MathNotations

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I decided to post a video solution of the Twitter problem I posted on 8-25-10:

4 red, 2 blue cards; 4 are chosen at random. What is the probability that 2 of the cards will be red?

Because of the 140 character restriction on Twitter, the questions are often highly abbreviated and I actually consider it a "fun" challenge to write the question both concisely and clearly. Of course, as we all know about human interpretation of word problems, "clear" is in the eye of the beholder!

There's no doubt that the question above needs some fleshing out and might appear on the SAT and other standardized tests something like this:

A set of six cards contains four red and two blue cards. If four cards are chosen at random, what is the probability that exactly two of these cards will be red?

I'm sure my astute readers can improve on this wording but we'll leave it at this.

A few questions naturally pop up:

(1) Could this really be an SAT/Standardized Test question? Well, as I state in the video below, a question quite similar to this appeared on the College Board website the other day as the Question of the Day.

(2) For whom is the video intended? Everyone who happens upon it! I certainly wrote it to be helpful to students who will be taking the PSAT/SAT in the near future. Rather than simply presenting a single quick efficient solution, I demo'd 2-3 methods and indicated some important strategies and reviewed key pieces of knowledge to be successful on these harder probability questions. By the way, someone who is comfortable with probability will surely not find this question so formidable, but we're talking here about high school students or even undergraduates who struggle mightily with these.

(3) I'm hoping that the video will also serve as a catalyst for dialog in your math department. From the inception of this blog, I've never even intimated that a suggested way of explaining a concept, skill or a problem solution is in any way prescriptive. I encourage you to continue using whatever instructional methods have worked for you and to share these with our readers! However, for novice teachers or those who wish to see other approaches, I hope it will have some benefit. Of course, the video is not in a classroom. There are no students asking or being asked questions. There are no interruptions and I have a captive audience (except for my dogs who bark incessantly!).

SOME KEY STRATEGIES/TIPS/FACTS FOR PROBABILITY QUESTIONS

(1) It is highly recommended that students begin by listing 2-3 possible outcomes and to include at least one that is NOT one of the desired outcomes! This will help you to decide on a plan: organized list vs more advanced counting/probability methods. Further, you can ask yourself the key question in all counting/probability problems: DOES ORDER COUNT!

(2) Although it appears difficult for most test-takers to be systematic when making a list under test-taking conditions, preparation is critical here. If one practices several of these in the weeks leading up to the test, the chances of success improve dramatically. Did I just suggest preparation and practice could make a difference!

Where do you find these problems? Any SAT/ACT review book or my Twitter Problems of the Day or my upcoming SAT Challenge Quiz book to name a few sources...

(3) The basic definition of probability should always be in the forefront of your mind:

P(an event) = TOTAL NUMBER OF WAYS FOR THAT EVENT TO OCCUR DIVIDED BY TOTAL NUMBER OF OUTCOMES.

As indicated in the video, one can and should think of this ratio as TWO SEPARATE COUNTING PROBLEMS! Do the denominator first, i.e., the TOTAL number of possible outcomes. In the Twitter problem it is 15 if order is disregarded. Whether you arrive at 15 by listing/counting or by combinations methods, the denominator is 15 and is a completely separate question from "How many ways are there to get 2 red and 2 blue cards?"

(4) Finally, there are other methods for solving this probability question using Laws of Probabilities and/or permutation methods. I was going to make a 2nd video but I'm not so sure about that now.

An important point about the video below: I used 4 Blue and 2 Red cards, the opposite of the original Twitter problem but that won't change the final result!




[埋込みオブジェクト:http://www.youtube.com/v/305z8R9d56k?fs=1&hl=en_US]



Look for my other videos on my YouTube channel MathNotationsVids . Look for all of my Twitter SAT Problems on twitter.com/dmarain .

As I develop my Facebook page further, I may start posting these questions there as well as my videos. Facebook allows up to 20 minutes videos, much less restrictive than YouTube's 10 minute limit.

Posted by Dave Marain at 10:49 AM 5 comments

Labels: counting problems, math videos, mathnotationsvids, probability, SAT strategies, SAT-type problems, systematic counting, twitter problem of the day

Please note correction to 2nd problem in the video. The correct answer is 4096 "real" values. The original answer, 13, applies to rational solutions only. Thanks to Nick Hobson for pointing out my careless error. Haste makes waste!!


Please vote in the poll at the right. Be candid in your opinion of these videos. It will guide me in the future to improve. Don't hesitate to share your opinions on MathNotationsVids and rate each video there as well. If you subscribe to my feed, please vote directly on the site. Only a few days left...


The title says it all so here is the video as promised:

Note: See above correction to 2nd problem! The video has not been corrected so beware!

[埋込みオブジェクト:http://www.youtube.com/v/-SCSVCtVBec&hl=en_US&fs=1&]


Comments on 2nd problem:


If x is greater than or equal to 0 and less than or equal to 3, for how many values of x will 16^x be an integer?

