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 8th roll” means “two
‘sixes’ on the first 7 rolls” and “a ‘six’ on the 8th 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 8th
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 nCx px qn–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)
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!
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