PHIL 105 Unit 3

Inductive strength comes in degrees.

True. Feedback: An inductive argument can be stronger or weaker, so it can have various degrees of strength. The reason is that inductive strength depends on probability, and probability comes in degrees.

Deductive validity is not defeasible.

True. Feedback: A standard is defeasible if adding new premises can turn an argument that meets the standard into an argument that does not meet the standard. Adding new premises cannot ever turn an argument that is valid into an argument that is invalid. If it was not possible for all of its premises to be true when its conclusion is false BEFORE adding the new premises (which is what it means to be valid), then it will still not be possible for all of its premises (including the original premises before the addition) to be true when its conclusion is false AFTER adding the new premises. Thus, the deductive standard of validity is not defeasible.

If A is positively correlated with B, then

any of the other answers might be true. Feedback: Correlation (positive or negative) does not by itself show causation or direction of causation.

An argument of the form "X percent of Fs are Gs, a is an F, so a is probably a G" is

defeasible. This argument is not valid, because it is possible for its premises to be true when its conclusion is false. It is also not a syllogism, because its premises and conclusion are not in the form of categorical propositions (A,E,I,O). But it is defeasible because adding further premises can turn a strong argument of this form into a weak argument of this form. For example, this argument would become very weak if we added the premise that a is an F* for some F* where no F*s are G.

Arguments that generalize from samples are:

inductive. Arguments that generalize from samples to a whole class were one of the kinds of inductive argument. What makes them inductive is that they are not intended to be valid, their strength varies in degrees, and they are defeasible.

Arguments that apply generalizations to particular cases are:

inductive. Feedback: Arguments that apply generalizations to particular cases were one of the kinds of inductive argument discussed in the lectures. What makes them inductive is that they are not intended to be valid, their strength varies in degrees, and they are defeasible.

Indicate which explanatory virtue is the main one that is lacking from this explanation: The cake did not come out right, because every one of the ingredients we used was wrong.

modesty. You do not need to claim that ALL of the ingredients were wrong in order to explain why the cake did not come out right. It would be enough to claim that we used at least some essential wrong ingredients. This explanation is falsifiable (because we can find out that we used SOME of the right ingredients), conservative (because it is compatible with the common observation that using the wrong ingredients can ruin cooking), powerful (because many other cooking failures can be explained by using the wrong ingredients), and deep (because it does not raise new explanatory questions, assuming that we used the wrong ingredients by mistake). Notice that this explanation does not raise the question of WHICH ingredients were wrong, because it says that they ALL were.

Being a bird is

necessary but not sufficient for being a swan. Feedback: The correct answer is "Necessary but not sufficient for being a swan." Every swan is a bird, so anything that is not a bird is not a swan. That makes being a bird necessary for being a swan. However, some birds (such as geese) are not swans, so being a bird is not sufficient for being a swan.

Consider this argument from analogy: I have visited many public gardens, and I almost always enjoyed walking in them. I just moved to a new town with a public garden. I have not visited it yet, but I know that it is similar in many ways to other public gardens that I have visited. So I will probably enjoy walking in the new public garden as well. Would this argument from analogy become stronger, weaker, or neither if we added a premise that the new public garden is a block north of my house, whereas all of the gardens that I visited before were south of my house?

neither stronger nor weaker. Feedback: Which direction the garden is from your house does not affect whether you enjoy walking in it (other things being equal).

What is the expected financial value of a bet where you will win $2 if you draw either a diamond or a club?

$0 Feedback: The expected financial value of a bet is the probability of winning times the net gain of winning minus the probability of losing times the net loss of losing. There are 13 diamonds and 13 clubs out of 52 cards in a standard deck, so the probability of winning this bet is 13/52 + 13/52 = 26/52 = 1/2. The gross gain from winning is $2, and playing costs $1 (which is not refunded if you win), so the net gain from winning this bet is $1. The probability of losing is 1 - (1/2) = 1/2. The net loss of losing is $1. Thus, the expected financial value of playing this game is: [(1/2) x $1] - [(1/2) x $1] = $0.

What is the expected financial value of a bet where you will win $4 if you draw a diamond?

$0 Feedback: The expected financial value of a bet is the probability of winning times the net gain of winning minus the probability of losing times the net loss of losing. There are 4 suits and 13 diamonds out of 52 cards in a standard deck, so the probability of winning this bet is 13/52 = 1/4. The gross gain from winning is $4, and playing costs $1 (which is not refunded if you win), so the net gain from winning this bet is $3. The probability of losing is 1 - (1/4) = 3/4. The net loss of losing is $1. Thus, the expected financial value of playing this game is: [(1/4) x $3] - [(3/4) x $1] = $0.

