Probability Basic Concepts

Here, you will learn probability basic concepts and definitions of probability.

Let’s begin –

Probability gives us a measure of likelihood that something will happen. However probability can never predict the number of times that an occurrence actually happens.

Probability Basic Concepts and Definitions

(a) Experiment :

An action or operation resulting in two or more well defined outcomes.

Example : tossing a coin, throwing a die, drawing a card from a pack of well shuffled playing cards.

(b) Sample Space :

A set S consist of all possible outcomes of an random experiment is called a sample space.

Example : in an experiment of “throwing a die” , following sample spaces are possible:

(i) {even number, odd number}

(ii) {1,2,3,4,5,6}

(c) Event :

An event is defined as an occurrence or situation.

Example : in a toss of a coin, it shows a head.

(d) Compound Event :

If an event has more than one sample points it is called Compound Event.

(e) Complement of an Event :

The set of all outcomes which are in S but not in A is called complement of the event A. It is denoted by \(A^c\), \(A^`\).

(f) Mutually Exclusive Event :

Two events A and B are said to be Mutually Exclusive if probability of A intersection B is zero i.e \(P(A \cap B)\)=0.

Example : choosing numbers at random from the set {3,4,5,6,7,8,9,10,11,12}. If,

Event A is the selection of a prime number.
and Event B is the selection of an odd number.
Event C is the selection of an even number.

Then A and C are mutually exclusive as none of the numbers in this set is both prime and even and B and C are also mutually exclusive.

(g) Equally Likely Events :

Events are said to be Equally Likely when each event is as likely occur as any other event. Note that the term ‘at random’ or ‘randomly’ means that all possibilities are equally likely.

(h) Exhaustive Events :

Events A,B,C…..N are said to be Exhaustive Events if no events outside this set can result as an outcome of an experiment.

Note :

(i) 0<=P(A)<=1

(ii) P(A)+P(\(A^c\))=1

(iii) If x cases are favourable to A & y cases are favourable to \(A^c\) then P(A)=\(x\over(x+y)\) and \(P(A \cap B)\)=\(y\over(x+y)\). We say that Odds in Favour of A are x:y & Odds Against A are y:x.

Example : If the letters of INTERMEDIATE are arranged, then odds in favour of the event that no two ‘E’s occur together, are-

Solution : Let 3’E’s, Rest 9 letters,
First arrange rest of the letters = \(9!\over {2! 2!}\)
Now 3’E’s can be placed by \(^{10}C_3\) ways, so favourable cases = \({9!\over {2! 2!}}\times\)\(^{10}C_3\) = 3 \(\times\) 10!
Total cases = \(12!\over {2! 2! 3!}\) = \({11\over 2} \times\)10!
Non-favourable cases = (\(11\over 2\) – 3)\(\times\)10! = \({5\over 2}\times\)10!

Odds in favour of the event = \(3\over {5/2}\) = \(6\over 5\) Ans.

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