As mentioned above, Nick pointed out my error. I should have restricted x to be of the form a/b, where a and b are integers, b ≠ 0. Normally, SAT questions avoid use of the term rational so they would spell it out. This problem however is very questionable for SATs. If real solutions were sought, this question would be more appropriate for a math contest. Here's one way of explaining why the answer is 4096 for real solutions:

16^x = k, k an integer → 2^(4x) = k
3 ≥ x ≥ 0 → 12 ≥ 4x ≥ 0 → 4096 ≥ 2^(4x) ≥ 1 since the exponential function 2^(4x) is increasing. This argument is reversible, so there are 4096 solutions for x, one of each integer value of k from 1 to 4096 inclusive. This solution could be written more concisely using log base 16 or log base 2 as Nick did, but I wanted to show a method without the log symbol.

Again, the video solution is WRONG as it shows only rational solutions! Well, at least i was thinking "rationally!"

I fully realize that the school year is over for some and about to end for others but these SAT Problems will be around for you or your students in perpetuity! Let me know if you like the questions. They are now appearing in the right sidebar of my blog so you will need to visit the page to see them.
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"All Truth passes through Three Stages: First, it is Ridiculed... Second, it is Violently Opposed... Third, it is Accepted as being Self-Evident." - Arthur Schopenhauer (1778-1860) You've got to be taught To hate and fear, You've got to be taught From year to year, It's got to be drummed In your dear little ear You've got to be carefully taught. --from South Pacific

Posted by Dave Marain at 10:14 AM 0 comments

Labels: advanced algebra, math videos, mathnotationsvids, SAT strategies, SAT-type problems

The video below presents a more challenging 3-dimensional geometry problem which would be at the upper end of SAT I or SAT II - Subject Tests (Math I/II). The key here is to challenge students' assumptions about a quadrilateral being a square because it has 4 congruent sides, a common error. This question will also review a considerable amount of geometry: Pythagorean Theorem, Volume of cube, spatial reasoning, 45-45-90 triangles, area of a rhombus, etc.

Posted by Dave Marain at 8:14 AM 4 comments

Labels: 3-dimensional geometry, cubes, geometry, mathnotationsvids

The full version in one of its many many variations:

If a hen and a half can lay an egg and a half in a day and a half, how many eggs can three hens lay in three days? Assume that all hens are a-laying at the same rate.

Putting aside the silliness of the riddle, there really is some serious mathematics going in these kinds of rate/ratio/proportion problems. Rather than solve the "hen" problem for you, I'll leave it to my readers to solve it by their own favorite methods. By the way, the answer to this riddle is in the description of the video below on my YouTube channel. Sorry 'bout that!!

Instead, the video below, which appears on my YouTube channel, MathNotationsVids, presents a developmental approach to a more complicated ratio problem for middle schoolers and beyond. I'm far more interested in your thoughts about the teaching strategies than I am about the problem itself. Please understand, further, that I am not suggesting the method shown in the video is efficient nor would it make much sense for the upper level math or science student. See comments below the video for further discussion of this.


The Problem in the Video Below:


If 10 workers can build 3 houses in 60 days, how many workers are needed to build 5 houses in 40 days? Assume all workers build at the same rate.

[埋込みオブジェクト:http://www.youtube.com/v/P_VCYl0zdts&hl=en_US&fs=1&rel=0]


More Advanced and Efficient Algorithms


(1) We assume from the "constant rate" assumption in the problem that the number of houses (H) which can be built varies jointly as the number of workers (W) and the number of days (D).
Thus, H = kWD.

Substituting, H=3, W=10 and D=60, we obtain:
3 = k(10)(60) or k = 1/200. Note that the units of k are Houses/(Workers x Days).
We can interpret k to mean that 1/200 of a house can be built by 1 worker in 1 day. Thus, k is not only a constant but actually represents a rate. Another way of expressing this rate is
(1 House)/(200 Worker-Days) or the reciprocal version:
(200 Worker⋅Days)/(1 House)

Substituting the new set of values into the relationship H = (1/200)WD, we obtain:
5 = (1/200)(W)(40) or W = 25 workers.

(2) This can be made even more efficient using the "factor-label" (dimensional analysis, etc.) format:

(200 Worker⋅Days)/(1 House)) x (5 Houses)/(40 Days) = 25 Workers!

(3) I could also exploit the inverse variation between W and D, but that's for my readers to bring up or for another video!

I see these efficient methods as "black box" methods for some students. Developing a deeper understanding of direct and inverse variation is far more important for the younger student.



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"All Truth passes through Three Stages: First, it is Ridiculed... Second, it is Violently Opposed... Third, it is Accepted as being Self-Evident." - Arthur Schopenhauer (1778-1860)

Posted by Dave Marain at 2:37 PM 3 comments

Labels: average rates, math videos, mathnotationsvids, middle school, proportions, ratios, work problems