What is the expected financial value of a bet where you will win $2 if you do not draw an Ace?

$11/13 Feedback: The expected financial value of a bet is the probability of winning times the net gain of winning minus the probability of losing times the net loss of losing. There are 4 Aces out of 52 cards in a standard deck, so the probability of drawing an Ace and winning this bet is 4/52 = 1/13. By the rule for negation, the probability of not drawing an Ace is 1 - 1/13 = 12/13. That is the probability of winning this bet. The gross gain from winning is $2, and playing costs $1 (which is not refunded if you win), so the net gain from winning this game is $1. The probability of losing is 1 - (12/13) = 1/13. The net loss of losing is $1. Thus, the expected financial value of playing this game is: [(12/13) x $1] - [(1/13) x $1] = $11/13.

What is the expected financial value of a bet where you will win $2 if you do NOT draw a King, a Queen, or a Jack?

$7/13 Feedback: The expected financial value of a bet is the probability of winning times the net gain of winning minus the probability of losing times the net loss of losing. There are 4 Jacks, 4 Queens, and 4 Kings out of 52 cards in a standard deck, so the probability of drawing a King, Queen, or Jack and winning this bet is 4/52 + 4/52 + 4/52 = 12/52 = 3/13. By the rule for negation, the probability of NOT drawing a King, Queen, or Jack is 1 - 3/13 = 10/13. That is the probability of winning this bet. The gross gain from winning is $2, and playing costs $1 (which is not refunded if you win), so the net gain from winning this game is $1. The probability of losing is 1 - (10/13) = 3/13. The net loss of losing is $1. Thus, the expected financial value of playing this game is: [(10/13) x $1] - [(3/13) x $1] = $7/13.

In the following games, you lay down $1 to bet that you will pick a certain card in a fair draw from a standard deck. If you lose, then you lose your $1. If you win, then you collect the gross amount indicated, so your net gain is $1 less. What is the expected financial value of a bet where you will win $3 if you draw a club?

-$1/4 Feedback: The expected financial value of a bet is the probability of winning times the net gain of winning minus the probability of losing times the net loss of losing. There are 4 suits and 13 clubs out of 52 cards in a standard deck, so the probability of winning this bet is 13/52 = 1/4. The gross gain from winning is $3, and playing costs $1 (which is not refunded if you win), so the net gain from winning this bet is $2. The probability of losing is 1 - (1/4) = 3/4. The net loss of losing is $1. Thus, the expected financial value of playing this game is: [(1/4) x $2] - [(3/4) x $1] = - $1/4.

In the following games, you lay down $1 to bet that you will pick a certain card in a fair draw from a standard deck. If you lose, then you lose your $1. If you win, then you collect the gross amount indicated, so your net gain is $1 less. What is the expected financial value of a bet where you will win $2 if you draw a diamond?

-$2/4 Feedback: The expected financial value of a bet is the probability of winning times the net gain of winning minus the probability of losing times the net loss of losing. There are 4 suits and 13 diamonds out of 52 cards in a standard deck, so the probability of winning this bet is 13/52 = 1/4. The gross gain from winning is $2, and playing costs $1 (which is not refunded if you win), so the net gain from winning this bet is $1. The probability of losing is 1 - (1/4) = 3/4. The net loss of losing is $1. Thus, the expected financial value of playing this game is: [(1/4) x $1] - [(3/4) x $1] = - $2/4.

Assume that you and a friend both enter a beauty contest with three prizes (first, second, and third place). You have a 20 percent chance of winning one of the prizes. Your friend has a 30 percent chance of winning one of the prizes. However, if you win a prize, then your friend has less chance of also winning a prize. Your friend has only a 25 percent chance of winning one of the prizes, given that you win one of the prizes. What is the probability that both you and your friend will win a prize in this contest?

.05 Feedback: Because your winning a prize affects the probability that your friend will win a prize, according to the Question, these are not independent events. If two events are not independent, then the probability of both events occurring is the product of the probability of the first event times the conditional probability of the first event given the second event. The Question stipulates that the probability of you winning a prize is 20 percent = 0.2, and the probability of your friend winning a prize, given that you win a prize is 25 percent or 0.25. The product of 0.2 x 0.25 is 0.05. Thus, the probability that both you and your friend will win prizes is 0.05.

