Thinking Skills: Critical Thinking and Problem Solving

Answers to assignments 335

The inner quartered square shows the

fast hydrofoil services, the outer square

the slow steamboats. The Xs mark

the cells that are empty (represent no

service). These are ferries going from

Waigura to anywhere other than Dulais

and fast hydrofoil services to anywhere

other than Dulais. All other cells may

contain services.

We can now answer the statements:

A Hydrofoils from Nooli to Dulais are

represented by the inner, top-right box

and are possible. So this statement

cannot be concluded.

B As the inner, top-right box is possible,

this statement cannot be true;

hydrofoils could leave from Nooli.

C This is not true – hydrofoils from

Nooli to places other than Dulais are

represented by the inner, bottom-right

box, which is empty.

D Steamboats from Waigura to Dulais

are represented by the outer, top-left

box, so this statement is possible;

but it cannot be concluded from the

data, as it could be that all the

ferries from Waigura to Dulais are

hydrofoils.

E This is true, since no hydrofoils from

Waigura go elsewhere.

4    The diagram shows the arrival times of

Anna and Bella.

Anna’s arrival time

8 a.m.9 a.m.10 a.m.11 a.m.12 a.m.1 p.m.2 p.m.3 p.m.4 p.m.

Be

lla

’s arrival ti

me

8 a.m.

9 a.m.

10 a.m.

11 a.m.

12 p.m.

1 p.m.

2 p.m.

3 p.m.

4 p.m.

The shaded area represents the times

when the two girls coincide. For example,

if Anna arrives at 12, she will meet Bella

if Bella arrives any time between 11.15

and 1.00; the area between these, and

such equivalent times, is shaded. The

probability required is the area of the

shaded portion divided by the whole area

of the graph. The large white triangles

have areas of:^772 × = 24.5 units (upper) and

725725

2

..× = 26.3 units (lower). The whole

graph has an area of 8 × 8 = 64 units. Thus

the shaded portion has an area of 64 –

24.5 – 26.3 = 13.2 units, so the chances of

them meeting are 13 2 64. = 0.206 or 20.6%.

This problem would be very difficult to

solve without a graphical method.

6.3 Probability, tree diagrams

and decision trees

1    This can be solved using a tree diagram

(see page 336).

The asterisked combinations give two

matching pairs.

There are 8 possibilities and the

probabilities of all but the last are the

same. The probabilities need to be

worked out with a calculator and are as

follows:

7 × 0.0699 + 0.0150 = 0.5043

(The 0.0699 is the result of the first 7

asterisked calculations:^814 ×^713 ×

(^612) × (^511) , and the .0150 is the 8th.)
Thus the chance of drawing two pairs is
approximately 50%.
2 The first two digits are 11 or 12. The
second two digits can be 11–19 or 21–29
(regardless of the first two digits) or 31
(but only if the first two digits are 12,
there being 31 days in December but not
in November). There are 37 possibilities,
so the chances of getting it right the
first time are^137. The chances of getting
it right the second time are^136 and the
third time^135. In order to calculate the
overall probability we need to add the