Conditional probability:
Let A and B be two dependent events, then the probability of occurrence of an event A when it is given that the event B has already occurred is known as conditional probability of A. It is denoted P(A/B). In the same way, the conditional probability of the event B is given A has already occurred, is denoted by P(B/A). P(A/B) and P(B/A) are calculated by the following formula:
$$\;P(A/B)\;=\;\frac{P(A\;\cap\;B)}{P(B)}\;,\;P(B)\;\neq\;0$$
$$\;P(B/A)\;=\;\frac{P(A\;\cap\;B)}{P(A)}\;,\;P(A)\;\neq\;0$$
$$\;where\;P(A\cap\;B)\;is\;the\;probability\;of\;the\;simultaneous\;occurrence\;of\;the\;events\;A\;and\;B\;$$
Two Basic Laws of Probability:
There are two basic theorems that are the basic laws of probability:
Addition Theorem (Theorem of Total Probability)
If A and B are two events with their respective probabilities P(A) and P(B).The addition theorem in the Probability concept is the process of determination of the probability that either event ‘A’ or event ‘B’ occurs or both occur. The notation between two events ‘A’ and ‘B’ the addition is denoted as 'U' and pronounced as Union.
The result of this addition theorem generally written using Set notation,
$$\;P(A\cup\;B)\;=\;P(A)+P(B)-P(A\cap\;B)$$
Where,
$$P\;(A\cup\;B)\;=\;probability\;of\;occurrence\;of\;event\;‘A’\;or\;event\;‘B’.$$
$$P\;(A\;\cap\;B)\;=\;probability\;of\;occurrence\;of\;event\;‘A’\;or\;event\;‘B’.$$
Addition theorem probability can be defined and proved as follows:
Let ‘A’ and ‘B’ are Subsets of a finite non-empty set ‘S’ then according to the addition rule
$$\;P(A\cup\;B)\;=\;P(A)+P(B)-P(A\cap\;B)$$
$$\;P(A\cup\;B)\;=\;P(A)+P(B)-P(A).\;P(B)$$
On dividing both sides by P(S), we get
$$\;\frac{P(A\cup\;B)}{P(S)}\;=\;\frac{P\;(A)}{P(S)}\;+\;\frac{P(B)}{P(S)}–\frac{P(A\cup\;B)}{P(S)}\;.\;.\;.\;.\;.\;.(1).$$
If the events ‘A’ and ‘B’ correspond to the two events ‘A’ and ‘B’ of a random experiment and if the set ‘S’ corresponds to the Sample Space ‘S’ of the experiment then the equation (1) becomes
$$\;P(A\cup\;B)\;=\;P(A)+P(B)-P(A).\;P(B)$$
This equation is known as the addition theorem in probability.
or
Let n be the total number of equally likely and exhaustive outcomes of a random experiment. If u, v and w be the number of cases favourable to the events A,B and common to A and B respectively, then
$$\;P(A)=\;\frac{u}{n}\;,\;P(B)\;=\;\frac{v}{n}\;and\;P(A\cap\;B)=\frac{w}{n}$$
Since, the outcomes w are common to both events A and B, so the cases favourable to the event A U B are (u+v-w)
$$\;Now,P(A\cup\;B)\;=\frac{(u+v-w)}{n}$$
$$=\frac{u}{n}\;+\ frac{v}{n}\;-\frac{w}{n}\;$$
$$\;=\;P(A)+P(B)-P(A\cap\;B)$$
Case 1: If A and B are mutually exclusive events then $$\;P(A\cup\;B)\;=\;P(A)+P(B)$$
Case 2 : If A, B and C are three events, then the probability of occurrence of at least one of these events is,
$$\;P(A\cup\;B\cup\;C)\;=\;P(A)+P(B)+P(C)-P(A\cap\;B)-P(B\cap\;C)-P(A\cap\;C)+P(A\cap\;B\cap\;C\;)$$
If the events are mutually exclusive then$$\;P(A\cup\;B\cup\;C)\;=\;P(A)+P(B)+P(C)$$
Multiplication Theorem (Theorem of Compound Probability)
If two events A and B are independent , then the probability of their simultaneous occurrence is equal to the product of their individual probabilities. Let n_{1 }and n_{2 }the total number of possible cases for events A and B respectively, have m_{1 }and m_{2 }as their respective favourable cases. Then,
$$\;P(A)\;=\;Probability\;of\;occurrence\;of\;an\;event\;A=\frac{m_1}{n_1}$$
$$\;P(B)\;=\;Probability\;of\;occurrence\;of\;an\;event\;B=\frac{m_2}{n_2}$$
Since, the total number of possible cases for the events A and B are n_{1 }and n_{2} respectively and since we can combine each of n_{1 } possible cases of the event A with each of the n_{2} possible cases of the event B, so the total number of possible cases for their simultaneous occurrence is n_{1}, n_{2. }In the same way, the total number of favourable cases for the simultaneous occurrence of the events A and B is m_{1}, m_{2}.
$$Now,\;P(A\cap\;B)\;=\;Prob.\;of\;simultaneous\;occurrence\;of\;the\;events\;A\;and\;B\;$$
$$\;=\frac{m_1.m_2}{n_1.n_2}\;=\;\frac{m_1}{n_1}.\frac{m_2}{n_2}\;=\;P(A).P(B)$$
Case 1: If A, B and C are three independent events, then the probability of their simultaneous occurrence of the events is given by $$\;P(A\cap\;B\cap\;C)\;=\;P(A).P(B).P(C)
Case 2: The probability that the event A will not happen is 1-P(A). Therefore the probability that now of the two independent events A and B will happen is [1-P(A)].[1-P(B)].
$$\therefore\;the\;probability\;that\;atleast\;one\;event\;will\;happen\;$$
$$=1-[1-P(A)].[1-P(B)]$$
Taken reference from
( Basic mathematics Grade XII and A foundation of Mathematics Volume II and Wikipedia.com )
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