Assume that both you and your unrelated neighbor are having a child, each of you has a 50 percent chance that your child will be a girl, and whether one of you has a girl does not affect the probability that the other will have a girl. What is the probability that both you and your neighbor will have girls?

.25 Feedback: The question stipulates that you and your neighbor having girls are independent events in the sense that whether one occurs does not affect the probability of whether the other will occur. If two events are independent, then the probability of both events occurring is the product of the probability of the first event times the probability of the second event. The Question stipulates that the probability of you having a girl and of your neighbor having a girl are both 0.5. The product of 0.5 x 0.5 is 0.25. Thus, the probability that both you and your neighbor will have girls is 0.25.

Imagine that there is a 50 percent chance of rain today in Hong Kong, and a 10 percent chance of rain today in Sydney. Also assume that the probability of rain today in Hong Kong is independent of the probability of rain today in Sydney. What is the probability that it will rain today in either Hong Kong or Sydney (that is, in at least one of these cities)?

.55 Feedback: It might rain in both cities, so these events are not mutually exclusive. If two events are not mutually exclusive, then the probability of one or the other occurring is the sum of the probability of the first event plus the probability of the second event minus the probability of both events together. (We would not have to subtract the probability of both occurring together if these events were mutually exclusive, because then the probability of both occurring together would be 0. However, these events are NOT mutually exclusive, so here we DO need to subtract the probability of both occurring together.) The Question stipulates that the probability of rain today in Hong Kong is 50 percent = 0.5 and the probability of rain today in Sydney is 10 percent = 0.1. The Question also stipulates that these events are independent, so the probability of both occurring together is 0.5 x 0.1 = 0.05. Thus, the probability of rain today in either one or the other of these cities is 0.5 + 0.1 - 0.05 = 0.55.

Assume that you and your neighbor will drive separately to work today. You have a 1 percent chance of getting into an accident on the way to work today, and your neighbor has a 2 percent chance of getting into an accident on the way to work today. Whether one of you has an accident does not affect the probability that the other will have an accident (because you drive opposite directions, the weather is not conducive to accidents, and so on). What is the probability that both you and your neighbor will have an accident on the way to work today?

0.0002 Feedback: The question stipulates that you and your neighbor having accidents on the way to work today are independent events in the sense that whether one occurs does not affect the probability of whether the other will occur. If two events are independent, then the probability of both events occurring is the product of the probability of the first event times the probability of the second event. The question stipulates that the probability of you having an accident on the way to work today is 1 percent = 0.01, and the probability of your neighbor having an accident on the way to work today is 2 percent or 0.02. The product of 0.01 x 0.02 is 0.0002. Thus, the probability that both you and your neighbor will have an accident on the way to work today is 0.0002.

Assume that you and a friend both enter a race with three prizes (first, second, and third place). You have a 30 percent chance of winning one of the prizes. Your friend has a 20 percent chance of winning one of the prizes. However, if you win a prize, then your friend has less chance of also winning a prize. Your friend has only a 10 percent chance of winning one of the prizes, given that you win one of the prizes. What is the probability that both you and your friend will win a prize in this contest?

0.03 Feedback: Because your winning a prize affects the probability that your friend will win a prize, according to the question, these are not independent events. If two events are not independent, then the probability of both events occurring is the product of the probability of the first event times the conditional probability of the first event given the second event. The question stipulates that the probability of you winning a prize is 30 percent = 0.3, and the probability of your friend winning a prize, given that you win a prize is 10 percent or 0.1. The product of 0.3 x 0.1 is 0.03. Thus, the probability that both you and your friend will win prizes is 0.03.

If the probability of a method of contraception working to prevent conception is 0.9, then what is the probability that it will not work?

0.1 Feedback: The probability of an event will not happen is one minus the probability that the event will happen. Here the event is the method of contraception working. The probability of that event is 0.9, according to the question. One minus 0.9 is 0.1. Hence, the probability of the contraception not working is 0.1.

Imagine that there is a 10 percent chance of rain today in Mumbai, India, and a 20 percent chance of rain today in Toronto, Canada. Also assume that the probability of rain today in Mumbai is independent of the probability of rain today in Toronto. What is the probability that it will rain today in either Mumbai or Toronto (that is, in at least one of these cities)?

0.28 Feedback: It might rain in both cities, so these events are not mutually exclusive. If two events are not mutually exclusive, then the probability of one or the other occurring is the sum of the probability of the first event plus the probability of the second event minus the probability of both events together. The question stipulates that the probability of rain today in Mumbai is 10 percent = 0.1 and the probability of rain today in Toronto is 20 percent = 0.2. The question also stipulates that these events are independent, so the probability of both occurring is 0.1 x 0.2 = 0.02. Thus, the probability of rain today in either one or the other of these cities is 0.1 + 0.2 - 0.02 = 0.28.

