Thinking Skills: Critical Thinking and Problem Solving

5.1 Combining skills – using imagination 207

add this back on to the 80¢. In the example
above the calculations reduce to:

20 × 2.5¢ = 50¢

50¢ – 6¢ = 44¢

44

19

26

19

¢¢=

Once again, experience and a lot of practice is
the way to become efficient at solving the
harder problems. The more different types of
problem you see, the more you will be able to
build on your skills and combine skills you
have previously learned into techniques for
solving new types of problem.
The activity below uses simple probability,
something we have encountered very little so
far. Once again, Chapter 6.1 gives some help if
you are not familiar with the mathematics. An
alternative way of answering the problem
using permutations is also shown below. The
question itself is probably harder and longer
than anything you will encounter in a
thinking skills examination.

My local supermarket has a promotional offer.

It gives a coloured token with every spend

over $50. There are three colours: red, blue

and yellow. When you qualify for a token, you

take a random one from a large bag which

contains equal numbers of each colour. When

you have collected one of each, you get a $20

rebate from your next shopping bill.

In order to maximise her chances, Helga

makes sure she spends just over $50 each

time she shops and plans on shopping four

times in the two weeks the promotion will run.

She is sure she will then have one of each.

What is her percentage chance of getting

a full set (to the nearest 1%)?

A 2% B 10% C 11%

D 44% E 100%

Activity

Commentary

As noted in the introduction, there are two ways

of approaching this problem. Using probability,

we can say that it doesn’t matter what colour she

gets on her first visit. The chances of her getting

a different colour on the second visit are^23.

There are a lot of dead alleys here, so we

need to concentrate on the routes which lead

to success. These are (where ‘different’ means a

colour she hasn’t had before):

1 Any – different – different – repeat

2 Any – repeat – different – different

3 Any – different – repeat – different

There must be two ‘differents’ and the repeat

can be anywhere in the sequence. We can now

look at the probability of these three winning

combinations:

1 1 ×^23 ×^13 × 1 =^29

2 1 ×^13 ×^23 ×^13 =^227

3 1 ×^23 ×^23 ×^13 =^427

Adding these I get ()^624

27

++ = 12

27 = 44%^

(to the nearest 1%). D is the correct answer.

We now look at an alternative way of

solving this problem using permutations. In

total there are 3^4 (= 243) orders in which she

can get her four tokens. However, all of these

will include at least one repeat so we must be

careful. Of these, any combination including

ABC (e.g. CABA) will do. All of these

combinations will have one repeat, so we can

list the winning combinations.

Listing those with two reds (Rs) we have:

RRBY RRYB RBRY RYRB RBYR RYBR BRRY

YRRB BRYR YRBR BYRR YBRR

This is 12 in total.

(For those familiar with permutations, this is

4! ÷ 2!1!1! Here the exclamation mark means

‘factorial’ and means multiplying together

all the integers up to the one shown, so

4! = 1 × 2 × 3 × 4.)

There will be the same number with two Bs

and with two Ys, making 36 which fulfil the

requirement.