Thanks, Textbooks gives examples of awkward, weird and just plain bad writing in textbooks. (Send examples to Karl at thankstextbooks@gmail.com.)

This problem had me stumped:

Thinking and Linking by Joanne Jacobs

August 7, 2012 by Joanne

Thanks, Textbooks gives examples of awkward, weird and just plain bad writing in textbooks. (Send examples to Karl at thankstextbooks@gmail.com.)

This problem had me stumped:

Filed Under: Education Tagged With: textbooks, word problems, writing

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That’s really funny. It’s a

random“choice”.It’s a strange question… but not completely awful. Take out the phrase “patrons a choice of” and it’s perfectly clear what they’re asking.

At least I think it is. I assume that the answer is whatever 3 divided by 22 is. Since that’s just a little smaller than 1/7, it’s probably somewhere just under 14%.

22?

Ketchup yes or no. Given that choice, onions yes or no. Given those choices, mustard yes or no. Given the first three choices, pickles yes or no. 2*2*2*2 is 16, not 22.

Seems like number of combinations calculations. Abbreviating catsup, mustard, pickles, onion as CMPO then:

1 condiment only on a burger: C, M, P, O = 4

2 condiments: CM, CP, CO, MP, MO, PO = 6

3 condiments: CMP, CMO, CPO, MPO, = 4

4 condiments: CMPO = 1

So I count 15 combinations. Probability of any 3 of those are 1/15.

Good warmup for the new year!

Don’t forget 0 condiments, which would make 16 combinations.

That will teach me to do mental math when I’m not really paying attention.

I did permutations instead of combinations for the sets of 2. (But I remembered zero!)

It depends on how the choice process is structured.

Among other things, does the “random fashion” have to have a stopping point? If the chef picks randomly from the 4 condiment jars but does it 10 times, the chances of getting “one with everything” are pretty high. If the chef can pick one, two, three, or four times, nothing is stopping her from picking a fifth time.

And, of course, if she is picking randomly, Mike’s 15 events are not equally likely. She might well pick catsup or pickles 4 times out of 4. If she won’t pick catsup again because she has already picked it once, she is not picking randomly.

Again, if she is picking randomly, it matters how easy it is to pick each condiment. If there are 20 equally accessible condiment dispensers and 12 hold catsup, 5 hold pickles, 2 hold pickles, and 1 holds onions, the chances of getting catsup on any one pick are going to be 12 times greater than the chances of getting onions.

Several people have rewritten the question as follows:

A burger joint has 4 condiments to add to its burgers: catsup, mustard, pickles, and onions. It offers burgers with none, one, two, three, or four of the condiments. If the joint prepares equal numbers of each kind and then hands the burgers out randomly, what is the the probability that a customer will get one of the following: catsup and onions, mustard and pickles, or one with everything?

That may well be what the question-asker is looking for. But it is not what was asked. Being able to translate the question shows that the student is becoming good at getting into the question-asker’s head. It does not show that the student is understanding random processes.

Actually, those two situations are indeed the same.

Well, since I HATE pickles on my burgers, and there isn’t any bacon, lettuce, cheese, mushrooms being offered, I’d say this place mentioned in the textbook just plain sucks 🙂

I LOVE good burgers…crappy ones aren’t worth eating (except perhaps by the family dog who eats just about everything) 🙂

A former textbook writer/editor laments the state of math textbooks:

http://open.salon.com/blog/annie_keeghan/2012/02/17/afraid_of_your_childs_math_textbook_you_should_be

(HT: Stuart Buck)

I think she misses some deeper issues. She can tell if a book is grammatically correct, if it makes logical errors, or gives wrong answers to problems. She knows what, by her standards, is a “quality” product. She makes a good case that lots of books that are sold today are not quality products. However, neither she nor anyone else knows which math books are most helpful to students, or even if one book is good for one group of students and a different book good for a different group. And that, after all, is what really matters.

It’s sad but true, but the easiest guide you can use is that, the more modern the K-12 textbook, the worse it’s going to be. Modern textbooks are so concerned with being PC that they often forget to even try to teach the core content anymore…

I agree the question is worded a little poorly, but I have seen a lot worse.

It appears to be a straight forward problem on independent trials:

(1) Catsup (50%) or no catsup (50%)

(2) Mustard (50%) or no mustard (50%)

(3) Pickles (50%) or no pickles (50%)

(4) Onions (50%) or no onions (50%)

The problem is equivalent to flipping a fair coin 4 times, and asking what is the probability that you obtain HTTH (where order matters).

