[Pri20150306MOD] A Telescope Too Far?

Question

Introduction

     This is a nasty multiple-choice question to be set for a primary 6 (~ grade 6) pupil for a test.  It could also be a time trap as the pupil might spend a lot of time to no avail just to try to score that miserable mark.  With advanced knowledge, we can solve this using telescoping sums (a.k.a. the method of differences), where many terms cancel and the sum can be shortened, much like a telescope.

     However, this solution looks like over-kill.  Is there a solution that is more accesible to a primary 6 pupil?  Read on!

A Simpler Solution?

     We observe that a denominator of  5  is common.  In fact, since every term has an even factor, we have a common factor of 2 in the denominator also.  That means every term in the sum can be expressed as ¹/₁₀ of something [ Heuristic H09 ].  We can quickly work out the first few partial sums and express each of them as  ¹/₁₀ of something.  [ Heuristic H04 ]

     Tabulating the results [ H02 & H03 ], we observe that each partial sum is  ¹/₁₀ of something slightly less than ½.  So we try to express them as ½ minus something.  How?  We can subtract to find out.  [ H05 ]  For example,  if  ⁵/₁₂ = ¹/₂ – ???   then   ??? = ¹/₂ – ⁵/₁₂ = ¹/₁₂.

     Later on, we notice   [ H04 again ] another pattern: the factors in the denominators of the subtracted quantity matches the last two denominators of the last term.  If we considered the full sum, then what would those factors be?  (See the part highlighted in yellow).  These would be 20 and 21, as per the last term of the series.  Using this pattern, we work out the required sum as follows

That is the answer!

Heuristics Used

H02. Use a diagram / model

H03. Make a systematic list

H04. Look for pattern(s)

H05. Work backwards
H09. Restate the problem in another way

H2Maths VJC/2007/P2/Q9 The Power of Rephrasing

Introduction

     Today we discuss a question taken from one of the top junior colleges in Singapore’s 2007 preliminary examinations (final school internal examinations before the actual A-levels).  The seemingly difficult problems dissolve quickly using the right heuristics (rules-of-thumb / guidelines / problem-solving tactics).  Once the correct approach is determined, the calculation is simply a matter pressing buttons on the graphing calculator (e.g. TI-84 Plus or Casio fx-9860G).  This question can be tackled using heuristics.

     Try to guess what heuristic(s) will be useful in solving the above problem.  In a typical examination question, they will not tell you what topic or concepts are being tested in that question.  Try to determine what concepts are involved.

     Let’s tackle the question part by part.  As usual, we use the 5 step problem-solving process with metacognition (self-prompting, self-monitoring).

Part (i)

 

Part (i) Stage 1:  Understanding the Problem

What concept(s) is being tested?

     That seems to be something to do with probability …

Do you understand the question?  Can you put it in your own words?

When Ai Wan (爱玩 “loves to play”?) rolls the die for the eight time, he gets exactly three ‘6’ to win the prize.  What are the chances of this event happening?

Part (i) Stage 2:  Planning the approach

Is there a heuristic you can use?

     How about “9. Restate the problem in another way” …

So how do you rephrase the question?

     “win on 8^th roll” means “two ‘sixes’ on the first 7 rolls” and “a ‘six’ on the 8^th roll”.

How will you proceed?

Break down the problem into parts (“11. Solve part of the problem”).

     (1) find probability of “two ‘sixes’ on the first 7 rolls”

     (2) find probability of “a ‘six’ on the 8^th roll” (this is easy.  Obviously  ¹/₆)

Then multiply them together, since these are independent.

How do you write out probability of “two ‘sixes’ on the first 7 rolls” in symbols?

P(X = 2)

Think backwards:  What is X?  How do you define it?

X  is a random variable.  It is the number of ‘sixes’ in the first 7 rolls.

What distribution does it follow?  How do you know?

It follows a Binomial Distribution: Because there are only two outcomes: either you get a ‘six’, or you don’t.  There is a fixed number of trials and these are independent, giving a constant probability. 

So what’s the number of trials (n)?

There are 7 rolls, so seven.  Each roll is a trial.

What is the probability of “success” (p)?

One out of six.

