Question

Simulate the experiment described in Exercise 3.7 using any five identically shaped objects, two of which are one colour and the three another colour. Mix the objects, draw two, record the results, and then replace the objects. Repeat the experiment a large number of times (at least 100). Calculate the proportion of time events A, B, and C occur. How do these proportions compare with the probabilities you calculated in Exercise 3.7? Should these proportions equal the probabilities? Explain.

Short answer

$\mathbf{A}\mathbf{=}$Getting two red balls drawn

The probability of getting two red balls drawn according to the simulation run is

$\mathbf{B}\mathbf{=}$A red and a green ball is drawn.

$\mathbf{P}\left(\mathbf{B}\right)\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{39}$according to the simulation

$\mathbf{C}\mathbf{=}$Getting two green balls

$\mathbf{P}\left(\mathbf{C}\right)\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{16}$

Step-by-step solution

Identifying an identical object and taking a sample of  100

Assume there are 5 balls. 2 of them are red in colour and the others are green.

Red balls	1	2
Green balls	3	4	5

A = Two red balls are drawn

B = A red and a green ball is drawn

C = Two green balls are drawn

Event	1st ball	2nd ball	n(A)	n(B)	n (C)
1	1	5	1	1	0
2	4	5	2	1	2
3	2	2	0	0	0
4	4	1	1	0	1
5	3	2	0	0	1
6	5	1	1	0	1
7	4	2	1	0	1
8	2	2	0	0	0
9	3	3	0	0	2
10	4	3	1	0	2
11	3	2	0	0	1
12	2	4	1	1	1
13	2	3	0	0	1
14	1	2	0	0	0
15	2	4	1	1	1
16	5	5	2	1	2
17	4	1	1	0	1
18	4	5	2	1	2
19	4	5	2	1	2
20	5	5	2	1	2
21	2	1	0	0	0
22	4	1	1	0	1
23	3	4	1	1	2
24	2	1	0	0	0
25	4	2	1	0	1
26	2	4	1	1	1
27	5	3	1	0	2
28	2	4	1	1	1
29	5	4	2	1	2
30	4	3	1	0	2
31	2	3	0	0	1
32	4	1	1	0	1
33	5	5	2	1	2
34	4	3	1	0	2
35	4	1	1	0	1
36	4	3	1	0	2
37	3	5	1	1	2
38	3	3	0	0	2
39	5	4	2	1	2
40	3	5	1	1	2
41	5	2	1	0	1
42	5	1	1	0	1
43	3	2	0	0	1
44	2	2	0	0	0
45	4	4	2	1	2
46	5	4	2	1	2
47	1	5	1	1	1
48	2	5	1	1	1
49	2	2	0	0	0
50	1	2	0	0	0
51	5	4	2	1	2
52	4	4	2	1	2
53	4	1	1	0	1
54	2	2	0	0	0
55	3	2	0	0	1
56	5	5	2	1	2
57	1	1	0	0	0
58	5	1	1	0	1
59	2	1	0	0	0
60	5	4	2	1	2
61	2	1	0	0	0
62	4	4	2	1	2
63	4	1	1	0	1
64	5	2	1	0	1
65	3	2	0	0	1
66	5	1	1	0	1
67	1	1	0	0	0
68	4	4	2	1	2
69	5	1	1	0	1
70	4	3	1	0	2
71	1	4	1	1	1
72	1	5	1	1	1
73	4	4	2	1	2
74	5	4	2	1	2
75	1	3	0	0	1
76	4	2	1	0	1
77	1	1	0	0	0
78	4	5	2	1	2
79	4	4	2	1	2
80	5	1	1	0	1
81	1	3	0	0	1
82	1	3	0	0	1
83	2	3	0	0	1
84	2	5	1	1	1
85	5	5	2	1	2
86	4	3	1	0	2
87	4	1	1	0	1
88	1	2	0	0	0
89	5	5	2	1	2
90	3	4	1	1	2
91	5	3	1	0	2
92	3	1	0	0	1
93	2	1	0	0	0
94	4	3	1	0	2
95	4	2	1	0	1
96	3	2	0	0	1
97	3	2	0	0	1
98	4	1	1	0	1
99	4	5	2	1	2
100	3	2	0	0	1

$\mathbf{n}\left(\mathbf{A}\right)\mathbf{=}\mathbf{23}\mathbf{,}\mathbf{n}\left(\mathbf{B}\right)\mathbf{=}\mathbf{37}\mathbf{,}\mathbf{n}\left(\mathbf{C}\right)\mathbf{=}\mathbf{17}\mathbf{,}\mathbf{n}\left(\mathbf{S}\right)\mathbf{=}\mathbf{100}$

Therefore, $\mathbf{P}\left(\mathbf{A}\right)\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{23}\mathbf{,}\mathbf{P}\left(\mathbf{B}\right)\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{37}\mathbf{,}\mathbf{P}\left(\mathbf{C}\right)\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{17}$

Defining the simulation run experiment

The probabilities calculated in exercise 3.7 for events A, B and C were $\mathbf{P}\left(\mathbf{A}\right)\mathbf{=}\frac{\mathbf{1}}{\mathbf{8}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{125}\mathbf{,}\mathbf{P}\left(\mathbf{B}\right)\mathbf{=}\frac{\mathbf{3}}{\mathbf{8}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{375}$and $\mathbf{P}\left(\mathbf{C}\right)\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{5}$.

Doing the simulation run for drawing two balls from 5 balls sample space, one gets $\mathbf{P}\left(\mathbf{A}\right)\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{11}\mathbf{,}\mathbf{P}\left(\mathbf{B}\right)\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{39}$ and $\mathbf{P}\left(\mathbf{C}\right)\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{16}$ .

Here, one observes that probabilities calculated in exercise 3.7 and the probabilities calculated here are near to each other. That is because of the Law of Large Numbers.

Law of large number states that when an experiment is conducted repeatedly, the possibility of the number of times an event will occur approaches the theoretical probability of the outcome.

Repeating the simulation run experiment

Repeating the experiment, say 100 times gives the possibility for events A, B and C as close to its theoretical probability which is calculated in exercise 3.7.

In exercise 3.7, the probability of events A, B and C were $\mathbf{P}\left(\mathbf{A}\right)\mathbf{=}\frac{\mathbf{1}}{\mathbf{8}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{125}\mathbf{,}\mathbf{P}\left(\mathbf{B}\right)\mathbf{=}\frac{\mathbf{3}}{\mathbf{8}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{375}$ and $\mathbf{P}\left(\mathbf{C}\right)\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{5}$.

While after running the simulation 100 times we get the possibility of events A, B and C happening as, $\mathbf{P}\left(\mathbf{A}\right)\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{11}\mathbf{,}\mathbf{P}\left(\mathbf{B}\right)\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{39}$and $\mathbf{P}\left(\mathbf{C}\right)\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{16}$. If one simulates let’s say 1000 times, the possibility of these events happening will get closer to the theoretical probability.