Imagine that there is a 10 percent chance that you eat dinner tonight before 6 pm and a 30 percent chance that you eat dinner tonight after 9 pm. Assume that you eat dinner tonight only once. What is the probability that you will eat dinner tonight either before 6 pm or after 9 pm?

0.3 Feedback: On the assumption that you eat dinner only once, eating dinner tonight before 6:00 p.m. and eating dinner tonight after 9:00 p.m. are mutually exclusive. If two events are mutually exclusive, then the probability of one or the other occurring is the sum of the probability of the first event plus the probability of the second event. The Question stipulates that the probability of eating dinner tonight before 6 pm is 10 percent = 0.1 and the probability of eating dinner tonight after 9 pm is 30 percent = 0.3. Thus, the probability of eating dinner tonight either before 6 pm or after 9 pm is 0.1 + 0.3 = 0.4.

If there is a 30 percent chance of rain this afternoon, then what is the probability that it will not rain this afternoon?

0.7 Feedback: The probability of an event will not happen is one minus the probability that the event will happen. Here the event is rain this afternoon. The probability of that event is 30 percent or 0.3, according to the question. One minus 0.3 is 0.7. Hence, the probability that it will not rain this afternoon is 0.7.

Imagine that you are offered two jobs. There is a 50 percent chance that you will take the first job and a 40 percent chance that you will take the second job. Assume that you cannot take both jobs. What is the probability that you will take either one job or the other job?

0.9 Feedback: On the assumption that you cannot take both jobs, taking one job and taking the other job are mutually exclusive. If two events are mutually exclusive, then the probability of one or the other occurring is the sum of the probability of the first event plus the probability of the second event. The Question stipulates that the probability of taking the first job is 50 percent = 0.5 and the probability of taking the second job is 40 percent = 0.4. Thus, the probability that you will take either one job or the other is 0.5 + 0.4 = 0.9.

What is the probability of rolling eleven at least once in a series of three fair rolls of two six-sided dice in each roll?

1 - (34/36 × 34/36 × 34/36) Feedback: First, we calculate the probability of rolling 11 on a single fair roll of two six-sided dice. There are two ways to get 11: (a) 6 on the first die and 5 on the second die and (b) 5 on the first die and 6 on the second die. There are 36 possible outcomes. Thus, the probability of rolling 11 is 2/36. Second, we use the rule for negations to calculate the probability of not rolling 11 on a single fair roll of two six-sided dice. That probability is 1 - 2/36 = 34/36. Third, we use the rule for conjunctions to calculate the probability of not rolling 11 on the first roll and then again on the second roll and then again on the third roll—so not on any of three consecutive rolls—of two fair six-sided dice. The results of fair rolls are independent, so we can use the simple version of the rule for conjunctions of independent events. Using this rule, the probability of not getting 11 in any of three fair rolls of two six-sided dice is 34/36 x 34/36 x 34/36 = (34/36)3. That result is the negation of getting 11 at least once in three fair rolls of two six-sided dice. Thus, fourth, we apply the rule for negations to calculate the probability of getting 11 at least once in three fair rolls of two six-sided dice: 1 - (34/36 x 34/36 x 34/36) = 1 - (34/36) 3. This calculation in effect applies the rule for calculating probabilities in a series that was discussed in the lectures.

What is the probability of rolling twelve at least once in a series of three fair rolls of two six-sided dice in each roll?

1- (35/36 x 35/36 x 35/36) Feedback: First, we calculate the probability of rolling 12 on a single fair roll of two six-sided dice. There is only one way to get 12 (6 on one die and 6 on the other die), and there are 36 possible outcomes, so the probability of rolling 12 is 1/36. Second, we use the rule for negations to calculate the probability of not rolling 12 on a single fair roll of two six-sided dice. That probability is 1 - 1/36 = 35/36. Third, we use the rule for conjunctions to calculate the probability of not rolling 12 on the first roll and then again on the second roll and then again on the third roll—so not on any of three consecutive rolls—of two fair six-sided dice. The results of fair rolls are independent, so we can use the simple version of the rule for conjunctions of independent events. Using this rule, the probability of not getting 12 in any of three fair rolls of two six-sided dice is 35/36 x 35/36 x 35/36 = (35/36)3. That result is the negation of getting 12 at least once in three fair rolls of two six-sided dice. Thus, fourth, we apply the rule for negations to calculate the probability of getting 12 at least once in three fair rolls of two six-sided dice: 1 - (35/36 x 35/36 x 35/36) = 1 - (35/36)3. This calculation in effect applies the rule for calculating probabilities in a series. It might seem simpler just to add 1/36 + 1/36 + 1/36, but that calculation yields a different result. Moreover, that method of calculation cannot be right, since then the probability of rolling twelve at least once in a series of FIFTY fair rolls of two six-sided dice in each roll would be 50/36, but probabilities cannot ever be greater than 1.

