Thought Exercises

[Dr T]

Consider this simple syllogism. '(1) All artists are beekeepers. (2) Some beekeepers are chemists.' If (1) and (2) are both true, then which of the following must be true?

1 Some artists are chemists.

2 None of the other options is correct.

3 Some chemists are artists.

4 Some chemists are not artists.

[pushka]

No. 1?

Hmm...on 2nd thoughts, maybe no.4!! lol.

[Brother Dorsey]

Consider this simple syllogism. '(1) All artists are beekeepers. (2) Some beekeepers are chemists.' If (1) and (2) are both true, then which of the following must be true?

1 Some artists are chemists.

2 None of the other options is correct.

3 Some chemists are artists.

4 Some chemists are not artists.

2) None of the other options are correct......because, just because all artists are beekeepers does not mean all beekeepers are artists and who the heck knows if any of the beekeepers are chemists....(1 & 3) Saying that some artists are chemists and vice versa is an opinion and has no basis in fact...(4) Saying that some chemists are not artists is close but again not fact as it hasn't been established that any chemists are artists so when you say some are not you are implying that some are...which again is opinion.

[kaytee]

#1 ONLY ..the question SHOULD have been worded better!!!

[CaptainTux]

2 None of the other options is correct.:)

That is not a guess.

[EricM]

#2. And yes, I decided before reading other people's posts .

[Princess3dward]

No. 1?

[Dr T]

Well done Bro D, Tux, and Eric! The correct answer was None of the other options is correct.

Thanks for your other replies. Here is the explaination of why the above is correct:

It is possible that all of the beekeepers who are chemists are also artists. It is possible that none of the beekeepers who are chemists are also artists. There is no justifiable conclusion about the overlap or non-overlap of chemists and artists.

Let us use the following convention: an 'A' is an artist, a 'B' is a beekeeper, a 'C' is a chemist, an 'AB' is an artist who is also a beekeeper, a 'BC' is a beekeeper who is also a chemist, an 'ABC' is an artist who is also a beekeeper and a chemist, and so on.

Here is one set that satisfies both premises of the syllogism: AB, AB, ABC, ABC. In this set, every A is also a B (premise 1), and some Bs are also Cs (premise 2). In addition, some As are also Cs, and all Cs are also As.

But here is another set that also satisfies both premises of the syllogism: AB, AB, ABC, ABC, BC, BC. Here, too, every A is also a B (premise 1), and some Bs are also Cs (premise 2). But here, while some As are Cs, as above, this time some Cs are not As.

And here is yet another set that also satisfies both premises of the syllogism: AB, AB, B, B, BC, BC. Again, every A is also a B (premise 1), and some Bs are also Cs (premise 2). But in this set, no A is also a C (and no C is also an A).

Because each of these possible sets (among many others) satisfies both premises of the syllogism, it is not justified to reach any conclusion about the relationship between Cs and As: some, all or none of the Cs could be As, and some, all, or none of the As could be Cs.

[CaptainTux]

I can see how this is a question that would be answered incorrectly more than correctly. It is not the wording, but you have to go through the logical process thoughtfully as opposed to going on first instinct.

[pushka]

Dr. T, your house has too many premises...I'm afraid I've given up on reading the way in which to solve this puzzle...thanks! :)

[Dr T]

OK, try this:

Three cards are in a hat. One is red on both sides, one is white on both sides, and one is red on one side and white on the other. I draw a card from the hat, and drop it on the table. The upward-facing side is red. What are the odds that the downward-facing side is also red?

1 Fifty percent.

2 Two out of three.

3 One out of three.

4 None of the other options is correct.

Dr. T

[prisonchaplain]

#2. And yes, I decided before reading other people's posts .

#2, and no I didn't decide before reading other people's posts. I'm piggy-backing on Eric's and Captain Tux's thoughts and confidence.

[Dr T]

Good choice P.C.

[CaptainTux]

2 out of 3 probability. In vegas I would say 2/3

Hm...either I am smart or I worked at that darn casino too long in my youth.

[Dr T]

Tux,

It looks like you are smart! Two out of three is correct.

