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

If interested in purchasing my NEW 2012 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/Math I/II Subject Tests and Daily/Weekly Problems of the Day. Includes both multiple choice and constructed response items.
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!).

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!








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.

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

PI Day, More Videos on Counting, "Odds and Evens"

Since pi day fell on a Sunday this year, we should still be celebrating it today. Besides, March should be declared pi-Month!


It is always fascinating to see how readership (or should I say one-time viewership) always picks up around March 14th every year! I feel obligated to add another pi Day activity or exploration in addition to those I've posted the past three years. By the way, the pi Day Scavenger Hunt is the most popular post by far and I'm not even the one who thought of that idea!


Despite the title of this post, I did not upload a video for this activity. However, there is another video on the MathNotationsVids Channel on YouTube.

Here is an investigation/exploration/activity for middle and secondary:

Part (A)
(i) List all ordered pairs of positive integers (m,n) such that
(1) 1 ≤ m ≤ 10 and 1 ≤ n ≤ 10
(2) m and n are divisible by the same prime p

For example, (m,n) could be (6,9) since 6 and 9 are each divisible by the prime 3.

(ii) Should (9,6) also be counted?

(iii) Another way of expressing Condition (2) is:
The _______________ of m and n is ________  one.
Answer: gcf; not equal to or greater than

(iv) If you listed and counted correctly, you should have found there are 37 ordered pairs which satisfy both conditions. If not, have a partner check your list. Each of you should be checking each other's lists routinely.

Part (B)
(i) Explain, using the multiplication principle, why there are 100 ordered pairs which satisfy Condition (1) above.

(ii) ) What % of all the possible ordered pairs from Condition (1) are relatively prime. If you have immediate access to the internet, research this term before asking your teacher what it means!

(iii) In probability terms, you could say:

If one of the 100 ordered pairs (m,n)  from Part (A) is selected at random, the probability that
m and n are relatively prime is ____%.

Part (C) (more advanced)

If you have access to a graphing calculator, such as the TI-84 or TI-Inspire, enter the following program into memory (call it RELPRIME):

:ClrHome
:Prompt N
:0 → K
:For (X,1,N)
:For (Y,1,N)
:If gcd(X,Y) ≠ 1
:K+1 → K
:End
:End
:Disp K
:Stop

Using this program, complete the following table:

N..........Total # ord. prs..........# of not rel prime prs........% rel prime prs

10.........100.............................37....................................63%

20.........400............................ 145.................................

30

40

50

100

Notes:
K represents the count of ordered pairs which are not relatively prime
N represents the greatest value for the integers
gcd is found by going to MATH, then NUM, then 9:gcd(
The program slows down considerably as N increases. For N = 10, it checks 100 ordered pairs which may take only 2-3 seconds. For N = 100, it checks 100^2 pairs, which could take up to 4-5 minutes. Be patient!!

Conclusion: So what does all of this have to do with π ?
Well, as N increases without bound in the program, the probability that a randomly chosen ordered pair of positive integers (with values up to an including N) will be relatively prime approaches 60.7% rounded.

From out of the blue, compute 6/π²...
Want to know why? Well, that requires some advanced machinery involving infinite products, infinite series, and the Riemann Zeta Function! Perhaps, I'll do an informal development in a video. I love this stuff...


----------------------------------------------------------------------------------
"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

Dorothy Revisited -- Another View...

Mathmom contributed some insightful thoughts about how most middle school students might feel about the probability investigation from the other day. I agree with her that some would be able to compute the results or even devise a general formula but "proving" it in the general case might be too ambitious. In my reply, I suggested there might be another way of deriving the formula 1/N for the probability of losing the game. Here's what I came up with. It still requires some careful development to show that the outcomes are equally likely but I will indicate how it could be done in the particular case where N = 10.

Brief Explanation of Method:
There are N equally likely (to be shown) ways for the game to end (i.e., when the red card is selected). Of these, only one will result in a loss -- when the red is the last card chosen. Therefore, the probability of losing is 1/N, hence the probability of winning is 1 - 1/N or (N-1)/N.

