Contents
Probability
Probability is a measure of belief about an event that will occur in the future, based on historical data. In other words, the probability of occurrence of a specific target event among the total from past data can be defined as probability.
Probability has the following axiom:
- A value between 0 and 1 $$ 0 \,\le\, P(x) \,\le\, 1$$
- The sum of all probability is 1 $$\begin{align}&\sum_{x \in S} P(x) =1\\ & S=\text{all possible spaces} \end{align} $$
- A: Event(s) included in S $$A \subset S, P(X \in A) = \sum_{x \in A}P(X=x)$$
The concept of the probability that a particular event will occur in an experiment can be applied and interpreted in many ways. For example, for a forecast that there is a 70% chance of rain tomorrow, you can analyze the data of past climatic conditions and interpret it as a result of 70% rain under conditions similar to today's weather. Or it can be interpreted as a result of the forecaster's subjective thoughts. In the former case, the method of calculating the probability from the frequency of target events in existing data is called frequency interpretation, and in the latter case, it is called subjective interpretation as a result of subjective reasoning. As such, probability can be interpreted in two ways, and there are basic terms used in all cases.
- Random experiment (trial): A trial in which all events are assumed to have the same probability.
- Tossing a coin: The probability of getting heads and tails is the same
- Change in stock price: exact results are not predictable and each value has the same probability of generation
- Element or sample point: a possible basic outcome in a probability experiment.
- Sample space: the set of all sample points appearing in a probability experiment
- All sample points of the dice trial are 1, 2, 3, 4, 5, and 6, and the set of all elements becomes the sample space S.
S={1, 2, 3, 4, 5, 6}
- All sample points of the dice trial are 1, 2, 3, 4, 5, and 6, and the set of all elements becomes the sample space S.
- Event: A set of elements of interest in a probability experiment. It is a set of specific element(s) in the sample space, and the probability of an event is calculated as follows. $$\displaystyle \text{probability}=\frac{\text{event}}{\text{sample}}$$
- Mutually exclusive or independent : If the intersection of two events is the empty set ($A \cap B =\varnothing$ ), it is said to be mutually exclusive. This term is used synonymously with independent events.
The relationship of specific events in a sample space can be clearly expressed using sets.
- Universal set: represents the sample space.
- Empty set: A set with no elements ($\varnothing$)
- Union set: Each target set, i.e. the set of all elements in the event ($A\; \text{or} \; B= A \cup B$), and its probability is calculated as follows.
P(A ∪ B)=P(A)+P(B)-P(A ∩ B) - Intersection: A set of element(s) that belong to several sets in common($A \; \text{and} \; B = A \cap B$). If two events are independent, the probability is calculated as follows.
P(A ∩ B)=P(A) × P(B)
If the above expression does not hold, the two events are not independent and indicate the existence of an interaction that influences each other.
P(A ∪ B)=P(A)+P(B)-P(A ∩ B)
Example 1)
Find the probability of getting two or more heads in a probability experiment (trial) of tossing three coins.
import numpy as np import pandas as pd
#head:1, tail:0 x=[1, 0] Smpl=np.array([(i,j,k) for i in x for j in x for k in x]) Smpl
array([[1, 1, 1],
[1, 1, 0],
[1, 0, 1],
[1, 0, 0],
[0, 1, 1],
[0, 1, 0],
[0, 0, 1],
[0, 0, 0]])
n=Smpl.shape[0] #.shape:the shape of the array, (# of rows, # of columns) n
8
The implementation of the coin is independent. That is, one coin cannot influence the outcome of another. Therefore, the probability that each coin comes up heads or tails is the same as $\displaystyle \frac{1}{2}$. Since we try 3 coins, the probability of each element in the sample space above is:
$$\displaystyle \frac{1}{2}\cdot \frac{1}{2} \cdot \frac{1}{2}=\frac{1}{8}$$In the code, 1 for heads and 0 for tails are substituted, so if they are all heads, the sum is 3. The probability for the part where the sum of all trials is 3 is
H3=len(np.where(np.sum(Smpl, axis=1)==3)) H3
1
prob_H3=H3/len(Smpl) prob_H3
0.125
from sympy import * prob_H3=Rational(H3,n) prob_H3
$\displaystyle \frac{1}{8}$
In the code above, the np.where(condition) function returns the index that meets the condition in the object and uses the Rational() function of the sympy package to display the result in fraction form.