This problem uses three cards (R/R, R/W, W/W), each with two sides. Each time a card is drawn there is a 1-in-3 chance that any particular card will be drawn, but from a formal standpoint that is a red herring, because, relevantly, there is a 1-in-6 chance that any particular SIDE will come up. That is, there is a 1-in-3 chance that the R/R card will be drawn, and then, given that R/R has been drawn, there is a fifty-fifty chance that the "A" side will come up and a fifty-fifty chance that the "B" side will come up. (The two sides are functionally identical, but, again, that fact is a red herring from a formal standpoint.) Let us call side "A" of the R/R card R/R:A and side "B" of that card R/R:B.

It is helpful to look at this as though we were making a series of observations of the problem, as though it were an experiment. Suppose we "ran" the problem 600 times. (We would expect to see red come up 300 times and white come up 300 times, but that observation would be meaningless from the formal standpoint. Nevertheless, it is probably this fact that triggers many people's incorrect intuitions about the problem.) Relevantly, we would expect the following: R/R to be drawn 200 times (of which we would expect R/R:A 100 times and R/R:B 100 times); R/W to be drawn 200 times (of which we would expect the red side 100 times and the white side 100 times); and W/W to be drawn 200 times (of which we would expect W/W:A 100 times and W/W:B 100 times).

So the problem tells us that on this particular instance a red side came up, meaning that we can throw out all observations where a white side came up. We are left with: the 100 times we saw R/R:A, the 100 times we saw R/R:B and the 100 times we saw R/W:A. Of the approximately 300 draws culminating in a red side, approximately 200 of them would have been draws of the R/R card and 100 would have been draws of the R/W card.

Thus, if we continued to "run" the "experiment," and every time we saw a red side come up we predicted that we had drawn the R/R card, the prediction would be accurate approximately 2/3 of the time. That is to say, there is a 2/3 probability on any draw that there will be red on the other side of any red draw.

[Dr T]

These are word games: Take letter and add it to the answer to get the word that answers the second part of the problem. These are fun.

For example:

1. P + Length of life = One leaf of a book P = age = one leaf of a book (page)

2. S + Sound of an owl = To discharge a weapon

3. P + To pull or carry with force or effort = To fill a gap or stop up a hole

4. J + To cause pain or to feel unwell = A prison

5. C + A standard quantity of paper consisting of 500 sheets = The fatty part of milk which rises to the surface

6. C + A swift stroke with a whip = To conflict or disagree

7. P + Small insects often kept in "farms"(Plural) = Trousers

8. S + To plow = Motionless or stationary

9. E + A long easy stride or to canter as a horse = To run off secretly to be married

10. T + To decay = A gait between a walk and a run (of a horse)

good luck

[Maureen]

2. S + Sound of an owl = To discharge a weapon - Shoot

3. P + To pull or carry with force or effort = To fill a gap or stop up a hole - Plug

4. J + To cause pain or to feel unwell = A prison - Jail

5. C + A standard quantity of paper consisting of 500 sheets = The fatty part of milk which rises to the surface - Cream

6. C + A swift stroke with a whip = To conflict or disagree - Clash

7. P + Small insects often kept in "farms"(Plural) = Trousers - Pants

8. S + To plow = Motionless or stationary - Still

9. E + A long easy stride or to canter as a horse = To run off secretly to be married - Elope

10. T + To decay = A gait between a walk and a run (of a horse) - Trot

M.

[Dr T]

Very well done, M. :)

[CaptainTux]

Thank goodness I did not see that one before it was answered. I would have blown it.

[Dr T]

The same 4-letter word can be placed in front of each of the following words to make a new word: END, BOLT, LOCK, WOOD.

[Guest MrsS]

The same 4-letter word can be placed in front of each of the following words to make a new word: END, BOLT, LOCK, WOOD.

Deadend, Deadbolt, Deadlock, Deadwood

[Dr T]

Good job MrsS

[Dr T]

The enemy camp had been observed, the message was written and encoded, but the spy was caught. We had to let him go when we could not decipher the innocent message he was carrying: 'AL: The trucks arrive, Charles knows when every day. Do answer when necessary.' What day of the week will the enemies attack?

1. Friday

2. Wednesday

3. Monday

4. Any time between Monday and Thursday

[CaptainTux]

Wednesday

[Dr T]

Hey Tux,

That is correct, Thinker! Will you share how you came to that conclusion?

Thanks,

Dr. T