Definition and Properties of Probability

Definition of Probability

Intuitive Definition of Probability

Probability (Intuitive Definition)

Probability is a numerical value describing how likely a random event is to occur, and it ranges over [0,1][0, 1].

Statistical Definition of Probability

Probability (Statistical Definition)

Let AA be an event of a random experiment. If in nn repeated trials event AA occurs mm times, then the ratio mn\frac{m}{n} is called the frequency of the occurrence of event AA in nn trials. When nn is sufficiently large, the frequency mn\frac{m}{n} stabilizes near some constant pp, then pp is called the probability of event AA, denoted by P(A)=pP(A) = p.

Classical Probability Model

Definition of the Classical Probability Model

Classical Probability Model

If a random experiment satisfies the following conditions:

  1. The sample space Ω\Omega is a finite set
  2. Every sample point is equally likely to occur

then the random experiment is called a classical probability model.

Probability Formula for the Classical Probability Model

Probability Formula for the Classical Probability Model

In the classical probability model, the probability of event AA is: P(A)=∣A∣∣Ω∣=A sample pointstotal number of sample points in the sample spaceP(A) = \frac{|A|}{|\Omega|} = \frac{A \text{ sample points}}{\text{total number of sample points in the sample space}}

Examples of the Classical Probability Model

Example 1: tossing a fair coin

  • Sample space: Ω={heads,tails}\Omega = \{\text{heads}, \text{tails}\}
  • P({heads})=12P(\{\text{heads}\}) = \frac{1}{2}
  • P({tails})=12P(\{\text{tails}\}) = \frac{1}{2}

Example 2: rolling a fair die

  • Sample space: Ω={1,2,3,4,5,6}\Omega = \{1, 2, 3, 4, 5, 6\}
  • P({an even number appears})=36=12P(\{\text{an even number appears}\}) = \frac{3}{6} = \frac{1}{2}
  • P({a number greater than 3 appears})=36=12P(\{\text{a number greater than 3 appears}\}) = \frac{3}{6} = \frac{1}{2}

Example 3: randomly drawing rr balls from nn balls

  • Sample space: all possible combinations of rr balls
  • ∣Ω∣=Cnr=n!r!(n−r)!|\Omega| = C_n^r = \frac{n!}{r!(n-r)!}
  • P(A)=∣A∣CnrP(A) = \frac{|A|}{C_n^r}

Geometric Probability Model

Definition of the Geometric Probability Model

Geometric Probability Model

If the sample space Ω\Omega of a random experiment is a geometric region and every sample point is equally likely to occur, then the random experiment is called a geometric probability model.

Probability Formula for the Geometric Probability Model

Probability Formula for the Geometric Probability Model

In the geometric probability model, the probability of event AA is: P(A)=m(A)m(Ω)P(A) = \frac{m(A)}{m(\Omega)}

where m(A)m(A) and m(Ω)m(\Omega) denote the geometric measures (length, area, volume, etc.) of event AA and the sample space Ω\Omega, respectively.

Examples of the Geometric Probability Model

Example 4: choosing a point at random on the interval [0,1][0, 1]

  • Sample space: Ω=[0,1]\Omega = [0, 1]
  • P(landing in [a,b])=b−a1=b−aP(\text{landing in } [a, b]) = \frac{b-a}{1} = b-a

Example 5: choosing a point at random inside the unit circle

  • Sample space: all points inside the unit circle
  • P(a point whose distance from the center is less than r is obtained)=πr2π=r2P(\text{a point whose distance from the center is less than } r \text{ is obtained}) = \frac{\pi r^2}{\pi} = r^2

Example 6: choosing a point at random inside the unit square

  • Sample space: all points inside the unit square
  • P(landing in the first quadrant)=14P(\text{landing in the first quadrant}) = \frac{1}{4}

Basic Properties of Probability

Non-negativity

Property 1: for any event AA, 0≤P(A)≤10 \leq P(A) \leq 1.

Normalization

Property 2: P(Ω)=1P(\Omega) = 1, P(∅)=0P(\varnothing) = 0.