Events with two or more heads are:
moreH2=len(np.where(np.sum(Smpl, axis=1)>=2.0)[0]) moreH2
4
morepH2=Rational(moreH2, n) morepH2
$\displaystyle \frac{1}{2}$
Example 2)
A and B play a game, and A's average record is $\displaystyle \frac{2}{3}$. If two games are played under these conditions, what is the probability that A will win at least once?
The prior information that can be known in this problem is the respective probabilities of A and B.
$$P(A)=\frac{2}{3},\; P(B)=\frac{1}{3}$$In the code, A and B are replaced with 1 and 0, respectively. In two games,
#win of A: 1, lose of A:0 x=[1, 0] S=np.array([(i,j) for i in x for j in x]) S
array([[1, 1],
[1, 0],
[0, 1],
[0, 0]])
p1=Rational(2, 3) p0=Rational(1, 3) p1, p0
(2/3, 1/3)
Applying each of the probabilities given in the problem, the probabilities for each outcome of the code above are:
ps1=p1*p1 # win:2
ps2=p1*p0 # win:1
ps3=p1*p0 # win:1
ps4=p0*p0 # win:0
print(f'win0:{ps4}, win 1: {ps2+ps3}, win 2: {ps1}')
win0:1/9, win 1: 4/9, win 2: 4/9
The event that A wins at least once is ps1+ps2+ps3 in the code above.
morepA1=ps1+ps2+ps3 morepA1
$\displaystyle \frac{8}{9}$
Example 3)
Under the conditions of Example 2, what is the probability that A will win 2 out of 3 matches before B?
x=[Rational(2, 3), Rational(1, 3)] S=np.array([(i,j,k) for i in x for j in x for k in x]) S
array([[2/3, 2/3, 2/3],
[2/3, 2/3, 1/3],
[2/3, 1/3, 2/3],
[2/3, 1/3, 1/3],
[1/3, 2/3, 2/3],
[1/3, 2/3, 1/3],
[1/3, 1/3, 2/3],
[1/3, 1/3, 1/3]], dtype=object)
trg=np.zeros((1,3))
for i in S:
if i[0]==Rational(2,3) and i[1]==Rational(2,3):
trg=np.concatenate((trg,i.reshape(1,3)), axis=0)
trg=trg[1:, :]
trg
array([[2/3, 2/3, 2/3],
[2/3, 2/3, 1/3]], dtype=object)
np.sum(np.prod(trg, axis=1))
$\displaystyle \frac{4}{9}$
Permutation & Combination
Probability calculations are based on the sample space, but in many cases calculating the sample space can be very cumbersome. The concept of permutations and combinations greatly reduces this hassle.
For example, if two dice are rolled, the number of possible outcomes is as follows:
$$\begin{align} 1 \rightarrow & \begin{bmatrix}1\\\vdots\\6\end{bmatrix} \Rightarrow \begin{bmatrix}(1,1)\\\vdots\\(1,6)\end{bmatrix}\\ 2 \rightarrow & \begin{bmatrix}1\\\vdots\\6 \end{bmatrix}\Rightarrow \begin{bmatrix}(2,1)\\\vdots\\(2,6)\end{bmatrix}\\ & \qquad \vdots\\ 6 \rightarrow & \begin{bmatrix}1\\\vdots\\6 \end{bmatrix}\Rightarrow \begin{bmatrix}(6,1)\\\vdots\\(6,6)\end{bmatrix}\end{align} $$S=np.array([[i, j] for i in range(1, 7) for j in range(1, 7)]) S[:3,:]
Probabilityarray([[1, 1],
[1, 2],
[1, 3]])
Sn=len(S) Sn
36
In an experiment in which two dice are thrown, the trials of each dice do not affect each other. These trials are called independent trials or mutually exclusive trials. The number of events generated when one die is rolled is six. Therefore, for two dice, the number of all possible cases is 6‧6=36. As such, in the case of independent events, the number of cases can be calculated using the multiplication law.