Demonstrating "Equally Likely" for N = 10:
P(game ending after one card) = 1/10
P(game ending after 2 cards) = P(black selected followed by red) = (9/10)(1/9) = 1/10
P(ending after 3 cards) = P(black,black,red) = (9/10)(8/9)(1/8) = 1/10
etc...

The general case is similar using N in place of 10. I do think that students with some understanding of algebra could follow it but deriving it on their own is another story!

I also indicated that I might provide a program for the TI-83 or -84 which could be used to simulate the game. The programming skills needed are not that advanced and some high schoolers or even middle schoolers can pick up on the code and begin writing their own programs - I've seen it happen! Here it is...











T represents the number of times the game is played with 3 cards. I entered 100 for the number of trials. K stores the number of times Dorothy won when playing 100 times. Can you make sense of the rest of the code?

The experimental probability of 0.68 is reasonably close to the theoretical probability of 2/3. I often feel more confident of my reasoning in difficult probability problems when my simulation approximates my answer. This doesn't prove anything but it does have value IMO. There is also the opportunity to demonstrate some important stat concepts by running the program several times and having students plot the experimental probabilities and observing their distribution.

Please Help Dorothy Go Home - A Probability Fantasy for Middle School and Beyond

Summer vacation is an appropriate time for fantasy. Enjoy the hiatus!



The following investigation is not intended to be a math contest challenge. It reviews fundamental principles of probability and you might want to bookmark it for the fall. We can also simulate the first problem using the programming capabilities of a graphing calculator. I may post a simple program for this later on.


The wizard will let Dorothy go home if she can pass three challenges.

He shows Dorothy 3 playing cards, 2 of which are black and one is red. He shuffles them and turns them face down. "Dorothy, here's your first challenge."

"You will pick a card. If it's red the game ends, you win the game. If it's black, I will remove the card and you will pick a card from the remaining two. If it's red you still win! Ah, but if it's black again you and Toto and your weird friends will remain here for at least one more month."

Well, Dorothy won the game and said, "Now, I want to go home!" But the crafty wizard said, "You weren't listening carefully, Dorothy. I never said you can go home if you won the game. You've only passed the first challenge. You must still pass two more." "That's not fair!" Dorothy protested but the wizard makes his own rules in Oz.

"Alright, Dorothy, you won the game but you knew the odds were in your favor since you had two chances to win. Here's your next challenge:

"What was the probability of your winning and you must give me two correct but different methods?"

Dorothy asked, "These are the remaining challenges, so if I get them right, I can go home, yes??"
"I will not lie to you, Dorothy. This is your 2nd challenge. There will still be one more."

Dorothy was upset but knew she had no choice but to trust him. She thought about the problem for a minute and replied, "The probability of my winning was 2/3. I know I'm right!"
"Very good, Dorothy, but you must explain that answer two different ways." Fortunately, Dorothy was a very responsible middle school student back in Kansas and had learned the methods of compound probabilities and the idea of complementary events (this is a fantasy after all!).

Dorothy was able to provide two correct methods. Can you?

"Very good, Dorothy! You only have one more challenge to conquer and you can go home.
This time there are N cards, one of which is red while the remaining cards are black. N is a positive integer greater than 1. Same rules as before. The cards are shuffled and laid out face down. You pick a card. If it's red the game is over and you win. If it's black, the card is removed and you try again. The game continues until you pick the red card. The only way to lose the game is if you pick all the black cards and the last card remaining is red."

"In terms of N, what is the probability that you will win? Oh, yes, you again have to show two different methods in detail on this magic board over here."

This time, Dorothy needs your help. She can guess the formula but she needs our help to show two ways to derive it. Please help Dorothy go home!

Math Contest Reminder and a Probability Paradox??

There's still time to register for MathNotation's First Math contest for Grades 7-12 to be held on Tue Feb 3rd. I've decided to extend the registration to Thu Jan 29th. We've had interest expressed from high schools, middle schools, homeschooling teams, even a chapter of an honorary math fraternity! I'd like to see 2-3 more teams compete but I understand that many students and teachers are overextended at this time of year and this was on short notice. Look here for how to register.
----------------------------------------------------------------------------------

So what's the paradox in the title? To someone with a firm grasp of probability there won't be one, but the following series of questions may lead to a surprise for some students.

