In earlier classes, probability was studied as a measure of uncertainty of events in a random experiment, using the axiomatic approach formulated by Russian mathematician A.N. Kolmogorov . In this chapter, we extend these concepts to conditional scenarios where the occurrence of one event affects the probability of another.
If $E$ and $F$ are two events associated with the same sample space $S$ of a random experiment, the conditional probability of the event $E$ given that $F$ has occurred, denoted by $P(E|F)$, is defined as:
$$P(E|F) = \frac{P(E \cap F)}{P(F)}, \quad \text{provided } P(F) \neq 0$$Alternatively, in terms of elementary events:
$$P(E|F) = \frac{n(E \cap F)}{n(F)}$$In particular, if $A$ and $B$ are disjoint events, then:
$$P((A \cup B)|F) = P(A|F) + P(B|F)$$The multiplication rule of probability allows us to find the probability of the simultaneous occurrence of two or more events.
For any two events $E$ and $F$ associated with a sample space $S$:
$$P(E \cap F) = P(E) P(F|E) = P(F) P(E|F)$$provided $P(E) \neq 0$ and $P(F) \neq 0$.
If $E$, $F$, and $G$ are three events of a sample space, then:
$$P(E \cap F \cap G) = P(E) P(F|E) P(G|(E \cap F)) = P(E) P(F|E) P(G|EF)$$This can be generalized analogously for $n$ events.
Two events are said to be independent events if the occurrence or non-occurrence of one does not affect the probability of the occurrence of the other.
provided $P(E) \neq 0$ and $P(F) \neq 0$.
Three events $A$, $B$, and $C$ are said to be mutually independent if all the following conditions hold simultaneously:
If at least one of these is not true, the events are not independent.
This section covers the partition of sample spaces, the theorem of total probability, and the formulation of Bayes' Theorem for reverse probability calculations.
A set of events $E_1, E_2, \dots, E_n$ is said to represent a partition of the sample space $S$ if:
Let $\{E_1, E_2, \dots, E_n\}$ be a partition of the sample space $S$, and suppose that each event $E_i$ has non-zero probability of occurrence. Let $A$ be any event associated with $S$, then:
$$P(A) = P(E_1) P(A|E_1) + P(E_2) P(A|E_2) + \dots + P(E_n) P(A|E_n) = \sum_{j=1}^{n} P(E_j) P(A|E_j)$$.If $E_1, E_2, \dots, E_n$ are $n$ non-empty events which constitute a partition of a sample space $S$, and $A$ is any event of non-zero probability, then the posteriori probability of the hypothesis $E_i$ given that event $A$ has occurred is:
$$P(E_i|A) = \frac{P(E_i) P(A|E_i)}{\sum_{j=1}^{n} P(E_j) P(A|E_j)}, \quad \text{for any } i = 1, 2, \dots, n$$Here, $P(E_i)$ are called the priori probabilities , and $P(A|E_i)$ are the likelihoods.
Below are step-by-step traces of key probability problems solved using the principles of conditional probability, total probability, and Bayes' Theorem.
Mathematical Problem Statement (Example 4): In a school, there are 1000 students, out of which 430 are girls. It is known that out of 430, 10% of the girls study in class XII. What is the probability that a student chosen randomly studies in Class XII given that the chosen student is a girl?
Step-by-Step Algorithmic Solution Trace:
- Let $F$ be the event that the randomly chosen student is a girl, and $E$ be the event that the student studies in Class XII.
- Compute the probability of choosing a girl:$$P(F) = \frac{430}{1000} = 0.43$$
- Compute the probability of choosing a girl who studies in Class XII:$$P(E \cap F) = \frac{10\\% \text{ of } 430}{1000} = \frac{43}{1000} = 0.043$$
- Apply the conditional probability formula:$$P(E|F) = \frac{P(E \cap F)}{P(F)} = \frac{0.043}{0.43} = 0.1$$
Final Evaluation Output: $P(E|F) = 0.1$
Mathematical Problem Statement (Example 16): Bag I contains 3 red and 4 black balls while another Bag II contains 5 red and 6 black balls. One ball is drawn at random from one of the bags and it is found to be red. Find the probability that it was drawn from Bag II.
Step-by-Step Algorithmic Solution Trace:
- Let $E_1$ be the event of choosing Bag I, $E_2$ be the event of choosing Bag II, and $A$ be the event of drawing a red ball.
- Since the bags are chosen at random:$$P(E_1) = P(E_2) = \frac{1}{2}$$
- Calculate the conditional probabilities of drawing a red ball from each bag:$$P(A|E_1) = \frac{3}{7}, \quad P(A|E_2) = \frac{5}{11}$$
- Apply Bayes' Theorem to find $P(E_2|A)$:$$P(E_2|A) = \frac{P(E_2) P(A|E_2)}{P(E_1) P(A|E_1) + P(E_2) P(A|E_2)}$$ $$P(E_2|A) = \frac{\frac{1}{2} \times \frac{5}{11}}{\frac{1}{2} \times \frac{3}{7} + \frac{1}{2} \times \frac{5}{11}} = \frac{\frac{5}{22}}{\frac{3}{14} + \frac{5}{22}} = \frac{35}{68}$$
Final Evaluation Output: $P(E_2|A) = \frac{35}{68}$
Mathematical Problem Statement (Example 21): A man is known to speak truth 3 out of 4 \times. He throws a die and reports that it is a six. Find the probability that it is actually a six.
Step-by-Step Algorithmic Solution Trace:
- Let $S_1$ be the event that six occurs, and $S_2$ be the event that six does not occur. Let $E$ be the event that the man reports a six.
- Determine the prior probabilities:$$P(S_1) = \frac{1}{6}, \quad P(S_2) = \frac{5}{6}$$
- Determine the likelihoods based on the man's truthfulness:$$P(E|S_1) = P(\text{speaks truth}) = \frac{3}{4}$$ $$P(E|S_2) = P(\text{does not speak truth}) = 1 - \frac{3}{4} = \frac{1}{4}$$
- Apply Bayes' Theorem to find $P(S_1|E)$:$$P(S_1|E) = \frac{P(S_1) P(E|S_1)}{P(S_1) P(E|S_1) + P(S_2) P(E|S_2)}$$ $$P(S_1|E) = \frac{\frac{1}{6} \times \frac{3}{4}}{\frac{1}{6} \times \frac{3}{4} + \frac{5}{6} \times \frac{1}{4}} = \frac{\frac{3}{24}}{\frac{3}{24} + \frac{5}{24}} = \frac{3}{8}$$
Final Evaluation Output: $P(S_1|E) = \frac{3}{8}$
Historical Chronicle: Evolution of Probability Theory