Saturday, August 28, 2010

Video Solution and Discussion of Twitter SAT Probability Question from 8-25-10

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Price is $9.95. Secured pdf will be emailed when purchase is verified. DON'T FORGET TO SEND ME AN EMAIL (dmarain "at gmail dot com") FIRST SO THAT I CAN SEND THE ATTACHMENT!

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!).


(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:


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!

Look for my other videos on my YouTube channel MathNotationsVids.  Look for all of my Twitter SAT Problems on  

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.

If interested in purchasing my new Math Challenge Problem/Quiz book, click on BUY NOW at top of right sidebar.  175 problems divided into 35 quizzes with answers at back. Suitable for SAT/Math Contest practice or Problems of the Day/Week.
Price is $9.99 and secured pdf will be emailed when purchase is verified. DON'T FORGET TO SEND ME AN EMAIL FIRST SO THAT I CAN SEND THE ATTACHMENT!

"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


Anonymous said...

So, you want 2 and 2? I can help. I'll grab a red card and another red card, and you can take the rest.

Hmmph. If I were grabbing at random, what would the probability of both cards being red be?

4 out of 6 for the first, times 3 out of 5 for the second.

So 12/30 of the time I'll leave you with 2 and 2.

(it's hard for me not to do. I was not taught to do this as a counting problem)


Dave Marain said...

Nice, Jonathan! In Part II, I was going to demonstrate other methods using multiplication of probabilities but none of them comes close to the simplicity and elegance of your thinking. At this point I doubt I will do a Part II.

In the end, I'm trying to help students who struggle in math learn to approach these questions using sound principles and basic methods/strategies. Listing the possible outcomes may not be practical for the more advanced problems but it usually suffices on the SAT. For students, this is a matter of survival. Many math teachers are appalled when students are shown test-taking strategies which circumvent the "real" math in the problem. They need to remember what it felt like when they were confronted with a math problem on a standardized test and they felt overwhelmed and lost...

Anonymous said...

So, I do appreciate that making a list is a far better strategy than many of us acknowledge.

But 2 qualifications:

Listmaking, that is, organized listmaking, is as much a skill as many of the other things we teach, and some kids have a knack for it, and others struggle to stay organized... Because it seems simpler to us does not mean it will actually be simpler for the kids who have the hardest time.

And my razzle-dazzle hid what I was doing: finding the probability of the complement. Conceptually tough? Absolutely. But it so often makes a hard problem easier that it is really worth knowing, teaching, practicing.

My dialogue with myself often looks like:
Hmm, probability, cool!
Oops, that looks involved. Is the complement simpler?

I get to it that quickly. And fairly often. And it has that indirect feel that almost translates to other branches...


Dave Marain said...

As usual, your comments are insightful and on target. You are right in stating that making an organized list does not come naturally for most students, but it is a skill that can be practiced. For those individual who are not analytical sequential thinkers this is difficult but one can improve. This process should start in the elementary grades!

Yes, your "complementary" thinking is a well-known and powerful construct. However, students more often see it and use it in the context of "What is the probability of getting at least one six in 5 rolls of a die?" Your approach actually changed the experiment to selecting TWO cards instead of the original FOUR. This seems as obvious to you as 6C4 equals 6C2 but the equivalence is not transparent IMO. I do think that once it's explained to a group, most would really like it and even prefer it to other methods.

Anonymous said...

So let them, while they list, or maybe better after they list, let them create a list of what's left over. Tell them to. And if they notice nothing, just let it go. But odds are, one kid's going to ask why they didn't just start with the things we don't want, the easier list.

And then you've got them thinking about complements.