Overview of Problem
We have two scenarios in this investigation:
A set of five 4-choice multiple-choice questions and a set of five 5-choice multiple-choice questions. Of course the latter is typical of most standardized tests like SATs so this discussion may have relevance to many juniors right now!

Instructional Suggestion
For the following questions, ask students to first make educated guesses before attempting any calculations. The idea is to get them to trust their intuition which often is more accurate than their mathematical procedures!

Background
We know that the probability of correctly guessing, at random, the answer to a 4-choice question is 1/4 which is greater than the chance (1/5) of correctly guessing, at random, the answer to a 5-choice question. That was easy, right? When we ask questions about more than one question the situation becomes more complicated and a deeper understanding of probability concepts is needed: Multiplication of probabilities of independent events, binomial probabilities, etc...

The Investigation

(a) Which of the following is more likely? Randomly guessing all 5 wrong on a 5-choice multiple choice quiz or randomly guessing all 5 wrong on a 4-choice multiple choice quiz?
By intuition (no calculation, respond in 10 sec or less): _________________
Explanation of Intuitive Guess (this may be worthy of class discussion):
Now compute each probability and compare result to your intuitive answer.

(b) Which is more likely? Randomly guessing at least one right out of five on a 5-choice multiple-choice quiz or on a 4-choice multiple-choice quiz?
By intuition: ______________
Explanation of intuitive guess:
By calculating:

(c) How's your intuition doing so far?
Let's try this one:
Which is more likely:
Randomly guessing exactly one right out of 5 on a 5-choice quiz or on a 4-choice quiz?
By intuition:
By calculating:

Any surprises? In case your results don't agree with mine, I will tell roughly you what I got (actual probabilities below). The probability of guessing exactly one right out of five on a 5-choice quiz is slightly more than the probability of guessing exactly one right on a 4-choice quiz! A paradox? An anomaly of the arithmetic involved? Logical? Can you explain it? Try!

(d) Back to normalcy? Compute the probabilities of getting exactly two right out of five on a 5-choice quiz and on a 4-choice quiz. Has the order of the universe been restored!


Selected Answers (not the norm for this blog):
(b) Approx 67.2% on a 5-choice quiz; 76.3% on a 4-choice
(c) Approx 40.96% on a 5-choice quiz; 39.55% on a 4-choice
(d) 20.48% on a 5-choice quiz; approx 26.37% on a 4-choice
Pls check these results for accuracy!!

What are the fundamental concepts in this investigation? What are the learning benefits of this series of questions? Please understand that my intent on this blog is to suggest instructional methods, never to impose. You may find far more effective ways to convey the essential concepts here but, from my experience, there's only sure way to perfect our craft. Keep experimenting and asking questions!!

A Preview of the Contest: Probability Investigation with Replacement

Don't forget to email me if you want your students to participate in the first MathNotations online math contest on Tue Feb 3rd. There is still time! Look here for info.


There may not be a probability question on the first contest but the following gives you a flavor of the type of multi-part question I'm talking about -- an investigation in more depth.

You will find many variations of the following problem in texts. From experience we know that the student needs to have numerous experiences with these. How do many students do on this topic when the exam question is slightly different from the ones reviewed in class!

THE PROBLEM STATEMENT
Five cards are numbered 1 through 5 (different number on each card). Typical scenario, right?
George chooses cards randomly one at a time. After he selects a card, he marks a dot on the card, then puts it back (replacement!) in the pile of 5 cards, reshuffles them and draws the next card and so on. The game continues until he selects one of the "marked" cards.

INSTRUCTIONAL STRATEGY
Before a technical analysis of this experiment (sample space, random variable, specific probabilities, expected value), I would typically ask students a broad intuitive question or ask them to suggest questions one might ask about this "game".

Intuitively, I might ask:
In the long run, how many draws would you "expect" it to take for the game to end?

With five cards, what do you think most students would guess? Draw three? Draw four? I think asking this initial question is crucial. In most cases, we want the mathematical result to be reasonable and to roughly agree with our intuition (not always of course, there are paradoxes in math which are counterintuitive!).

