The term probability reminds us of uncertainty or randomness. Probability is a measure of the likelihood of an event, yet the axiomatic explanation gives a formal, logical way of defining it. Rather than specifying experiments or observations, the axiomatic approach defines probability by means of a set of mathematical rules, called axioms, described by Andrey Kolmogorov.
These axioms lay the foundation for modern probability theory, ensuring that when people calculate probabilities, the principles applied will be consistent.
Axiomatic Approach
Perform a random experiment whose sample space is S, and P is the probability of the occurrence of any random event. This model assumes that P should be a real-valued function with a range between 0 and 1. The domain of this function is defined to be the power set of the sample space. If all these conditions are satisfied, then the function should satisfy the following axioms:
Axiom 1: For any given event X, the probability of that event must be greater than or equal to 0. Thus,
0 ≤ P(X)
Axiom 2: We know that the sample space S of the experiment is the set of all the outcomes. This means that the probability of getting some outcome from the sample space is 100 percent, i.e P(S) = 1. Intuitively this means that whenever the experiment is performed, the probability of obtaining one of the possible outcomes is certain.
Axiom 3: For the experiments where we have two outcomes A and B. If A and B are mutually exclusive,
P(A ∪ B) = P(A) + P(B) and P(A ∩ B) = 0
Here, ∪ stands for union, ∩ stands for the intersection of two sets. This can be understood as if saying, “If A and B are mutually exclusive outcomes, the probability that either one of these events will happen is the probability of A happening plus the probability of B happening”.
These axioms are also called Kolmogorov's three axioms. The third axiom can also be extended to several outcomes, given that all are mutually exclusive.
Let's say the experiment has A1, A2, A3, and ... An. All these events are mutually exclusive. In this case, the three axioms become:
Axiom 1: 0 ≤ P(Ai) ≤ 1 for all i = 1,2,3,... n.
Axiom 2: P(A1) + P(A2) + P(A3) +.... = 1
Axiom 3: P(A1 ∪ A2∪ A3 ....) = P(A1) + P(A2) + P(A3) ....
Applications Of Axiomatic Probability
- Axiomatic probability is the backbone of statistical inference. It allows us to define the normal, binomial, and Poisson distributions, among others, that model data and arrive at predictions in statistics, data science, and AI.
- The probability axioms are fundamental in the models where outcomes are not equally probable. That is, quantum mechanics, thermodynamics, and genetics depend on measurable probability laws.
- The axiomatically defined probability spaces give the foundation to entropy, randomized algorithms, and many error-detecting and error-correcting schemes, ensuring results that make sense.
- The axiomatic framework in finance, insurance, and project management quantifies uncertain outcomes. It offers a structured way of calculating the probability of losses or returns towards objectives, which enables data-driven decisions on risk.
Sample Problems on Axiomatic Approach to Probability
Question 1: Find out the probability of getting a number 3 when a die is tossed.
Answer:
We know that possible outcomes when a die is tossed are,
{1, 2, 3, 4, 5 and 6}
We want to calculate the probability for getting a number 3.
Number of favorable outcomes = 1
Total Number of outcomes = 6.So, the probability of getting a number 3, P(3) =
\frac{\text{Number of favourable Outcomes}}{\text{Total number of possible outcomes}} P(3) =
\frac{1}{6}
Question 2: Find out the probability of getting an even number when a die is tossed.
Answer:
We know that possible outcomes when a die is tossed are,
{1, 2, 3, 4, 5 and 6}
We want to calculate the probability for getting an even number. Even number are {2,4,6}
Number of favorable outcomes = 3
Total Number of outcomes = 6.
So, the probability of getting an even number, P(Even) =
\frac{\text{Number of favourable Outcomes}}{\text{Total number of possible outcomes}} P(Even) =
\frac{3}{6}
⇒ P(Even) =\frac{1}{2}
Question 3: Let's say we have an urn with 5 red balls and 3 black balls. We want to draw balls from this bag. Find out the probability of picking a red ball.
Answer:
Let's define the experiment as “Drawing a ball from the bag”. Now it is required to calculate the probability for getting a red ball.
Number of favorable outcomes = 5
Total Number of outcomes = 8.So, P(red) =
\frac{5}{8}
Question 4: Let's say a class is choosing its class captain through a random draw. The class has 30% Indian students, 50% American students, and 20% Chinese students. Calculate the probability that the chosen captain will be an Indian.
Answer:
Let's define an event A: Chosen captain is Indian. We know that there are only 30% Indian students in class.
Measure of favorable outcome = 0.3
Total Number of outcomes = 1
So, P(A) =
\frac{\text{Number of favourable Outcomes}}{\text{Total number of possible outcomes}}
P(A) =\frac{0.3}{1}
P(A) = 0.3So, there is a 30% probability that an Indian student will be chosen as class captain.
Question 5: Find out the sample space “S” for a random experiment involving the tossing of three coins.
Solution.
We know that tossing a coin gives us either Heads or Tails. Tossing three coins will give us either triplets of either heads or tails. So, the possible outcomes can be,
HHH, HHT, HTH, HTT, ....
All these outcomes will constitute the sample space.
S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
Question 6: For the above experiment, verify that the probability of getting a red ball and the probability of getting a black ball follow the axioms of probability mentioned above.
Answer:
Let's define two events,
R = Red ball is picked
B = Black Ball is pickedCalculate the probability for getting a red ball in the previous example,
P(R) =
\frac{5}{8} Similarly,
P(B) =
\frac{3}{8} Now notice that both P(R) and P(B) lie between 0 and 1. So they satisfy axiom 1. Let's verify it for second axiom.
P(R) + P(B)
⇒P(R) + P(B) =\frac{5}{8} +\frac{3}{8}
⇒ P(R) + P(B) = 1Thus second axiom is also satisfied.
We know that both of these events are mutually exclusive.
So, P(R ∪ B) = P(Getting either a Red Ball or Black Ball)
⇒ P(R ∪ B) = P(R) + P(B)
⇒ P(R ∪ B) =\frac{5}{8} +\frac{3}{8}
⇒ P(R ∪ B) = 1Thus, all three of these axioms are satisfied. Thus, above experiment follows the axioms of the probability.
Practice Problems on Axiomatic Approach to Probability
Question 1: In a bag of 10 balls, 4 are blue and 6 are green. What is the probability of drawing a green ball?
Question 2: A deck of cards contains 52 cards, 26 of which are red and 26 are black. What is the probability of drawing a red card?
Question 3: Alice has five toys, which are identical, and one of them is underweight. Her sister, Sesa, chooses one of these toys at random. Find the probability for Sesa to choose an underweight toy?
Question 4: A box contains 3 red and 4 blue socks. Find the probability of choosing two socks of the same colour.
Question 5: In a group of 50 students, 30 students like basketball, 20 like football, and 10 like both basketball and football. What is the probability that a randomly selected student likes either basketball or football (or both)?
Question 6: For two events A and B, find the probability that exactly one of the two events occurs.
Question 7: Aliya selects three cards at random from a pack of 52 cards. Find the probability of drawing:
- 3 spade cards.
- One spade and two knave cards
- one spade, one knave, and one heart card.
Question 8: For three events A, B, and C, show that
- P (at least two of A, B, C occur) = P(A∩B) + P(B∩C) + P(C∩A) − 2P(A∩B∩C)
- P (exactly two of A, B, C occur) = P(A∩B) + P(B∩C) + P(C∩A) − 3P(A∩B∩C)
- P (exactly one of A, B, C occurs) = P(A) + P(B) + P(C) − 2P(A∩B) − 2P(B∩C) − 2P(C∩A) + 3P(A∩B∩C)