has the same information as a tree diagram. True or False?

This is a false statement. A probability tree contains the probability of an event, not just which events occur.

The probability of only one event can be determined from a frequency table. True or False?

This is a false statement. The probabilities of many events can be determined from a frequency table.

The probability of two independent events both occurring is P(A)+P(B) . True or False?

This is a false statement. The probability of 2 independent events both occurring is P(A)×P(B) .

The sum of the probabilities on a probability tree do not equal 1 . True or False?

This is a false statement. The sum of probabilities on all levels is 1 .

Two events can be independent and dependent. True or False?

This is a false statement. Two events can be either independent or dependent.

Independence can be determined from any of the four given statements below: P(A and B)=P(A)⋅P(B) P(A|B)=P(A) P(B|A)=P(B) P(B|A)=P(B|not A)

This is a true statement. Any of the four cases determine if the two events are independent.

The law of total probability is the sum of the product of all sequences of possible outcomes in a sample space is 1 . True or False?

This is a true statement. In a probability tree, multiply each branch of the tree and the sum of the products is 1 .

A frequency table can have a different number of rows and columns. True or False?

This is a true statement. The frequency table can have an unequal number of rows and columns.

The totals in a row or column can be used to find the probability of an event in a frequency table. True or False?

This is a true statement. The probability of an event can be determined by using the totals of a row or column.

A probability tree can be used to compare the probabilities of different events. True or False?

This is a true statement. The probability tree can compare different probabilities.

Frequency tables can estimate the probability of an event. True or False?

This is a true statement. The table can be used to estimate the probability of an event.

Determining Independence from Probabilities

Thus far we have been told (or inferred) whether two events are independent. Now we will look at the probabilities of events to determine whether they are independent or not. Remember that P(A and B) for independent events* is P(A)×P(B) , while P(A and B) for two dependent events* is P(A)×P(B|A) . It is worth determining whether two events are dependent or independent in order to determine whether you must find the more complicated probability of B|A .

Getting a beach ball and a brown item, using P(A|B)=P(A) .

dependent is Correct × The answer is dependent. Let A= getting a beach ball. Let B= getting a brown item. P(A|B)=0, as there are no brown beach balls. P(A)=212. P(A|B)≠P(A). Therefore, the events are dependent.

You select one card from a deck of cards, and do NOT place that card back in the deck. Then, you select a second card from the deck of cards. Determine if the following two events are independent or dependent: Selecting a queen and then selecting a king. a) Independent b) Dependent

Dependent Correct. The correct answer is b. The occurrence of the first affects the probability that the second event, therefore they are known as dependent events.

Dependent Events

Dependent Events Definition Two events for which the occurrence of the first affects the probability of the second. Examples It being cloudy outside, and it raining outside Being over age 65 and getting prostate cancer Probability of the intersection of two dependent events P(A)×P(B|A)

There are 12 items available for rewards: Green toy car Brown toy car Blue beach ball Green beach ball Brown toy lawnmower Green toy lawnmower Green stuffed animal Brown stuffed animal Green hockey stick Brown toy truck Orange toy truck Green toy truck For each set, determine if the events are dependent or independent according to the formula indicated. In each case, let A= the first event and B= the second event.

Getting a toy lawnmower and a green item, using P(B|A)=P(B) . independent is Correct × The answer is independent. Let A= getting a toy lawnmower. Let B= getting a green item. There are two lawnmowers, and one of them is green. Therefore, P(B∣∣A)=12. Meanwhile, the probability of choosing a green item is 612. Therefore, P(B|A)=P(B). According to this formula, the events are independent.

You roll a fair, six-sided die twice. Determine if the following two events are independent or dependent: Rolling a three and rolling a four. a) Independent b) Dependent

Independent Correct. The correct answer is a. Rolling a three on the first roll does not affect the probability that you will roll a four on the second roll. Therefore, they are independent events.

INDEPENDENT

Independent Events Definition Events, where the occurrence of one does not affect the probability that the other event(s) will occur. Examples Flipping heads on one coin flip and flipping heads on another coin flip Rolling a 3 on one die roll and rolling a 5 on another die roll Probability of the intersection of two independent events P(A)×P(B)

What do we call events where the occurrence of one event does not affect the probability that the other event will occur? a) Independent events b) Dependent events c) Mutually exclusive events d) All of the above

Independent events Correct. The correct answer is a. If the occurrence of one event does not affect the probability that the other event will occur, they are known as independent events.

to find probabilities

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

Determining Independence

Two Events are Independent If: P(A and B)=P(A)⋅P(B) P(A|B)=P(A) P(B|A)=P(B) P(B|A)=P(B| not A) Please note, if any of these conditions holds, the events are independent

Frequency Table Example

We can now use two-way tables to examine important kinds of probability. For example, a teenager considering quitting smoking might wonder what the probability of developing lung cancer is if he continues to smoke. Or a woman in her 30 s who is considering delaying having children might wonder what the probability if she can have two healthy children once she is above 40 years old. Many important probabilities change according to other conditions. In the table below, we consider a hypothetical study that tracked 1,000 men according to their smoking habits and rates of lung cancer. Smoking and Cancer Rates for Men Smoking Status: Never Smoked Current Smoker Current Heavy Smoker Totals Lung Cancer Status Cancer free Never 808 curren 127 heavey 29 total 964 Developed Cancer in Lifetime never 2 current 25 heavy 9 total 36 Totals never 810 current 152 heavy 38 total 1000

Using Probability Trees

You can use the tree you have built to calculate conditional probabilities. To do this, calculate the probabilities for each branch of the tree and then multiply the sequence of events you're interested in finding the probability for.

A and B are independent events. P(A)=0 and P(B)=0.2 . Calculate P(A|B) . a) 0 b) 0.2 c) 1 d) 1.2

a) 0 Correct. The correct answer is a. P(A|B)=P(A)=0 .

A and B are independent events. P(A)=0.3 and P(B)=0. Calculate P(B|A). a) 0 b) 0.3 c) 0.6 d) 1

a) 0 Correct. The correct answer is a. P(B|A)=P(B)=0 .

For independent events, what does P(A and B) equal? a) P(A)⋅P(B) b) P(A) c) P(B) d) P(B|not A)

a) P(A)⋅P(B) Correct. The correct answer is a. P(A and B)=P(A)⋅P(B) .

For independent events, what does P(B|A) equal? a) P(A)⋅P(B) b) P(A) c) P(B) d) P(A)+P(B)

c) P(B) Correct. The correct answer is c. P(B|A)=P(B) .

Getting a beach ball and a blue item, using P(A|B)=P(A) .

dependent is Correct × The answer is dependent. Let A= getting a beach ball. Let B= getting a blue item. P(A|B)=1, as there is only 1 blue item, and it is a beach ball. The P(A)=2./12=1/6 since there are 2 beach balls out of 12 total items. Therefore, P(A|B)≠P(A) and the events are dependent.