The number of possible cases in the trials of two independent events, m and n, is m ‧ n.
The number of cases in which 4 numbers are selected from the numbers 0~9 can be thought of as follows. Since there are 10 possible cases for each place, the number of all possible cases is:
However, if the number in each digit is different, the number in each digit is 10 in the first digit, 9 in the next digit, 8 in the third digit, and 7 in the last digit. Therefore, the number of cases that can be in 4 places is calculated as follows.
$$\frac{10 \cdot 9 \cdots \cdot 1}{6 \cdot 5 \cdots \cdot 1}=10 \cdot 9 \cdot 8 \cdot 7$$
In the left term of the above expression, the numerator can be expressed as 10! (factorial), and the denominator can be expressed as 5!. Therefore, it can be expressed as
$$\frac{10!}{(10-4)!}=10 \cdot 9 \cdot 8 \cdot 7$$The process of calculating the number of cases without duplicates considering the order as above is called permutation and is defined as Equation 1.
It is the number of cases in which k different items are selected out of n without duplicates in consideration of the order, expressed as nPk. $$\begin{equation}\tag{1} _nP_k=\frac{n!}{(n-k)!} \end{equation}$$
Factorial uses the factorial() function of the numpy.math class. You can also compute permutations directly with the perm() function in the scipy.special class.
from numpy import math (math.factorial(10))/(math.factorial(10-4))
5040.0
from scipy.special import perm perm(10, 4)
5040.0
The above permutation takes into account the order of the elements listed, so it includes both 1234 and 2134. However, if the order is not taken into account, the two cases are equivalent. That is, if the selected element contains 1,2,3,4, they are all considered equal events, so this condition must be taken into account. The result is obtained by dividing the result of the above permutation by the number of cases in which the selected numbers are arranged.
from numpy import math (math.factorial(10))/(math.factorial(10-4))/math.factorial(4)
210.0
This process can be generalized as in Equation 2 and is called combination.
A case of selecting k different items from among n items without considering the order, expressed as nCk. $$\begin{equation}\tag{2} _nC_k=\binom{n}{k} =\frac{n!}{k!(n-k)!} \end{equation}$$
Combinations can use the scipy.spectial module function comb().
from scipy import special special.comb(10, 4)
210.0
Equation 2 can be applied to determine the result of the expansion of a binomial such as (a+b)2=a2+2ab+b2. The coefficients on the right-hand side of this expression are equivalent to $\displaystyle \binom{2}{0},\; \binom{2}{1}$, and $\displaystyle \binom{2}{2}$. This can be generalized as Equation 3.
$$\begin{equation}\tag{3} (x+y)^n=\sum_{k=0}^n \binom{n}{k}x^{n-k}y^k \end{equation}$$For example, using Equation 3, the coefficient of x2y in (x+y)3 can be calculated as
from scipy import special special.comb(3, 2)
3.0
Example 4)
The number of cases where 20 students can go to a retreat (MT) and can allocate 4 rooms with a capacity of 6, 4, 5 or 5 can be calculated as follows:
from scipy.special import comb
x=[(20, 6), (14, 4), (10, 5), (5,5)]
p=1
for i in x:
p=p*comb(i[0], i[1])
p
9777287520.0
The above example is equivalent to the case where the total number of elements n consists of elements of n1 n2, … and nr respectively. This case can be calculated as Equation 4.
$$\begin{equation}\tag{4} \binom{n}{n_1\ n_2\, \cdots \, n_r}=\frac{n!}{n_1!n_2! \cdots n_r!} \end{equation}$$By applying Equation 4, the expanded form of the polynomial can be determined by Equation 5.
$$\begin{equation}\tag{5} (x_1+x_2+ \cdots +x_k)^n = \sum \binom{n}{n_1\ n_2\, \cdots \, n_r}x^{n_1}_1x^{n_2}_2 \cdots x^{n_k}_k \end{equation}$$Example 5)
A probability experiment in which a dice is tossed can determine:
- What is the probability that 1 will come out?