Assume that A and B are highly correlated. If B changes when you manipulate A, but A does not change when you manipulate B, then which of the following is most likely? Assume normal circumstances and no interfering factors.

A causes B. Feedback: Given the assumptions, if B causes A, then A would change when you manipulate B. Thus, that alternative is ruled out by the observation that A does not change when you manipulate B. Next, if a third thing causes both A and B, or if there were no causal relation between them, then B would not change when you manipulate A. Thus, these alternatives are ruled out by the observation that B changes when you manipulate A. The remaining alternative is that A causes B, so that alternative is most likely.

If a fair dealer deals you five cards out of a shuffled fair standard deck of cards, then which of the following hands is most likely?

All of these hands are equally likely. Feedback: If the deck and the dealer really are fair, as the question stipulates, then the probability of getting dealt any particular hand is the same as the probability of getting dealt any other particular hand. That is what it means to call them fair.

Assume that A and B are highly correlated. If A changes when you manipulate B, but B does not change when you manipulate A, then which of the following is most likely? Assume normal circumstances and no interfering factors.

B causes A. Feedback: Given the assumptions, if A causes B, then B would change when you manipulate A. Thus, that alternative is ruled out by the observation that B does not change when you manipulate A. Next, if a third thing causes both A and B, or if there were no causal relation between them, then A would not change when you manipulate B. Thus, these alternatives are ruled out by the observation that A does change when you manipulate B. The remaining alternative is that B causes A, so that alternative is most likely.

Indicate which explanatory virtue is the main one that is lacking from this explanation: My roof is leaking, because it wants to ruin my floors in order to get back at me for not fixing it.

Conservativeness. Feedback: The claim that roofs have wants or desires conflicts with well-established knowledge. Only living things have wants or desires, but roofs are not alive. This explanation might also seem to lack depth if it raises the further question of why a roof would want to ruin floors, but this question is answered: The roof wants to get back at me for not fixing it.

Deductive validity comes in degrees.

False. An argument is either valid or not valid. It cannot be partly valid or have a degree of validity. The reason is that deductive validity depends on possibility, and it is either possible or not possible (but not partly possible) that the premises are true when the conclusion is false.

Which experiment shows that the conjunction of the new computer and the new software is not sufficient for the system to run slowly?

Experiment 2. Experiment 2 used both the new computer and also the new software (X is present), and the system ran fast rather than slowly (Y is absent), so Experiment 2 refutes the claim that, whenever a system includes both the new computer and the new software (X is present), that system runs slowly (Y is present). Thus, Experiment 2 shows that the conjunction of the new computer and the new software fails the negative sufficient condition test, so this conjunction is not sufficient for running slowly. No other experiment had the conjunction of the new computer and the new software without running slowly, because the only other experiment with the conjunction of the new computer and the new software was Experiment 1, and the system ran slowly in Experiment 1. Thus, no other experiment shows that the conjunction of the new computer and the new software is not sufficient for running slowly.

Which experiment shows that the new computer is NOT NECESSARY for the system to run slowly?

Experiment 5. Experiment 5 used the old computer instead of the new computer (X is absent), and the system ran slowly (Y is present), so Experiment 5 refutes the claim that, whenever a system does not include the new computer (X is absent), that system does not run slowly (Y is absent). Thus, Experiment 5 shows that the new computer fails the negative necessary condition test, so the new computer is not necessary for running slowly. No other experiment ran slowly without the new computer, because the only other experiments without the new computer were Experiments 6, 7, and 8, and the system did not run slowly in any of those experiments. Thus, no other experiment shows that the new computer is not necessary for running slowly. Many students get confused about necessary conditions, because there are so many negations. You might have thought that a system that did include the new computer (such as Experiments 1 - 4) show that the new computer is not necessary for running slowly. However, when the question is about necessary conditions instead sufficient conditions, then the crucial test cases—the ones that might show that a candidate is not a necessary condition—are the experiments without the candidate condition (here = those that did not use the new computer). Those are the only systems that can be cases of running slowly WITHOUT the new computer.

