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For this solution we have to thank Andrei from Tudor Vianu
National College, Bucharest, Romania.
(a) Evaluating $(x + x^2 + x^3 + x^4)^2$, I obtained $$x^2
+2x^3+3x^4+4x^5+3x^6+2x^7+x^8$$
(b) The sum of two spinners labelled 1, 2, 3 and 4 varies from
2 to 8. There are 16 possibilities in total, giving the following
probabilities: \begin{eqnarray} &\text{Scores} &2 &3
&4 &5 &6 &7 &8 \\ &\text{Frequencies}
&1 &2 &3 &4 &3 &2 &1 \\
&\text{Probabilities} &0.0625 &0.125 &0.1875
&0.25 &0.1875 &0.125 &0.0625 \end{eqnarray}
(c) I observe that the frequency distribution is the same as
the coefficients from the expansion of the polynomial from
(a).
(d) Using the computer simulation, I took two spinners
labelled from 1 to 4, obtaining the following table of
frequencies:
| Exp.$\quad$ |
Relative frequency$\quad$ |
| 0 |
0 |
| 1 |
0 |
| 2 |
0.0631 |
| 3 |
0.1253 |
| 4 |
0.1872 |
| 5 |
0.2493 |
| 6 |
0.188 |
| 7 |
0.1248 |
| 8 |
0.062
|
The results obtained are very close to the theoretical
frequency distribution of the scores.
(e) $(x + x^2 + x^3 + x^4)^2$ could be factorised as follows:
$x^2 (1+x)^2 (1+x^2)^2$ and these factors can be re-written
as:
1) $(1+x)(x+x^2) (1+x^2)(x+x^3)$
2) $(x+x^2)^2 (1+x^2)^2$
3) $(1+x)^2(x+x^3)^2$
The first case corresponds to 4 spinners (0,1), (1,2),
(0,2),(1,3), the second to 4 spinners: (1,2), (1,2), (0,2), (0,2)
and the third to another 4 spinners (0,1), (0,1), (1,3), (1,3).\par
(f) Using the simulation, I obtained similar frequency
distributions as with two 1,2,3,4 spinners.
| Score$\quad$ |
Frequency$\quad$ |
distributions$\quad$ |
|
|
(1) |
(2) |
(3) |
| 2 |
0.3324 |
0.3312 |
0.3359 |
| 3 |
0.4996 |
0.5021 |
0.4982 |
| 4 |
0.1679 |
0.1665 |
0.1658
|
The theoretical probabilities would be $2/6 = 33.33 \%$, for
2, $3/6 = 50\%$ for 3, and $1/6 = 16.66\%$, which are very close to
the simulated values.
For a 2-spinner, there are equal probabilities of obtaining
the two numbers with which it is labelled. It corresponds to the
polynomial $x^m + x^n$. If $m$ and $n$ are equal, then the
probability ofobtaining that value is 1 ($100\%$).\par For a
3-spinner, I have 3 cases:
- All three numbers are different. There is a probability of 1/3
($33.33\%$) of obtaining any of the three numbers. It corresponds
to the polynomial $x^m + x^n + x^p$, with $m \neq n \neq p$
- Two of the three numbers are equal. There is a probability of
1/3 ($33.33\%$) of obtaining the unique number, and a probability
of $66.67\%$ of obtaining the number which appears twice. It
corresponds to the polynomial $x^m + 2x^n$, $m \neq n$.
- All 3 numbers are equal. The probability of obtaining this
number is $100\%$.
Two ordinary dice are equivalent to 2 6-spinners. The
theoretical probabilities and the simulated ones are:
| Number |
2 |
3 |
4 |
5 |
6 |
7 |
8 |
9 |
10 |
11 |
12 |
| Theoretical |
1/36 |
2/36 |
3/36 |
4/36 |
5/36 |
6/36 |
5/36 |
4/36 |
3/36 |
2/36 |
1/36 |
| Frequency |
0.0281 |
0.0556 |
0.083 |
0.1123 |
0.1397 |
0.1674 |
0.1381 |
0.1093
|
0.0842
|
0.0539 |
0.028 |
This case corresponds to the polynomial $(x + x^2 + x^3 + x^4
+ x^5 + x^6)^2$. The expansion of this polynomial is: $$x^2 + 2x^3
+ 3x^4 + 4x^5 + 5x^6 + 6x^7 + 5x^8 + 4x^9 + 3x^{10} + 2x^{11} +
x^{12}.$$ This polynomial could be factorised as follows: $$(x +
x^2 + x^3 + x^4 + x^5 + x^6)^2 = x^2 (1+x)^2 (1+x^2+x^4)^2$$ and
from this decomposition different combinations could be obtained,
all of them producing the same frequency distribution.