THE INVESTIGATION

Part I
What is the probability that George chooses a "marked" card on his second draw for the first time? On the 3rd draw for the first time? 4th draw? 5th draw? 6th draw?
Another way to ask these are: What is the probability that the game "ends" after 2 draws? 3 draws, etc.

Part II
"On average", how many cards would George need to draw to get one of the marked cards for the first time?

Note: In more technical language we are asking for the expected number of draws before the game ends?

Normally, I don't publish answers to these questions but, in this case I will give partial results. Please check for accuracy.

The probability the game ends after 3 draws is 8/25 or 32%.
The expected value for the number of draws for the game to end is approximately 3.51. What does this mean!

Teaching Probabilities and Strategies Via Games!

NCTM Teaching Standards:

Develop and evaluate inferences and predictions that are based on data

Understand and apply basic concepts of probability

Target Audience: Grades 7-12

Tools Needed: Graphing calculator with a random integer generator or an online random number generator (look here for example)

Classroom Organization
(After demo mode): Students working in groups of 3 (two opponents, the 3rd calls out the numbers and keeps score; roles are rotated)

Sample Classroom Scenario
Who thinks they can beat me at a game of chance? I will demo the game, then I will play against an opponent. If you beat me two out of three, you are the new champ and you can pick your opponent. After 10 minutes, you will be playing in small groups and recording the results.

The Play
Using a random integer generator we will generate random digits, one at a time, from 1 through 9, inclusive (no zeros). The object is to build a 5-digit integer which is greater than your opponent's by placing each 'called' digit into one of the five place-value positions. Once you place a digit you cannot change it!

Let's try it... Ok, Marissa, turn on the random integer generator, press Enter and call out the first integer. FOUR!
Ok, I'll place it here: ___ ___ 4 ___ ____
Call out the next integer: SIX!
I'll place it here: ___ 6 4 ___ ___
Next: TWO!
___ 6 4 ___ 2
Next: FOUR!
___ 6 4 4 2
Last digit! FIVE!
5 6 4 4 2
How did I do? Could I have used a better strategy? Do you think you could have beaten me?
Who wants to play! To win, you have to beat me two out of three. Ok, Dimitri, I will work on my paper and you work on yours. Remember, you cannot change a digit's position once you place it...

Brief Discussion of Strategy Based on Probability Arguments:
Suppose the first two digits called are 3 and 6 in that order. Would it be better to place the 6 in the thousands' place or the ten thousands' (leftmost) position? If you place the digits here:
___ 6 ___ 3 ___, what is the probability that at least one of the next three digits chosen will be 6,7,8, or 9. (Otherwise, your strategy would have backfired). To compute this, we look at the complementary condition, i.e., we determine the probability that the next 3 digits chosen will all be in the range 1 through 5. The probability of this is (5/9)(5/9)(5/9) or approximately 17%, so the probability that our strategy works is about 83%, odds that seem worth playing! Experienced game players often compute these probabilities mentally or have seen these situations so many times they know these probabilities by heart!

Notes
(1) Students may not know there is a Random Integer generator built into many graphing calculators. For example on the TI-84, press MATH, then PRB, then 5:randInt(.
From the home screen, Enter randInt(1,9), ENTER. Each time you press ENTER another "random" digit will be displayed. The person calling these out must be instructed to announce only ONE digit at a time!
(2) Why 5-digit numbers? This seems to make the game fairly interesting and moving at a good pace. Expect ties of course!

Perhaps this is a good activity before the holidays. Have fun and let me know how it goes!

Why not predict the election with a simple probability problem...

Up to now, I've chosen not to bring the election into this blog, however with the outcome to be decided, hopefully, in a few hours, perhaps we can make our predictions based on the following:

The election will be decided by a fair 8-sided die whose faces are numbered from 1 through 8.
The die will be rolled 5 times.

Candidate A wins if the number 7 appears at least once in the 5 rolls.
Candidate B wins otherwise.

You can decide the identity of each candidate! It would be interesting if the popular vote turns out to be this close...