In the sample space S={1,2,3,4,5,6}, event 1 is one of six, so the probability of its occurrence is expressed as follows.
$$p(X=1)=\frac{1}{6}$$Probability is the rate at which a particular event occurs out of all events, so if you roll a dice 6 times, each point must come up once. However, it is intuitively clear that this is highly unlikely. But what happens if you repeat this trial multiple times?
Simulation using Python code shows the following trends.
np.random.seed(2)
p={}
for i in range(1001):
x=np.random.randint(1, 7, size=i)
if 1 in x:
p[i]=round(len(np.where(x==1)[0])/i, 4)
else:
p[i]=0
p[0], p[1000]
(0, 0.159)
import matplotlib.pyplot as plt
plt.figure(figsize=(7, 4))
plt.plot(p.values(), label='P(1)')
plt.plot(np.repeat(1/6, 1000), '--', color='red', label=r'$\frac{1}{6}$')
plt.xlabel("# of trial", size=12, weight="bold")
plt.ylabel("Probability", size=12, weight="bold")
plt.text(0, -0.3, "Figure 1. The probability of occurrence of 1", size=15, weight="bold")
plt.text(0, -0.4, "when 1000 dice are executed.", size=15, weight="bold")
plt.legend(loc='best')
plt.show()
As shown in Figure 1, it can be seen that the probability converges to $\displaystyle \frac{1}{6} \approx 0.16667$ as the number of trials increases. This is called Law of Large Numbers. In general, probability refers to the average proportion observed over multiple trials. That is, the probability is equal to Equation 6.
$$\begin{equation} \tag{6} \text{probability}=\frac{\text{frequency of specific event(s)}}{\text{frequency of all events}} \end{equation}$$This probability converges to a constant value as the number of trials increases as shown in Figure 1.
In the case of certain trials as above, the case where no conditions are imposed on the results of each trial and all have the same probability is called **random process**.
- What is the probability of 1 or 2 in the above trials?
From the sample space S={1, 2, 3, 4, 5, 6}, the target set of events is E={1, 2} , so the probability of that event is the sum of the probabilities for each element. Since each event is independent, the intersection of the two events is ∅, so the union of the set of events E is computed as
$$\begin{aligned} &\text{p(X=1)}=\frac{1}{6}, \quad \text{p(X=2)}=\frac{1}{6}\\ &\text{p(X=1 or X=2)}=\frac{1}{6}+\frac{1}{6}=\frac{1}{3} \end{aligned}$$np.random.seed(3)
p={}
for i in [10, 100, 1000, 2000, 3000]:
x=np.random.randint(1, 7, size=i)
re=np.around((len(np.where(x==1)[0])+len(np.where(x==2)[0]))/i, 4)
print(f'number of trials :{i}, probability:{re}')
number of trials :10, probability:0.5
number of trials :100, probability:0.41
number of trials :1000, probability:0.323
number of trials :2000, probability:0.3395
number of trials :3000, probability:0.3323
- Conversely, consider the case where the probability of the trial is 1.
It means all possible cases when the dice are executed. That is, the set of events with a probability of 100\% is E=\{1,2, 3, 4, 5, 6\} equal to the sample space.
- What is the probability of getting one of all numbers except 2 in a trial of a die?
Excludes the probability of getting a 2 from the whole
$$1-\frac{1}{6}=\frac{5}{6}$$- What is the probability that two dice are rolled and the points on both dice are 3?
Since the trial of each dice is an independent event, the multiplication rule can be applied to the number of cases in which two dice are executed.
$$6 \cdot 6 =36$$In this trial, there is only one case where the dice is (3, 3), and since it is an independent event, it is calculated using the multiplication rule as follows:
$$\frac{1}{6} \cdot \frac{1}{6} =\frac{1}{36} $$
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