Imagine that you buy a new computer system with independent components including a new desktop computer (with a CPU and a graphics card), new software, and a new monitor. You want to play games on the new system, but it runs games very slowly. You assume that the keyboard and mouse are not creating the problem; so, to figure out what is making the system run so slowly, you experiment with combinations of your old equipment with the new equipment. Here are your experiments and results: Experiment 1: New computer, new software, and new monitor — and it runs slowly. Experiment 2: New computer, new software, and old monitor — and it runs fast. Experiment 3: New computer, old software, and new monitor — and it runs slowly. Experiment 4: New computer, old software, and old monitor — and it runs fast. Experiment 5: Old computer, new software, and new monitor — and it runs fast. Experiment 6: Old computer, new software, and old monitor — and it runs fast. Experiment 7: Old computer, old software, and new monitor — and it runs slowly. Experiment 8: Old computer, old software, and old monitor — and it runs fast. Based on this data, which experiment shows that the new monitor is not sufficient for the system to run slowly?

Experiment 5. Experiment 5 used the new monitor (X is present), and the system ran fast rather than slowly (Y is absent), so Experiment 5 refutes the claim that, whenever a system includes the new monitor (X is present), that system runs slowly (Y is present). Thus, Experiment 5 shows that the new monitor fails the negative sufficient condition test, so the new monitor is not sufficient for running slowly. No other experiment had new monitor without running slowly, because the only other experiments with the new monitor were 1, 3, and 7, and the system ran slowly in all of those experiments. Thus, no other experiment shows that the new monitor is not sufficient for running slowly.

Imagine that you buy a new computer system with independent components including a new desktop computer (with a CPU and a graphics card), new software, and a new monitor. You want to play games on the new system, but it runs games very slowly. You assume that the keyboard and mouse are not creating the problem; so, to figure out what is making the system run so slowly, you experiment with combinations of your old equipment with the new equipment. Here are your experiments and results: Experiment 1: New computer, new software, and new monitor—and it runs slowly. Experiment 2: New computer, new software, and old monitor—and it runs fast. Experiment 3: New computer, old software, and new monitor—and it runs slowly. Experiment 4: New computer, old software, and old monitor—and it runs fast. Experiment 5: Old computer, new software, and new monitor—and it runs slowly. Experiment 6: Old computer, new software, and old monitor—and it runs fast. Experiment 7: Old computer, old software, and new monitor—and it runs fast. Experiment 8: Old computer, old software, and old monitor—and it runs fast. Based on this data, which experiment shows that the new monitor is NOT SUFFICIENT for the system to run slowly?

Experiment 7. Experiment 7 used the new monitor (X is present), and the system ran fast rather than slowly (Y is absent), so Experiment 7 refutes the claim that, whenever a system includes the new monitor (X is present), that system runs slowly (Y is present). Thus, Experiment 5 shows that the new monitor fails the negative sufficient condition test, so the new monitor is not sufficient for running slowly. No other experiment had new monitor without running slowly, because the only other experiments with the new monitor were 1, 3, and 5, and the system ran slowly in all of those experiments. Thus, no other experiment shows that the new monitor is not sufficient for running slowly.

Which experiment shows that the new computer is not necessary for the system to run slowly?

Experiment 7. Experiment 7 used the old computer rather than the new computer (X is absent), and the system ran slowly (Y is present), so Experiment 7 refutes the claim that, whenever a system does not include the new computer (X is absent), that system does not run slowly (Y is absent). Thus, Experiment 7 shows that the new computer fails the negative necessary condition test, so the new computer is not necessary for running slowly. No other experiment ran slowly without the new computer, because the only other experiments without the new computer were Experiments 5, 6, and 8, and the system did not run slowly in any of those experiments. Thus, no other experiment shows that the new computer is not necessary for running slowly. Many students get confused about necessary conditions, because there are so many negations. You might have thought that a system that did include the new computer (such as Experiments 1 - 4) show that the new computer is not necessary for running slowly. However, when the question is about necessary conditions instead sufficient conditions, then the crucial test cases—the ones that might show that a candidate is not a necessary condition—are the experiments without the candidate condition (in this case, those that did not use the new computer). Those are the only systems that can be cases of running slowly WITHOUT the new computer.

Which experiment shows that the conjunction of the old computer and the new monitor is not sufficient for the system to run slowly?