2008 has 2 digits that are the same -- A Probability Investigation For Middle Schoolers And Beyond

As the school year is beginning...

Which would you conjecture is more likely:
No digits the same in a 3-digit number or no digits the same in a 2-digit number?

You have 30 seconds to choose one of these - - - - - - - - - -
NOW WRITE YOUR GUESS ON YOUR PAPER and compare with your partner. Take one minute to discuss your thoughts...

Alright, I know some of you take exception to wasting these 30 seconds. What could be gained from such 'blind' guessing without the time to really think it through and work it out. I often used device this to encourage youngsters to react instinctively and to learn to trust their intuition. How many times have all of us had the experience of not trusting ourselves, only to find later that we were right. If it turns out that this gut reaction is not supported by the data, then the mathematical researcher (or the experimental mathematician in this case) revises the hypothesis. Ultimately, one attempts to validate one's conjectures via logic (deduction, induction, etc.). If you're still not convinced this is worthwhile, it's only a suggestion...

Now we're past the prelims. Our goal is to have our students begin with solving a particular case of the problem above and then to develop a general relationship for:
The probability that an N-digit positive integer will have N different digits. Of course, N is restricted to be in the range 1..10. We would hope our students from middle school on would recognize that the probability for N = 1 is 100%, whereas the probability for N = 11,12,13,... would be zero! Yes, we would hope!

(1) Show that the probability a 2-digit positive integer has different digits is 90%.
Comments: This is a well-known and fairly basic problem, but this is just the jumping-off point for this investigation. Various methods are likely here, depending on the background of the student. The middle school student (and many secondary as well) would likely list or count the number of 2-digit numbers with different digits. Some would realize that it might be easier to count those with identical digits and subtract from the total. More advanced students may use more sophisticated approaches for this and the other parts below. One could use this activity to develop the multiplication principle, permutations, use of factorials, etc. However, there is much to be gained from 'first principles.' Careful counting and making an organized list never go out of style!

(2) Show that the probability a 3-digit positive integer has 3 different digits is 72%.

(3) Complete the following table up to N = 10:
Note: P(N...) denotes the probability of the indicated outcome.

Number of Digits N.....P(N different digits)

...........1......................100% or 1
...........2..................... 90% or 0.9
...........3......................72% or 0.72
...........4......................50.4% or 0.504......
.
.
..........10.....................................................


(4) Time to revisit your original conjecture.... Explain why the probabilities decrease as the number of digits increase.
Note: One could give a purely descriptive explanation here.

(5) For more advanced groups:
Develop a formula for P(N).

(6) For more advanced groups:
Enter your expression from (4) into Y₁ of the Y= menu in your graphing calculator. Set up a TABLE with Start value of 1, increment (Δ) = 1 and Auto for Indpnt and Depend. Display your table and check the values you found from your own table.

(7) [Optional]
Closure: Write 3 ideas, methods, strategies, mathematical principles, etc., you have learned from this activity.

Taking Two or More At a Time- What is the Probability You Don't Like These!

Only a few days left for your July MathAnagram. Either it's harder than I thought or no one is paying attention!

SIR "OY", E. NOET!

Look here for details.

Probability questions will forever addle the minds of students and adults alike. If all problems could involve selecting one object randomly, life would be good. Unfortunately, selecting 'more than one' is becoming common on standardized tests these days. Taking two or more objects immediately ratchets up the difficulty:
Does order count?
Multiple solution paths
Making a list
Combinations? Permutations? Multiplication Principle? Using "rules" of probability?

In what course do students receive sufficient instruction in this important area? Algebra 1? Algebra 2? Precalculus? A probability/statistics/discrete math class? AP Stat? IMO, the lack of standardization in secondary curriculum can lead to some topics getting short shrift.

I've come to the conclusion that middle schoolers should devote more time to some of this, since 4th graders are generally expected (in most states' standards) to solve the "select one object" type. What do you think? Since most readers enjoy the math challenges and not this kind of curriculum discussion, here are our offerings for today...

A bag contains eight coins: two each of pennies, nickels, dimes and quarters.

Question 1: If two coins are randomly selected, show that the probability that the two coins will total at least 20 cents in value is 1/2.

