In this lecture, we discuss Conditional Probability. In an earlier lecture, when we were discussing statistics, we also looked at this in a certain way and when we looked at categorical variables, we discussed the case of admission to a program through different training classes and so on. So, we take the same example and then we try to convert the proportions into probabilities to try to understand these concepts in conditional probability.
So, we have seen this before we looked at this data of getting into a
management school, imaginary data and let us assume that 3849 applicants were there and they belong to 3 preparation classes with the numbers given 2166, 1047, 636. And then we say that 65 people joined and this number is given here, those who joined and those who did not or could not join and then we also have this number of 2166, 1047 and 636.
We also looked at this and the contingency table that we drew; showed the counts of cases of one categorical variable contingent on the value of another. So, the 65 is divided into 37, 18, and 10, and 2166 now get distributed to 2129 and 37. So, out of 2166 who went to class 1, 2129 did not or could not join the program, while 37 dead and so on. So, we have already
seen that the cells of the contingency table are mutually exclusive and each case appears exactly in one cell. We also looked at the frequency distribution of the selected people; that is called the marginal distribution we have seen.
The next thing we did when we looked at contingency tables was we then started computing these percentages. Now, what are these percentages and how are they computed. We now look at this category under yes and then you looked at the category under class 3. So, then 10 students from class 3 joined the program, 10 out of 65 joined the program, but this 10 students out of 3849 who actually applied are 0.26 percent of the people who applied and
joined went to class 3 for earlier preparation.
Now, this is 1.57 students who went to class 3. So, class 3 had a total of 636 people who went and 10 of them joined; so, 1.57 percent. So, 1.57 percent of those who went to class 3 as a preparation class joined the program, and 15.38 percent of the students joined the program. So, out of the 65 who joined, 10 went to class 3. So, 15.38 percent of the people who joined the program went to class 3. So, there we saw the proportions, now we are going
to generalize these proportions as probabilities.
So, then we start writing this; out of the people who applied 0.02, totally joined, 0.98 did not or could not join, and then we represent this 0.2 as proportions of the other numbers. So, out of those who applied 0.56 came from class, 1, .272 from plus 2, and 0.166 from class 3, and that gets distributed in this manner. So, proportions become probabilities; the probability that a person who is applied and gone to class 1 and got selected or joined is 0.01 which is given here. So, each of these outcomes describes two attributes, joined and not join and the preparation class. So, it is the joint probability of joining the program and the preparation class. So, the probability of yes and class 1 is 0.01 that is people who join the program who went to class 1 is 0.01.
So, the joint probability is the probability of intersection, the marginal probability is the probability of observing an outcome with a single attribute regardless of other attributes. So, the probability of a person joining the program is 0.01 plus 0.005 plus 0.003 it's actually approximated to 0.02 because these are all got from numbers like 65, 3900, and so on. So, this is
probability of joining and going to class 1 or probability of joining and going to class 2 and probability or probability of joining and going to class 3 which is the sum of the probability of joining and class 1 plus the probability of joining and class 2 plus the probability of joining and class 3 which is this 0.02.
So, the probability of a person applied from class 1 is 0.56. So, of all the people who attended the interview or who applied so, that is the probability of people who joined and went to class 1 and the probability of people who did not or could not join and went to class 1. Therefore, it is the probability of join and class 1 plus the probability of not joined and class 1; 0.01 plus 0.55
which is equal to 0.56. So, we show all these computations this way. So, this is 0.01; now if you say what percentage of people who of the people who came from this went there. So, it is 0.01 by 0.56 which is 0.0178, and so on. So, we get the last table where we do the divisions and say that 0.0178
comes from 0.01, 0.56. So, what is the probability that if the person going to class 1 joins the program would be 0.01 by 0.56 and so on.
Now with a new sample space of class 1 what is the probability of yes and class 1, it is not 0.01 because it does not add up to one. The answer is 0.01 by 0.56 which is 0.0178 so that the sample space adds up to 1 here. So, the probability of joining given class 1 is probability of joining and went to class 1 divided by the probability of going to class 1.
So, the probability of joining and class 1 is given by this 0.0178 which is the probability of joining and going to class 1 here is 0.01 divided by 0.56. So, this is the probability of joining given that they went to class 1 is the probability of joining and went to class 1 divided by the probability of class 1.
So, the conditional probability of yes that is joining given the person is from class 1 is the probability of joining given the person is from class 1 is equal to the probability of joining and the person going to class 1 divided by the probability of person from class 1 which is 0.01 by 0.56 which is 0.178.
So, the probability of A given B is equal to the probability of A and B divided by the probability of B.