Experiment 7. Feedback: Experiment 7 used the old computer with the new monitor (X is present), and the system ran fast rather than slowly (Y is absent), so Experiment 7 refutes the claim that, whenever a system includes both the old computer and the new monitor (X is present), that system runs slowly (Y is present). Thus, Experiment 7 shows that the conjunction of the old computer with the new monitor fails the negative sufficient condition test, so this conjunction is not sufficient for running slowly. No other experiment had the conjunction of the old computer with the new monitor without running slowly, because the only other experiment with the conjunction of the old computer with the new monitor was Experiment 5, and the system ran slowly in Experiment 5. Thus, no other experiment shows that the conjunction of the old computer with the new monitor is not sufficient for running slowly.

In an argument of the form "X percent of Fs are Gs, a is an F, so a is probably a G," the reference class is

F. Feedback: X is the percentage, G is the attribute or target class, and a is the case, so F is the reference class.

Whenever the expected financial value of a bet is greater than $0, you should always take the bet, regardless of its expected overall value.

False. Feedback: There are other values besides money, so sometimes you should not risk those other values in order to get some chance of making more money even when the expected financial value of the bet is greater than $0.

If there is no case of a candidate (X) without the target (Y) in a data set, then that data provides reason to believe the conclusion that the candidate (X) is sufficient for the target (Y).

False. Feedback: This question asks about a positive conclusion that a candidate (X) IS sufficient for a target (Y). In order to justify such a positive conclusion, the data must meet the conditions in the positive sufficient condition test, not just the negative sufficient condition test. The positive condition test says that we have good reason to believe that X is a sufficient condition of Y if (a) we have not found any case where X is present and Y is absent, (b) we have tested a wide variety of cases, including cases where X is present and cases where Y is absent, (c) if there is any other feature that is never present where Y is absent, then we have tested cases where that other feature is absent but X is present as well as cases where that other feature is present but X is absent, and (d) we have tested enough cases of various kinds that are likely to include a case where X is present and Y is absent if there is any such case. The question asked here does not say that these other conditions are met by the data set in question, so that data does not provide a reason for the positive conclusion that the candidate (X) is sufficient for the target (Y).

Since it is good for a college student to own one computer, it is three times as good for a college student to own three computers.

False. This is an example f diminishing marginal utility.

Inductive strength is not defeasible.

False. A standard is defeasible if adding further premises can turn an argument that meets the standard into an argument that does not meet the standard. Adding further premises can turn an argument that is very strong into an argument that is very weak. For example, the argument "The sun has come up every day for thousands of years, so it will probably come up tomorrow" is very strong, but the same argument becomes very weak if we add the new premise that a meteor will destroy the earth this afternoon, because then it is unlikely that the sun will come up tomorrow. Thus, inductive strength is defeasible.

Which experiment shows that the conjunction of the old software and the new monitor is not sufficient for the system to run slowly?

None of these experiments. Feedback: In this data set from Question 12, Experiments 3 and 7 are the only experiments with the conjunction of both the old software and also the new monitor (X present). In both of those experiments, the system ran slowly (Y is present). Thus, there is no case in these data with old software and new monitor (X) present but slow running (Y) absent, so none of these experiments shows that the conjunction of the old software and the new monitor is not sufficient for the system to run slowly.

Which experiment shows that the conjunction of the new computer and the new monitor is not sufficient for the system to run slowly?

None of these experiments. Feedback: In this data set, Tests 1 and 3 are the only experiments with the conjunction of both the new computer and also the new monitor (X present). In both of those experiments, the system ran slowly (Y is present). Thus, there is no case in these data with new computer and new monitor (X) present but slow running (Y) absent, so none of these experiments shows that the conjunction of the new computer and the new monitor is not sufficient for the system to run slowly.

Which experiment shows that the conjunction of the new software and the new monitor is not sufficient for the system to run slowly?

None of these experiments. Feedback: In this data set, Experiments 1 and 5 are the only Experiments with the conjunction of the new software with the new monitor (X present). In both of those experiments, the system ran slowly (Y is present). Thus, there is no case in these data with new software and new monitor (X) present but slow running (Y) absent, so none of these experiments shows that the conjunction of the new software with the new monitor is not sufficient for the system to run slowly.

Which experiment shows that the conjunction of the new computer and the new monitor is not sufficient for the system to run slowly?

None of these experiments. In this data set, Experiments 1 and 3 are the only experiments with the conjunction of both the new computer and also the new monitor (X present). In both of those experiments, the system ran slowly (Y is present). Thus, there is no case in these data with new computer and new monitor (X) present but slow running (Y) absent, so none of these experiments shows that the conjunction of the new computer and the new monitor is not sufficient for the system to run slowly.