Finite Additivity

Property 3: if the events A1,A2,…,AnA_1, A_2, \dots, A_n are pairwise mutually exclusive, then: P(A1∪A2∪⋯∪An)=P(A1)+P(A2)+⋯+P(An)P(A_1 \cup A_2 \cup \dots \cup A_n) = P(A_1) + P(A_2) + \dots + P(A_n)

Probability of the Complement Event

Property 4: P(A‾)=1−P(A)P(\overline{A}) = 1 - P(A).

Monotonicity

Property 5: if A⊂BA \subset B, then P(A)≤P(B)P(A) \leq P(B).

Addition Formula for Probability

General Addition Formula

Addition Formula for Probability

For any two events AA and BB: P(A∪B)=P(A)+P(B)−P(A∩B)P(A \cup B) = P(A) + P(B) - P(A \cap B)

Generalization of the Addition Formula

Generalization: for any nn events A1,A2,…,AnA_1, A_2, \dots, A_n: P(A1∪A2∪⋯∪An)=∑i=1nP(Ai)−∑1≤i<j≤nP(Ai∩Aj)+∑1≤i<j<k≤nP(Ai∩Aj∩Ak)−⋯+(−1)n−1P(A1∩A2∩⋯∩An)P(A_1 \cup A_2 \cup \dots \cup A_n) = \sum_{i=1}^n P(A_i) - \sum_{1 \leq i < j \leq n} P(A_i \cap A_j) + \sum_{1 \leq i < j < k \leq n} P(A_i \cap A_j \cap A_k) - \dots + (-1)^{n-1} P(A_1 \cap A_2 \cap \dots \cap A_n)

Proof of the Addition Formula

Proof:

  1. A∪B=A∪(B−A)A \cup B = A \cup (B - A)
  2. AA and B−AB - A are mutually exclusive
  3. Hence P(A∪B)=P(A)+P(B−A)P(A \cup B) = P(A) + P(B - A)
  4. Also B=(B−A)∪(A∩B)B = (B - A) \cup (A \cap B)
  5. Hence P(B)=P(B−A)+P(A∩B)P(B) = P(B - A) + P(A \cap B)
  6. Therefore P(B−A)=P(B)−P(A∩B)P(B - A) = P(B) - P(A \cap B)
  7. Substituting gives P(A∪B)=P(A)+P(B)−P(A∩B)P(A \cup B) = P(A) + P(B) - P(A \cap B)

Examples

Example 7: tossing a coin twice

Problem: toss a fair coin twice and find the probability of the event that exactly one head appears.

Solution:

  1. Sample space: Ω={(H,H),(H,T),(T,H),(T,T)}\Omega = \{(\text{H},\text{H}), (\text{H},\text{T}), (\text{T},\text{H}), (\text{T},\text{T})\}
  2. Event A={exactly one head appears}={(H,T),(T,H)}A = \{\text{exactly one head appears}\} = \{(\text{H},\text{T}), (\text{T},\text{H})\}
  3. P(A)=24=12P(A) = \frac{2}{4} = \frac{1}{2}

Example 8: rolling a die

Problem: roll a fair die and find the probability of getting an even number or a number greater than 3.

Solution:

  1. Let A={an even number appears}={2,4,6}A = \{\text{an even number appears}\} = \{2, 4, 6\}
  2. Let B={a number greater than 3 appears}={4,5,6}B = \{\text{a number greater than 3 appears}\} = \{4, 5, 6\}
  3. A∩B={4,6}A \cap B = \{4, 6\}
  4. P(A)=36=12P(A) = \frac{3}{6} = \frac{1}{2}
  5. P(B)=36=12P(B) = \frac{3}{6} = \frac{1}{2}
  6. P(A∩B)=26=13P(A \cap B) = \frac{2}{6} = \frac{1}{3}
  7. P(A∪B)=P(A)+P(B)−P(A∩B)=12+12−13=23P(A \cup B) = P(A) + P(B) - P(A \cap B) = \frac{1}{2} + \frac{1}{2} - \frac{1}{3} = \frac{2}{3}

Exercises

Exercise 1

Toss a fair coin twice and find the probability of the event that exactly one head appears.

classical probability modelprobability calculationrandom experiment

Approach: Use the classical probability model to compute the probability.