Is this a probability distribution function (p.d.f. ) or a cumulative distribution function (c.d.f.)?  How do you calculate it?

This is just probability for a single value of  X, so it’s a p.d.f.  It can be found from the graphing calculator.  [ Or use the formula ⁿC_x p^x q^n–x = ⁷C₂ (¹/₆)²(⁵/₆)⁵ ]

Part (i) Stage 3:  Executing the plan

Part (i) Stage 4:  Evaluating the answer

Does this answer feel correct?  Is it believable?

     Yes.  It’s quite small, which tallies with my intuition, as getting three sixes in 8 rolls is highly unlikely.  If I get an answer like 0.4 something, I might smell a rat.  If I get a probability that is less than 0 or more than 1, then I know for sure it’s definitely wrong.

Part (ii)

Remark

     This problem is related to the Geometric Distribution, which is not in the syllabus.  Nevertheless it is solvable using the knowledge contained in the syllabus, so it is within the student’s reach.

Part (ii) Stage 1:  Understanding the Problem

What concept(s) is being tested?

     probably probability again … maybe Binomial or something related

Do you understand the question?  Can you put it in your own words?

Ai Ying (爱赢 “loves to win”?) already has got 2 ‘6’ and 2 ‘1’.  That means she needs just one more ‘6’.  She needs to throw exactly four more times (no more and no less).  What are the chances of this happening?

Part (ii) Stage 2:  Planning the approach

Is there a heuristic you can use?

     “9. Restate the problem in another way” … but the problem still seems complicated because of the initial conditions (two “sixes” and two “ones” ) …

Is there a way to simplify the problem or another way to think about it?  Is there a feature about the situation … ?

     Ah!  Since all the trials are independent, what happens in the past does not affect the future.  This is the memoryless or forgetfulness property of the (Bernoulli) trials.  That means I don’t need to worry about the two “sixes” and two “ones”.  I can forget the past!  It is as though I can just start from a clean slate!

So what does it mean to say “Ai Ying needs to throw exactly four more times”?

     It means after throwing three more times, she does not get any six, and then she gets a six on the next throw.

How do you write this out mathematically?

P(S’, S’, S’, S)

What do you mean by S?  What is S’ ?

S  is the event of getting a ‘6’ in a roll of the die.  S’  means not getting a ‘6’.

Now, how do you proceed?

The chain of events can be broken down.  (“Split the problem into smaller parts”)

P(S) = ¹/₆.  P(S’) = 1 – ¹/₆ = ⁵/₆.  Since the four events are independent, I can just multiply their probabilities  all together!  This is easy!

Part (ii) Stage 3:  Executing the plan

 

Part (ii) Stage 4:  Evaluating the answer

Does this answer feel correct?  Is it believable?

     Yes.  Again the answer is quite small, which tallies with my intuition.

Part (iii)

Part (iii) Stage 1:  Understanding the Problem

What concept(s) is being tested?

     Probability, Binomial Distribution and … er … rephrasing?

Do you understand the question?  Can you put it in your own words?

Ai Du (爱赢 “loves to win”?) cannot win in eight rolls, so he needs to roll some more.  What are the chances of this happening?

Part (iii) Stage 2:  Planning the approach

Is there a heuristic you can use?

     “9. Restate the problem in another way”

So how do you restate the question?

 “requires more than eight rolls of the die to win a prize” means “after 8 rolls, he does not get three ‘sixes’”.

Which means?

Which means “after 8 rolls, he gets only 2 or less ‘sixes’”.

How do you write the required probability mathematically?

P(D < 2)

Think backwards:  What do you mean by D?

D  is a random variable that counts the number of ‘sixes’ within 8 rolls.

What distribution does it follow?  Why?

Binomial Distribution, again.  Same reasons as in part (i).

So what’s the number of trials (n)?  What is the probability of “success” (p)?

n = 8,  p = ¹/₆.

Is this a probability distribution function (p.d.f. ) or a cumulative distribution function (c.d.f.)?  How do you calculate it?