Specify the main problem with the following generalization from a sample. There might be more than one problem, but indicate the most important one. I sampled thousands of people in Tokyo, and almost all of them speak Japanese. Therefore, most people in the world speak Japanese.

The sample is biased. Feedback: Tokyo is in Japan, and the percentage of people who speak Japanese is higher inside Japan than outside Japan. Thus, the fact that the sample was gathered in Tokyo makes it biased.

Specify the main problem with the following generalization from a sample. There might be more than one problem, but indicate the most important one. This Apple computer worked reliably for five years, so computers made by Apple are very reliable.

The sample is too small. A sample of one is sometimes big enough for a generalization, but only when we can assume uniformity in relevant respects throughout the sample. That assumption is questionable here. Apple makes many different kinds of computers, so some might be reliable while others are not. Moreover, even if 50 percent of Apple computers break down, the computer mentioned in the premise might be one of the other 50 percent that does not break down. Because there might be such variation in reliability, this sample is too small. Of course, it still might be true that Apple computers are reliable, but this one case does not provide strong evidence for that conclusion.

If two six-sided dice come up seven on seven fair rolls in a row, then what is the probability that these dice will come up seven on the eighth roll?

one sixth (1/6=6/36) Feedback: If the dice rolls really are fair, as the question stipulates, then the probability that they will come up heads on any flip is one sixth, because there are six ways of getting seven and thirty-six possible outcomes when rolling two fair six-sided dice. It does not matter how many times they have come up seven before. Each roll is independent in the sense that the result on one flip does not affect the probability of any result on any other flip.

Consider this argument: 80 percent of Italians like gnocchi, and Lucia is a Italian, so Lucia probably likes gnocchi. Would this argument be stronger or weaker if we added the information that 90 percent of Italians from Lucia's region like gnocchi?

stronger. Feedback: The alternative reference class "Italians from Lucia's region" does not conflict with the original argument. Instead, the argument becomes stronger with this new reference class, because 90 percent is more than 80 percent.

Being a swan is

sufficient but not necessary for being a bird. Every swan is a bird, so being a swan is sufficient for being a bird. However, some things that are not swans (such as geese) are still birds, so being a swan is not necessary for being a bird.

If two explanations are equal in all other respects, then

the more conservative explanation is better. Conservativeness is a standard explanatory virtue, so it makes explanations better, other things being equal.

If two explanations are equal in all other respects, then

the more modest explanation is better. Feedback: Modesty is a standard explanatory virtue, so it makes explanations better, other things being equal.

An argument that generalizes from a sample to a whole class is weaker (other things being equal) when

the sample is biased. Feedback: When the sample is biased, the argument commits the fallacy of biased sampling, which makes the argument weaker. In contrast, true premises and a larger sample make the argument stronger (instead of weaker) than when the premises are not true or the sample is too small.

An argument that generalizes from a sample to a whole class is stronger (other things being equal) when

the sample is larger. When the sample is too small, the argument commits the fallacy of hasty generalization. A larger sample helps the argument avoid this fallacy, so it makes the argument stronger. In contrast, false premises and a biased sample make the argument weaker (instead of stronger) than when the premises are true or the sample is unbiased.

If A is negatively correlated with B, then

there is no causal relation between A and B. Feedback: Correlation (positive or negative) does not by itself show causation or direction of causation.

Consider this argument from analogy: I have visited many public gardens, and I almost always enjoyed walking in them. I just moved to a new town with a public garden. I have not visited it yet, but I know that it is similar in many ways to other public gardens that I have visited. So I will probably enjoy walking in the new public garden as well. Would this argument from analogy become stronger, weaker, or neither if we added a premise that the new public garden is a rock garden, but none of the gardens that I visited before were rock gardens?

weaker. Whether the garden is a rock garden or a plant garden affects whether some people enjoy walking there, so this difference between previous gardens and the new garden is relevant to whether I will enjoy walking there. I might enjoy the rock garden even more than a plant garden, because of the novelty, but I might not. This uncertainty means that the analogies between the gardens give me less reason to believe that I will enjoy walking in the rock garden than if the new garden were also a plant garden like the gardens that I enjoyed before.

Consider this argument: 90 percent of Germans like sausage, and Fritz is a German, so Fritz probably likes sausage. Would this argument be stronger or weaker if we added the information that only 20 percent of Germans who are Muslims like sausage, and Fritz is a Muslim?

weaker. The class "German Muslims" is a conflicting reference class that undermines the strength of the argument.