1. Concepts & Definitions
1.1. Experiment, observation, and sample space
1.2. Sample space: Venn and Tree diagram
1.3. Simple and composite events
1.4. Three definitions of probability
1.5. Law of large numbers and its consequences
1.6. Frequency and empirical probability
2. Problem & Solution
2.4. Frequency of categories from tables
2.5. Simple and marginal probabilities
2.6. Conditional probabilities
The intersection of events A and B, denoted A ∩ B, is the collection of all outcomes that are elements of both of the sets A and B. It corresponds to combining descriptions of the two events using the word “and.”. To compute the probability of the intersection of events the next equation could be employed.
P(A∩B) = P(A) x P(B|A)
This equation could be interpreted as: “The probability of an intersection between event A and B is equal to the probability of happening event A and then, given that event A happened, the conditional probability to happen event B.”)
Imagine that you developed a system that tracks products and the routes of certain materials for a chosen port employing maps as described in Track 02 - section 2.7 like the one shown in the next figure.
Now, suppose an experiment where a certain port receives certain materials from three routes (R1, R2, and R3), and importation documentation could be in perfect order (P) or defective (D).
The detailed data about the problem is: three raw materials from routes R1, R2, and R3 have the proportions of 50% (P(R1)), 30% (P(R2)), and 20% (P(R3)), respectively.
It is known that an average:
• 5% of raw material documentation from route R1 is defective (P(D|R1)).
• 2% of raw material documentation from route R2 is defective (P(D|R2)).
• 1% of raw material documentation from route R3 is defective (P(D|R3)).
If a raw material is drawn at random from X’s importation area:
• What is the probability that it is defective (P(D))?
• What is the conditional probability that the raw material is from a route Ri, i = 1, 2, 3, given that it is defective (P(Ri|D))?
To solve this could be understood as a two-stage experiment where:
The first stage considered the event of randomly choosing a product that comes from a specific route;
The second stage consists of carrying out the decision from the first stage and considering the event of randomly choosing a product that is perfect or not.
The next tree figure helps to consolidate in a structured manner the problem data and the calculations that are necessary to compute the joint probability of finding a defective product in this port considering the three routes.
Since the process of selecting a product from a given route is disjunctive, the three joint probabilities for each route could be summed up using the following equation:
P(D) = P(R1 ∩ D) ∪ P(R2 ∩ D) ∪ P(R3 ∩ D) = P(R1)*P(D|R1) + P(R2)*P(D|R2) + P(R3)*P(D|R3), from the problem data consolidated in the tree figure, = 2.5% + 0.6% + 0.2% = 3.3%
The conditional probability that a product has a defective process coming from a country Ri, i = 1, 2, 3 leads to the computation of P(Ri|D), i = 1, 2, 3. The Eq. (9.10) could be modified as:
P(Ri|D) = P(Ri∩D) /P(D)
Given the information from the first item:
P(R1|D) = P(R1∩D) /P(D) = 2.5%/3.3% = 76%
P(R2|D) = P(R2∩D) /P(D) = 0.6%/3.3% = 18%
P(R3|D) = P(R3∩D) /P(D) = 0.2%/3.3% = 6%
A generalization of the probability calculations for the first item of the previous numerical example is given by the Total Probability Theorem whose definition is:
If events A1, A2, ..., Am form a complete partition that covers the sample space Ω and B is any event of this space, then the following equation holds:
P(B) = P(A1) × P(B|A1) + · · · + P(Am) × P(B|Am)
In the raw material experiment the values are: m = 3, Ai = Ri (Regions), and B = D (Defective).
This Theorem is important since using the same conditions enables the definition of Bayes Theorem (Bayes rule).
Given the same conditions from the Total Probability Theorem, then the following equation holds.
P(Ai|B) = P(Ai) × P(B|Ai)/P(B) , ∀i = 1, · · · , m.
The Bayes Theorem has the following properties:
• The fact that we know that event B has occurred may alter the expectations about the occurrence of Ai’s.
• It is possible to change the order of the conditioning and their associated probabilities.
The second item of numerical example better illustrates how the Bayes Theorem works. In particular, it employed that Ai = Ri (Regions), and B = D (Defective).