Probability Event

The probability event is a set of outcomes of an experiment (a subset of the sample space) to which a probability is assigned.  The particular output of sample space is called event

A single outcome may be an element of many different events, and different events in an experiment are usually not equally likely, since they may include very different groups of outcomes. An event defines a complementary event, namely the complementary set (the event not occurring), and together these define a Bernoulli trial.

Typically, when the sample space is finite, any subset of the sample space is an event . However, this approach does not work well in cases where the sample space is uncountably infinite. So, when defining a probability space it is possible, and often necessary, to exclude certain subsets of the sample space from being events. [Wikipedia]

Example:

(1) S={ H , T }

Possible Events$= 2^n = 2^2 = 4$

Solution:

Possible Events$= 2^n = 2^6 = 64$

Types of Probability Events

1. Impossible event
2. Sure/Certain event
3. Equally likely events
4. Mutually exclusive event
5. Exhaustive Event

(1) Impossible event

An event that neither occurs is called impossible event

It is denoted by ɸ

Probability of impossible event is zero

P(impossible event) = 0

P( ɸ ) = 0

(2) Sure/Certain event

An event which surely occur in any condition is called sure or certain event

It is denoted by ‘S’

P(sure event) = 1

P(S) = 1

The range of probability is between 0 & 1

0 ≤ P(E) ≤ 1

(3) Equally likely events

Two or more events are said to be equally likely events if they have an equal chance of occurrence

Example:

S={1, 2, 3, 4, 5, 6}

A={1, 3, 5} B={2, 4, 6}

Above A & B are equally likely event as

A∪B= {1, 2, 3, 4, 5, 6}

(4) Mutually Exclusive Events:

Two or ore events are said to be mutually exclusive events if they cannot occur simultaneously

Inset Notation:

If A and B are two events then AB= ɸ

Example:

S={1, 2, 3, 4, 5, 6}

A={1, 3, 5} B={2, 4, 6}

A∪B= { } = ɸ

(5) Exhaustive Events

Two or more mutually exclusive events are said to be exhaustive events if they constitute the sample space

Example: They are different sets but when combined make one set that is constitute of its sample space is known as exhaustive events

Probability of an event

If E is an event then probability of an event is denoted by P(E) is given as

$P(E) = \frac{Total\: favorable\: cases}{Total\: possible\: cases}$

OR

$P(E)= \frac{n(E)}{n(S)} = \frac{ No.\: of \: sample \: probability \: in \: E }{ No. \: of \: sample \: probability \:in \: S }$

P(E) = ɸ E= Impossible event , denoted by ‘ɸ’

P(E) = 1 E= Sure event denoted by ‘S’

Example1:

A card is drawn at random from a pack of 52 plane cards what is the probability that card is

1. Red
2. Black
3. Diamond
4. Pictured Cards
5. Divisible by 3
6. King
7. Red King
8. Black Queen
9. Jack of club
10. Pictured card of heart

Solution:

1.P(a red card) =$\frac{26}{52}\:= \frac{1}{2}$

2.P(a black card) = $\frac{26}{52}\:= \frac{1}{2}$

3.P(a diamond card) = $\frac{13}{52}\:= \frac{1}{4}$

4.P(a pictured card) = $\frac{12}{52}\:= \frac{3}{13}$

5.P(a card divisible by 3) = $\frac{12}{52}\:= \frac{3}{13}$

6.P(a king) = $\frac{4}{52}\:= \frac{1}{13}$

7.P(a red king) =$\frac{2}{52}\:= \frac{1}{26}$

8.P(black queen) = $\frac{2}{52}\:= \frac{1}{26}$

9.P(a jack of club) = $\frac{1}{52}$

10.P( pictured card of heart) = $\frac{26}{52}$

Example 2:

A coin is tossed 3 times what is the probability the 3 coin shows

Solution:

No. of possible outcomes =$n(S) = 2^3 =8$

S = { HHH, HHT , HTH , HTT , THH ,THT ,TTH ,TTT }

1.$E_1$ = { 1 Head }

={ HHT, THT ,TTH }

$P( E_1) =\frac{n(E_1)}{n(S)}=\: \frac{3}{8}$

2.$E_2$ = { 2 Head }

={ HHT, HTH, THH }

$P( E_2) =\frac{n(E_2)}{n(S)}=\: \frac{3}{8}$

3.$E_3$ = { 3 Head }

={ HHH }

$P( E_3) =\frac{n(E_3)}{n(S)}=\: \frac{1}{8}$

4.$E_4$= { At-least 1 Head }

= { HHH, HHT , HTH , HTT , THH ,THT ,TTH }

$P( E_4) =\frac{n(E_4)}{n(S)}=\: \frac{7}{8}$

5.$E_5$= { At-least 2 Head }

= { HHH, HHT , HTH , THH }

$P( E_5) =\frac{n(E_5)}{n(S)}=\: \frac{4}{8} =\: \frac{1}{2}$

6.$E_6$= { At-least 3 Head }

= { HHH }

$P( E_6) =\frac{n(E_6)}{n(S)}=\: \frac{1}{8}$

7.$E_7$= { At-most 1 Head }

= { HHH, THT, TTH, TTT }

$P( E_7) =\frac{n(E_6)}{n(S)}=\: \frac{4}{8} =\: \frac{1}{2}$

8.$E_8$= { At-most 1 Head }

= { HHT , HTH , HTT , THH ,THT ,TTH ,TTT }

$P( E_8) =\frac{n(E_8)}{n(S)}=\: \frac{7}{8}$

9.$E_9$= { At-most 3 Head }

= { HHH, HHT , HTH , HTT , THH ,THT ,TTH ,TTT }

$P( E_9) =\frac{n(E_9)}{n(S)}=\: \frac{8}{8}$