Probability

Total and Compound Probability and Mathematical Expectation

For real-world problems simultaneously occurring multiple events make it very difficult to find out probabilities. Total and compound probability and mathematical expectation are the concepts used while dealing with such situations involving joint events. In fact, in real-life we hardly come across simple events.

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Total Probability

Total probability or the law of total probability is a theorem which helps to calculate the total probability of an event. We calculate the total probability by taking into account several other distinct events that are disjoint from each other but are related to the event under consideration.

Compound Probability

In the figure above we notice:

  • B as the event under consideration.
  • A1 and A2 as the mutually exclusive events which are related to our event of interest i. e., B.

To solve problems like this we can use the knowledge of compound probability, conditional probability and subsequently the Bayes’ theorem.

Compound Probability

Compound probability which is also the joint probability is the probability of two events of an experiment occurring simultaneously. We can calculate the compound probability can be calculated for both independent as well as the dependent variable.

  • In the case of independent variables, we calculate the probability by multiplying the individual probabilities of the events.
  • For dependent variables, we use the concept of conditional probability specifically Bayes’ theorem.

Bayes’ Theorem

We derive Bayes’ theorem from the relation of conditional probability.

  • The conditional probability of A such that B has already occurred is

P(A/B) = P(A∩B)/P(A);  given P(A)>0.     —–(1)

  • Modifying the above reaction we get the probability

P(A∩B)=P(A/B)×P(A);   given P(A)>0.     ——(2)

Also, P(B/A)=P(B∩A)/P(B);  given P(B)>0.    ——(3)

⇒P(B∩A)=P(B/A)×P(B);  given P(B)>0.    ——(4)

Since P(A∩B)=P(B∩A)

from (2) and (3) we have,

P(B/A)×P(B)=P(A/B)×P(A)

⇒P(B/A)=(P(A/B)×P(A))/P(B);  given P(A),P(B)>0.    ——(Bayes’ Theorem)

Mathematical Expectation (μ)

It is the summation of all the values of a particular random variable multiplied with the corresponding values of probability. It is given as:

µ = E(x) = Σ xipi

Here, xi and pi are the values of random variables x and their corresponding probabilities p for an ith outcome.

Illustration: From a bag containing two balls coloured red and blue. Two balls are drawn one after another at one time. The possible outcomes are {RR, RB, BR, BB}. Table for the probability of occurrence of red is

xi 0 1 2
pi 1/4 2/4 =1/2 1/4

Here, xi is the random variable for denoting the number of times the red ball can occur in all the possible events and pi is its corresponding probability. A random variable can be either continuous or discrete and can be either finite or infinite.

Solved Examples on Total and Compound Probability

Question. Consider the following information about the population of a particular town about the types of regular check-ups they get for free: 40% diabetes check-up, 30% blood pressure check-up, 25% eye check-up, 23% both check-up diabetes and blood pressure, and 51% neither diabetes nor blood pressure nor eye check-up. In addition, 88/100 who get blood pressure will also get diabetes check-up, and 70/100 who get blood pressure will also get the eye check-up.

What is the probability that someone who has got the eye check-up also gets the blood pressure check-up?

Solution: Using Bayes theorem the solution becomes easy!

Given,

  • The probability of free diabetes check-up: P(D) = 40%
  • Free blood pressure check-up probability: P(B) = 30%
  • The probability of  free eye check-up: P(E) = 25%
  • The probability of  both free diabetes and blood pressure check-up P(D∩B) = 23%
  • Free check-up probability neither for Diabetes nor for blood pressure check-up and nor for the eye check up P(D∪B∪E)’ = 51%
  • The probability that the ones who got a free blood pressure check-up will also get the free diabetes check-up P(D/B) = 88%
  • The probability that the ones who got the free blood pressure check-up will also get free eye check-up P(E/B) = 70%

We need to find the probability that someone who has got the eye check-up will also get the blood pressure check-up. i.e., P(B/E)

By Bayes’ theorem, we can write

P(B/E) = P(E/B) . P(E))/P(B)

= (0.7/0.25)/0.3

= 73/25

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Kumar
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Kumar

There are so many errors in two of the lectures that I have watched. The flow of the lectures are also inappropriate. Firstly you never defined what an event is. For this lecture you can just say that an event is a subset of sample space. Therefore it can be any subset of sample space, even phi(empty set) or the whole sample space itself. You are confusing events with elements of sample space. There is a fundamental errors on tis page too. Like P(A|B) is probability of event A given that event B has (already) occurred. However the text in… Read more »

Mandeep P Shetty
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Mandeep P Shetty

@Kumar Thank you! I thought I was somehow wrong in my understanding. Another error is: “Even compound events (two events occurring at the same time) can be independent events. Ex. Tossing a coin and rolling a die. Sample space S = {(1,H), (2, H), (3, H), (4, H), (5, H), (6, H), (1, T), (2, T), (3, T), (4, T) (5, T) (6, T)}.” The example is still simple even and not a compound event. An actual example for a compound event will be ex: Tossing a coin and rolling a die and getting a 1 every time. The sample… Read more »

Mandeep
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Mandeep

@Kumar Thank you! I thought I was somehow wrong in my understanding. Another error is: “Even compound events (two events occurring at the same time) can be independent events. Ex. Tossing a coin and rolling a die. Sample space S = {(1,H), (2, H), (3, H), (4, H), (5, H), (6, H), (1, T), (2, T), (3, T), (4, T) (5, T) (6, T)}.” The example is still simple even and not a compound event. An actual example for a compound event will be ex: Tossing a coin and rolling a die and getting a 1 every time. The sample… Read more »

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