Because it is a “<” probability, it’s a c.d.f, which can be found from the graphing calculator, or checking tables.  [ Calculation by hand is possible, but tedious.  Also, there is a recurrence formula that can help a bit, but this is out of syllabus. ]

Part (iii) Stage 3:  Executing the plan

 

Part (iii) Stage 4:  Evaluating the answer

Does this answer feel correct?  Is it believable?

     Yes.  The answer is high, which tallies with my intuition.  Since it is hard to win, my gut feel is that you would very likely need more rolls of the die to win.

Stage 5:  Reflection

What did you learn from solving this problem?

     I learned that the memoryless (forgetfulness) property of Bernoulli trials (i.e. the type of trials involved in the Binomial distribution) is useful to allow me to disregard past events and simplify the problem.

     I learned that whenever I encounter a maths problem that seems difficult, I should not despair.  I should try the following heuristics:-

     * rephrasing the problem (heuristic 9)

     * breaking the problem into smaller parts (heuristic 11)

     * writing in mathematical notation (heuristic 13)

     * thinking backwards (heuristic 5)

     * simplify the problem (heuristic 10)

Rephrasing is particularly useful, as it helps to tackle all three parts of this exam question.  Even though part (ii) was on the fringes of the syllabus (just barely in the syllabus), we could still solve it by using heuristics, metacognition and using what we know already.

Conclusion

     Heuristics and metacognition will guide you when you seem to be in uncharted territory.  Use what you know (basic probability) to tackle what you do not know.  Do not be afraid.  Have faith.  May the Heuristics be with you!

 

JCCDQBHWHCB037(ii) Combinatorics : Grouping and Insertion Method

Introduction

     This question is from a source that does not credit the original source.  In the original question was poorly worded.  It did not have the words “the letters of the word”.  Instead of “each vowel must be separated”, it said “a vowel must be separated”, which is might mean there is just one such instance.  This is ambiguous.  I have taken the liberty to rephrase some parts of the question to make its meaning clearer.  In this article, I shall discuss only part (ii) of this question.

Stage 1:  Understanding the question

What is the given in the problem?  Can you organise the information?

     Though not absolutely necessary, it is helpful to draw a diagram that separate the letters into vowels and consonants and write the stack up the same letters in columns.

     There are  5  vowels (of which  O is repeated)  and  6  consonants (of which  N  and  S  are repeated).

Can you explain the problem in your own words?

     The letters of the word ‘CONNOISSEUR’ are re-arranged, which means that all the 11 letters are used.  Each vowel must be separated from another with exactly one consonant, which means that the letters must contain the pattern  “v c v c v c v c v” (where v = vowel, c = consonant).  Important: Note that the question does not say that the first letter must be a consonant.

Stage 2:  Planning

Have you seen a similar problem before?

     Yes, but this looks a bit more challenging.  There are more possibilities as first letter need not be a consonant.

What heuristics can you try?

     ·  Solve part of the problem

     ·  Split the problem into smaller problems

What topic-specific tactics can you try?

     The “v c v c v c v c v” pattern can be treated as a group (Grouping Method).  Since there are six consonants, there are two more “c”s (consonants) in the full pattern.  This looks like a problem that can use the Insertion Method.

Stage 3:  Execution

The number of ways to insert the group = ³C₁ = 3

     [these are the patterns “ccvcvcvcvcv”, “cvcvcvcvcvc” and “vcvcvcvcvcc”  ]

For each pattern,

     the vowels can be arranged in  ^5! / _2!  ways (division because there are 2 ‘O’)

     the consonants can be arranged in  ^6! / _{2! 2!}  ways (division because of 2 ‘N’ and 2 ‘S’)

Hence the total number of ways is

Stage 4:  Evaluation

Is the answer correct?

     Yes, the answer is correct.

Stage 5:  Reflection

What have we learned by solving this problem?

     We have learned once again that heuristics and metacognition are useful in solving mathematical problems.  Specifically, we have used the following heuristics:-

          ·  Drawing a diagram

          ·  Solve part of the problem

          ·  Split the problem into smaller problems

     We have also used the following techniques that are useful for combinatorical problems:-

          ·  Grouping Method

          ·  Insertion Method

          ·  Division Method (for dealing with repeated letters)

     It is also important to understand the problem correctly and not make unfounded assumptions.  If the wording is not clear, you may want to rephrase it.