This is the conditional probability equation probability of A given B. This line has to be read as given B this vertical line. So, the probability of A given B is equal to the probability of A and B divided by the probability of B the verticals line symbol in a given B means given phrases given that conditional on it known that they all indicate conditional probabilities.
Now, dependent events; two events A and B are independent if the probability that both occur is the product of the probabilities of the two events. So, we know that P(A and B) is equal to P(A) into P(B) for independent events. For example, if A represents customers who see an
advertisement and B identifies customers who buy the product, the probability of buying the product given seeing the advertisement this P(A and B) divided by P(A). But if A and B are
independent then P(A and B) is P(A) into P(B) which is, therefore, P(B gave A) is P(B), it does not depend on A.
So, in our case, the probability of joining given that they went to class 1 is 0.178 that we calculated while the probability of joining multiplied by the probability of going to class 1 is 0.02 into 0.56 which is 0.112 and therefore, we can say that these two are not independent and there is a dependency which is that. So, there is another way of checking whether events are
dependent or independent using conditional probabilities.
Now, let us look at this multiplication rule; joint probability of two events; A and B is the product of the marginal probability of 1 times the conditional probability of another. So, since the probability of B given A is the probability of A and B divided by the probability of A and B is equal to the probability of A multiplied by the probability of B given A is called the multiplication rule. So, the joint probability of two events A and B, P(A and B) is the product of the marginal probability of one which is P(A) multiplied by the conditional probability of the other P(B gave A).
Now, P(A given B) is P(A and B) divided by P(B). Therefore, P(A and B) is equal to P(B) into P(A given B). Remember, now P(A and B) is equal to P(A) multiplied by P(B given A), it is also equal to P(B) multiplied by P(A given B). So, the probability that events A and B both occur which is A and B is the probability of A times the probability of B given A or probability of B times probability of A given B occur. So, both are valid now let us look at
another question.
The probability of loan one defaulting is p 1 probability of loan to defaulting is p 2 and the probability of loan 3 defaulting is P 3. So, the probability of all defaulting is p 1, p 2, p 3. Are they independent? If they are borrowers, if the 3 borrowers are suppliers to the same company and because of some issue it is defaulting, then there is a dependency. Therefore, when we multiply unconditional probabilities check always whether they are independent.
So, only when they are independent, we can do this multiplication rule.
Now order in conditional probabilities it's very important to know this P(A given B) is not equal to P(B given A). So, the example probability of joining given class 1 is the probability of joining and from class 1 divided by class 1 which was 0.0178. The probability of class 1 given joining is the probability of class 1 and joining divided by the probability of joining which happened
to be 0.5. So, P(A given B) is not equal to P(B given A).
We now look at probability trees and try to solve some problems using probability trees. In fact, indirectly we have seen this in one of the examples where we looked at the batsman having to score 2 runs to win a match with 1 ball remaining. And we looked at a tree-like solution where if the person went for the clean hit, there is a probability, and then when the person does not, there is another probability and then there is an outcome, and so on.
The example where there is a 30 percent if they go for a clean hit and 50 percent if tie and another 50 percent of winning. Now we look at another example. On a Sunday evening, 50 percent of the people watch movies, 30 percent watch a cricket match, and 20 percent watch comedy shows. The percentage of people skipping ads are 20 percent, 25 percent and 15
percent respectively. Now, what happens? Now we have this probability tree. So, 0.5 movies, 0.3 cricket, 0.2 comedy; now within this percentage of people, skipping ads are 20 percent if they watch a movie.
So, people who watch a movie and see ads is 0.8, people who watch a movie and skip adds 0.2. Similarly, 0.75, 0.25 and 0.85, 0.15; so, we now realize that people who watch movies and see ads are 0.5 into 0.8 which is 0.4; the other one is 0.5 into 0.2 which is 0.1, and so on.
Therefore, people who watch cricket and see ads are the probability of people watching cricket multiplied by people watching cricket and seeing ads. So, this is 0.3 into 0.75 which is 0.22, and so on.
We also have this very important result called the Bayes’ rule. The conditional probability of A given B can be found from the conditional probability of B given A by using this formula.
So, the probability of A given B is the probability of A and B divided by P(B). So, this is B gave A into P(A) divided by the probability of B given A into P(A) plus the probability of B given A complement into the probability of A complement. So, now, B is divided into B is expanded into B given A into P of A plus B given A complement into P of A complement P of A and B is
expanded by the original multiplication rule formula; so, B of A into P of A.
So, we could use the Bayes’ rule and try to solve some problems. So, with this, we come to the end of this lecture and we will look at discussion questions on this lecture and continue our discussion on probability in the next lecture.
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