Important definitions, formulae and the Picture Method idea needed before solving probability and Bayes’ theorem problems.
This video should explain the definitions and formulae before students start solving probability problems.
A random experiment is an experiment whose particular outcome cannot be predicted in advance, although all possible outcomes are known.
The set of all possible outcomes of a random experiment is called the sample space. It is denoted by \(S\).
\[ S=\{H,T\}\quad \text{for one coin toss} \]
An event is any subset of the sample space.
Two or more events are mutually exclusive if they cannot occur at the same time.
\[ A\cap B=\varnothing \]
The word exhausted means completely used up. In probability, events are called exhaustive when they use the whole sample space completely.
That means at least one of the events must occur.
\[ E_1\cup E_2\cup E_3\cup\cdots\cup E_n=S \]
If events are both mutually exclusive and exhaustive, they fragment the sample space.
Mutually exclusive means the fragments do not overlap. Exhaustive means the fragments cover the whole sample space.
\[ E_i\cap E_j=\varnothing\quad (i\ne j) \]
\[ E_1\cup E_2\cup E_3\cup\cdots\cup E_n=S \]
Two events are independent if the occurrence of one event does not affect the probability of the other event.
\[ P(A\cap B)=P(A)P(B) \]
\[ P(A|B)=P(A),\qquad P(B|A)=P(B) \]
If all outcomes are equally likely, then the probability of an event \(A\) is:
\[ P(A)=\frac{n(A)}{n(S)} \]
Here, \(n(A)\) is the number of favourable outcomes and \(n(S)\) is the total number of possible outcomes.
\[ 0\leq P(A)\leq1 \]
The complement of \(A\), denoted by \(A'\) or \(\bar{A}\), means that event \(A\) does not occur.
\[ P(A)+P(A')=1 \]
\[ P(A')=1-P(A) \]
For any two events \(A\) and \(B\):
\[ P(A\cup B)=P(A)+P(B)-P(A\cap B) \]
If \(A\) and \(B\) are mutually exclusive, then \(P(A\cap B)=0\). Hence:
\[ P(A\cup B)=P(A)+P(B) \]
Conditional probability means probability of one event when another event is already known to have occurred.
\[ P(A|B)=\frac{P(A\cap B)}{P(B)},\qquad P(B)\ne0 \]
From conditional probability:
\[ P(A\cap B)=P(B)P(A|B) \]
Also:
\[ P(A\cap B)=P(A)P(B|A) \]
If \(A\) and \(B\) are independent, then:
\[ P(A\cap B)=P(A)P(B) \]
Let \(E_1,E_2,\ldots,E_n\) be mutually exclusive and exhaustive events with non-zero probabilities. If \(A\) is any event, then:
\[ P(A)=P(E_1)P(A|E_1)+P(E_2)P(A|E_2)+\cdots+P(E_n)P(A|E_n) \]
Let \(E_1,E_2,\ldots,E_n\) be mutually exclusive and exhaustive events with non-zero probabilities. If \(A\) is an event with \(P(A)\ne0\), then:
\[ P(E_i|A)=\frac{P(E_i)P(A|E_i)}{\sum_{j=1}^{n}P(E_j)P(A|E_j)} \]
Bayes’ theorem is used to find the probability of a cause when the effect is already known.