Mathematics

surface. Alternately two persons A and B draw one sphere each path without
replacement. The winner is the person who drew the white sphere. Which
person can take advantage of it?”
After a first short discussion students have to vote on which person they want
to be. Then students play this game some times. (It is easy to simulate the
game.) They have to vote again and they have to give reasons for their choice
and first explanations. To decide on the problem the teacher asks the students
if they can accept this as a model for that game. A and B draw until the box is
empty. Each sequence of moves corresponds to an arrangement of these 6
spheres. There are 6 arrangements exactly, for instance bbwbbb, and each A
and B wins the game 3 times. Therefore they have the same chance. But,
what about the case if students do not trust in that model. Then we have to
decide by experiments and the calculation of relative frequency whether the
model describes reality right or wrong and we have alternatives to discuss the
conditions and assumptions for the model.
After discussing the concepts frequency and probability Students were
instructed to develop a program at home which simulates throwing dice and
calculating the relative frequency. Students accepted immediately that a
computer is able to choose numbers of an interval randomly, because “a
computer is able to do all you want”. That programme needs instruction IF-
THEN-ELSE. It is the best way leaving students alone with the problem. At
least one student is able to solve it. He or she shows how it works and helps
the others to enter the programme. Of course you can show how the random
variable “relative frequency of throwing number 6” converges to 1/6, but it is up
to the teacher to do that.
Consideration about drawing spheres without replacement
We consider a box with 8 white and 12 black spheres in it and we draw 5 times
without replacement. (It is easy to generalize.) Each sphere can be drawn
equally likely. X is the random variable which indicates the number of white
spheres which are drawn. Thus the values of X are W(X) = {0, 1, ..., 5}. I would
like to display some representations of the probability of drawing 3 white
spheres, that is P(X = 3).
These terms (1) to (4) are derived from the following models.
(1) We imagine all spheres are different (w1, w2, …, w8, b1, …, b12) , but
each of it can be drawn equally likely. Therefore we have 8 over 3 = 8!:3!:5!
possibilities to take 3 white spheres out of 8 at a time. (We have to notice that
this is a different situation to what we have assumed before.) We get 12!:2!:10!
combinations of 12 black spheres taken 2 at a time for each combination of
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