Steps:

  1. Sample space: Ω={(H,H),(H,T),(T,H),(T,T)}\Omega = \{(\text{H},\text{H}), (\text{H},\text{T}), (\text{T},\text{H}), (\text{T},\text{T})\}
  2. Event A={exactly one head appears}={(H,T),(T,H)}A = \{\text{exactly one head appears}\} = \{(\text{H},\text{T}), (\text{T},\text{H})\}
  3. P(A)=24=12P(A) = \frac{2}{4} = \frac{1}{2}

Answer: 12\frac{1}{2}

Exercise 2

Given P(A)=0.4,P(B)=0.5,P(A∩B)=0.2P(A) = 0.4, P(B) = 0.5, P(A \cap B) = 0.2, find P(A∪B)P(A \cup B).

probability addition formulaunion of eventsprobability calculation

Approach: Compute with the addition formula.

Steps:

  1. P(A∪B)=P(A)+P(B)−P(A∩B)P(A \cup B) = P(A) + P(B) - P(A \cap B)
  2. =0.4+0.5−0.2=0.7= 0.4 + 0.5 - 0.2 = 0.7
  3. We subtract P(A∩B)P(A \cap B) because the intersection was counted twice in the direct addition.

Answer: P(A∪B)=0.7P(A \cup B) = 0.7.

Exercise 3

Roll a fair die and find the probability of getting an even number or a number greater than 3.

probability addition formulaunion of eventsprobability calculation

Approach: Compute with the addition formula.

Steps:

  1. Let A={an even number appears}={2,4,6}A = \{\text{an even number appears}\} = \{2, 4, 6\}
  2. Let B={a number greater than 3 appears}={4,5,6}B = \{\text{a number greater than 3 appears}\} = \{4, 5, 6\}
  3. A∩B={4,6}A \cap B = \{4, 6\}
  4. P(A)=36=12P(A) = \frac{3}{6} = \frac{1}{2}
  5. P(B)=36=12P(B) = \frac{3}{6} = \frac{1}{2}
  6. P(A∩B)=26=13P(A \cap B) = \frac{2}{6} = \frac{1}{3}
  7. P(A∪B)=P(A)+P(B)−P(A∩B)=12+12−13=23P(A \cup B) = P(A) + P(B) - P(A \cap B) = \frac{1}{2} + \frac{1}{2} - \frac{1}{3} = \frac{2}{3}

Answer: 23\frac{2}{3}

Exercise 4

Choose a point at random on the interval [0,1][0, 1] and find the probability of landing in [0.3,0.7][0.3, 0.7].

geometric probability modelcontinuous random variableprobability calculation

Approach: Use the geometric probability model to compute the probability.

Steps:

  1. Sample space: Ω=[0,1]\Omega = [0, 1]
  2. Event A=[0.3,0.7]A = [0.3, 0.7]
  3. m(Ω)=1m(\Omega) = 1
  4. m(A)=0.7−0.3=0.4m(A) = 0.7 - 0.3 = 0.4
  5. P(A)=0.41=0.4P(A) = \frac{0.4}{1} = 0.4

Answer: 0.40.4

Exercise 5

Prove that P(A‾)=1−P(A)P(\overline{A}) = 1 - P(A).

complement eventproperties of probabilityproof

Approach: Use the definition of the complement event and the properties of probability.

Steps:

  1. A∪A‾=ΩA \cup \overline{A} = \Omega
  2. A∩A‾=∅A \cap \overline{A} = \varnothing
  3. Hence P(A∪A‾)=P(A)+P(A‾)P(A \cup \overline{A}) = P(A) + P(\overline{A})
  4. Also P(A∪A‾)=P(Ω)=1P(A \cup \overline{A}) = P(\Omega) = 1
  5. Therefore P(A)+P(A‾)=1P(A) + P(\overline{A}) = 1
  6. So P(A‾)=1−P(A)P(\overline{A}) = 1 - P(A)

Answer: the proof is complete

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