That’s probably what the question-asker wants you to do. But that requires you to read a number of things into the question. It requires you to say, in effect, “I just learned to find probabilities for the flipping of a fair coin. I will assume this is just like that and use the same algorithm I used for coins.”

This is going to sound harsh but it requires you to NOT think deeply about the problem. It requires you to pretend that you can mechanically apply something because you have just learned it. It says, “I have just given you a hammer. Now pretend that everything you see here is a nail.” That bothers me.

Seems pretty clear to me, and I interpret it the same way jab does. The writing could be clearer, but the hamburger maker essentially flips a fair coin to decide on ketchup, flips again for mustard, flips again for pickles, flips again for onions. So each condiment choice is independent of each other condiment choice.

That means every four-condiment selection is equally likely. There are 16 (2*2*2*2) possible condiment selections. So a particular condiment selection has probability 1/16. You have a 1/16 chance of getting ketchup and onions only, 1/16 of getting mustard and pickles only, and 1/16 chance of getting everything. Adding those up, you’ve got a 3/16 chance of getting a burger with ketchup and onion only OR mustard and pickles only OR everything.

Hmm. Coin flips isn’t where my head went at all.

I just counted: how many zero combs, 1 combs, 2 combs, etc.

Then I added.

(Except I screwed up in counting the 2-combs).

I tend to approach math problems like this combinatorially rather than statistically, but either way is good.

But if the question were more complicated, then enumerating the possibilities would be even more error prone. Suppose you’re getting a random sandwich at Sandwich Express (“Have It Our Way”). The possibilities ar white, whole wheat or rye bread; egg salad, ham, corned beef or roast beef; swiss cheese , American cheese or no cheese; mustard or no mustard; lettuce or no lettuce. You can’t order what you want; the servers make all the choices randomly with equal probability.

You ask for a sandwich. You want egg salad on white with mustard, lettuce and no cheese. What’s the chance you get it? I can compute the answer to that question in thirty seconds, the amount of time it takes me to figure how many choices there are for each element and multiply them together. If you solve the problem by enumeration, it’ll take you a while.

I’m just going to say two things.

First, I am not talking about enumeration. I’m talking about mathematical

counting. You seem like a smart fellow. I’m sure you took combinatorics.Second, the problem you gave is structurally a different sort of problem, and therefore doesn’t require the same sort of work, and shouldn’t take any more than about 9 seconds, without writing anything down. 3 x 4 x 3 x 2 x 2? Please, we could both do that in our sleep.

Smart, maybe, fellow, no.

Of course I’ve studied combinatorics. I see. You figured out, what’s four choose one (4), what’s four choose two (6), what’s four choose three (4), what’s four choose four (1), then you added them up. I find that a little indirect, but obviously it works, if you remember to also consider four choose zero.

Of course, the only REAL correct answer is that you’d refuse to eat the burger, because livestock hurts the Earth. You’d have a tofu burger instead. To answer otherwise these days is risking being denied your student loans, and ending up on some ‘watch lists’!

Yes, the question is poorly written, but being able to figure out the thought behind an unclear question, in particular a technical/quantitative one, is actually an important job skill. Seriously–I remember when I first graduated from college I struggled for a few years because real world problems and job specifications so often include information that is incomplete or simply wrong.

4 times 3 times 2 equals 24 possible results. All things being equal, it’s simply a numerical permutation. There are 24 possible ways to serve a hamburger limited to four codiments; without tasing the same combination twice.

The way you did it, the order matters.

Plus, your solution assumes all four condiments are used (you did an ordering, not a choice).

I think.

Gee, maybe that means the question was poorly phrased.

I think we are closing in on agreement about this 🙂

This interpretation makes no sense if you look at what you’re supposed to answer. If all you’re concerned about is the order of the condiments, then how do you determine a probability for the mustard and pickles hamburger? In your scheme, all the burgers have mustard and pickles.

And what would the order of the condiments even mean? How would I even tell the difference between a hamburger with ketchup first and mustard second, versus a hamburger with mustard first and ketchup second?

The question is unanswerable without making assumptions about what “at random” means. It could be a 50/50 coin flip for each condiment, but it doesn’t have to be. Randomness is almost useless unless they specify its distribution.

I don’t believe that any complicating factors apply. The problem is pretty straightforward. There are 16 possible combinations of condiments, so each specific combo has a 1 in 16 chance of being served.