Question 2: If 4 coins are randomly selected, show that the probability of getting exactly one of each kind of coin is 8/35. (At least two methods please!)
Note: This result implies that the chances of getting at least one matching pair of coins among the 4 coins is greater than 75%!!

Question 3: Invent your own!

Comments:
There are many problems of this type on this blog (see probability, combinatorial math in the index). Further, there are many other excellent blogs and web sites that address these topics and provide wonderful challenges and explanations. Two of the best are Jonathan's over at jd2718 and Isabel over at God Plays Dice.

Probability and Counting Challenge -- You'll Need To Sit Down for This One

Five people with first initials A, B, C, D and E were seated randomly in a row in a movie theater with no spaces between them. What is the probability that A, B and C were adjacent to each other in some order? (For example: "DBACE")

A potential SAT problem or is it a level above? A math contest problem or not difficult enough? More importantly, how much experience do most of our students have with these kinds of combinatorial problems? I know that some of our educators who visit here do these with their classes, but is it the norm? My feeling is that students need to see many of these developed over time in more than one course.

What method (s) do you consider the most effective for solving these kinds of problems? For teaching? Are these 2 questions really the same?

After this question is discussed with the class, how does the instructor assess the learning? Give them another one to try immediately or give a worksheet of these (if the text does not provide enough practice)? How could one raise the bar even higher?


Suggested Extension #1: This time we have 10 people seated randomly in a row (no spaces). What is the probability that 4 of them, say A. B, C and D, would be adjacent to each other?

Suggested Extension #2: N people seated randomly in a row (no spaces). What is the probability that a particular subset of M of these people would be adjacent to each other?

Your thoughts...

1,2,2,3,3,3,4,4,4,4,... What is the 2008th term? SAT-type Questions vs. Math Contest Problems

Don't forget to submit the name of our Mystery Mathematician. Contest ends around 4-15-08. Thus far, only one correct submission (emailed of course!).

The problem in the title is the math contest version. Knowing the formula for the nth triangular number would be helpful (so might a calculator). Do all middle school and hs students become familiar with triangular and other figurate numbers? Should they? My vote: Yes!


The SAT -type would be:

What is the 56th term of the sequence 1,2,2,3,3,3,4,4,4,4,..., in which each positive integer N occurs N times?

Comment: This is considerably easier than the contest problem as one could do it by listing with or without a calculator. It also may reveal your strong number sense students who will see the idea fairly rapidly. Try it as a warm-up in class!

Another SAT-type (probability, counting) to help students prepare for the May Exam:

Let S be the set of all 3-digit positive integers whose middle digit is zero. If a number is chosen at random from S, what is the probability that the sum of its digits is even?

Note: I wrote this question to demonstrate basic principles of probability and counting. Although one could use the Multiplication Principle (aka, Fundamental Principle of Counting), students should also be encouraged to make an organized list and, by grouping, see why the answer is 1/2.

Drawing 3 cards one at a time vs. Drawing 3 cards simultaneously - Those nasty probability questions!

[For a detailed conceptual discussion of several methods, read the excellent comments by Mike and novemberfive.]

It's 'probably' too early in the year for this, but...
This is one of those typical probability questions that all students struggle with until they have enough experience to feel comfortable. When away from these for awhile, most everyone needs to revisit this kind of question and work through the theory again. It's so elusive...

Here's a typical version of this question:

Three cards are drawn at random from an ordinary deck of 52. What is the probability that they will consist of a king, a queen and an ace?

Do you feel totally comfortable when seeing these? Imagine how most students feel. They either have a strong concept here or they get that queasy feeling and express, "I don't like these!"

So here's the issue:
Is there one method you have found for these that makes sense to most students or should multiple methods be demonstrated (and shown to be equivalent)?

To get you started:
Method One: For the denominator, select all 3 cards 'simultaneously' in C(52,3) ways. The numerator is more interesting, so I'll leave that to you.
Method Two: Analyze one card at a time without replacement. Thus, the probability that the first card will be a king is 4/52, etc. Of course, most of you know the 'catch' here, but the method can still work if...
Method Three: Yours!