# Probability Theory: 7 Complete Quick Facts

Probability theory emerged from the concept of taking risk. there are many complication today that come from the game of chance, such as wining a football match, playing cards and throwing a coin or throwing a dice.

Probability theory is used in many different sectors and the pliability of probability theory furnishes tools for almost so many different requirements. Here we are going to discuss the probability theory and few samples with the help of some fundamental concepts and results.

## RANDOM EXPERIMENTS:

“Random experiment is a kind of experiments where the result cannot be predicted.”

## SAMPLE SPACE:

The set of all possible outcomes from the experiment is called the sample space, it is usually denoted by S and all test out comes are said to be a sample point.
Eg: Think about the random Experiment of tossing 2 coins at a time. There are  4 outcomes constitute a sample space denoted by, S ={ HH, TT, HT, TH}

## TRAIL & EVENT:

Each non-empty subset of A of the sample space S is called an event. Consider the experiment to throw a coin. When we throw a coin, we can find a head (H) or a tail (T). Here throwing a coin is the trail and getting a head or a tail is a an event.

## COMPOUND EVENTS:

Events acquired by combining two or more basic events are called compound events or decomposable events.

## EXHAUSTIVE EVENTS:

The total number of feasible results of any trail  is called exhaustive events.

Eg: In throwing a dice the potential results are 1 or 2 or 3 or 4 or 5 or 6. So we have a total of 6 events in throwing die.

## MUTUALLY EXCLUSIVE AND EXHAUSTIVE SYSTEM OF EVENTS :

Let S is sample space of random experiment,  If  X1, X2, …..Xn are the subsets of S and

(i) Xi ∩ Xj =Φ for ij and (ii) X1 ∪ X2 ………∪ Xn =S

Then this collection of  X1∪ X2 ………∪ Xn is said to create  a mutually exclusive and exhaustive system of events.

## What is Independence?

When we pull out a card in a pocket of well-adjusted cards and secondly we also extract a card from the rest packet of cards (containing 51 cards), then the second extracting hangs on the first. But if, on the other hand, we pull the second card out of the pack by inserting the first card drawn(replacing), the second draw is known as independent of the first.

Example:  Two coins are thrown. Let  the first coin having head be event X and the Y be the second coin showing tail after throw. Two events X and Y are basically independent.

Example:   Two fair dice are drawn. If odd number come on first die consider it as event X and for second die even number as event Y.

The two event X and Y are mutually independent.

Example:  A card is drawn from a pack of 52 cards. If A = card is of Hearts, B = card is an King and A ⋂ B = card is King of Hearts, then events A and B are dependent

FAVOURABLE NUMBER OF CASES: The number of cases which permit an event to be tried in a trial is the total number of primary events that the aspect of any of them ensures the occurrence of the event.

## What is meant by Probability

If a arbitrary demonstration results in n incongruous , equally likely and exhaustive outcomes, out of which m are agreeable to the occurrence of an event A, then the probability of happening of A is given by

Probability notation: P(X)=m/n

For two events X and Y,

(i) X′ or   or XC indicates for the non-occurrence or negation of X.

(ii) X ∪ Y means for the occurring of at least any one of X and Y.

(iii) X ∩ Y means for the concurrent occurrence of X and Y.

(iv) X′ ∩ Y′ means for the non-occurrence of one and the other X and Y.

(v) X⊆ Y means for “the happening of X indicates occurrence of Y”.

Example: A bucket contains 6 red & 7 black marbles. Find the probability of drawing a red color marbles.

Solution: Total no. of possible ways of getting 1 marble= 6 + 7

Number of ways of getting 1 red marble= 6

Probability = (Number of favorable cases)/(Total number of exhaustive cases) = 6/13

Example: From a pack of 52 cards, 1 card is drawn at random. Find the probability of getting a queen card.

Solution:  A queen card may be chosen in 4 ways.

Total number of ways of selecting 1 queen card = 52

Probability = Number of favorable cases/Total number of exhaustive cases = 4/52=1/13

Example: Find the probability of throwing:

(a) getting 4 , (b) an odd number, (c) an even number

with an ordinary die (six faced).

Solution: The problem is dice problem

a) When throwing a die there is only one way of getting 4.

Probability = Number of favorable cases/Total number of exhaustive cases = 1/6

b) Number of ways of falling an odd number is 1, 3, 5 = 3

Probability = Number of favorable cases/Total number of exhaustive cases = 3/6=1/2

c) Number of ways of falling an even number is 2, 4, 6 = 3

Probability = Number of favorable cases/Total number of exhaustive cases = 3/6=1/2

Example: What is the possible chance of finding a king and a queen, when 2 cards are drawn from a pack of 52 playing cards?

Solution:  2 cards can be drawn from a pack of 52 cards = 52C2 (52 choose 2) ways

52 C2 =52!/2!(52-2)!=(52*51)/2=1326

1 queen card can be picked from 4 queen cards = 4C1=4 ways (4 choose 1)

1 king card can be taken from 4 king cards = 4C1=4 ways (4 choose 1)

Favorable cases = 4 × 4 = 16 ways

P(drawing 1 queen & 1 king card) = Number of favorable cases/Total number of exhaustive cases = 16/1326=8/663

Example: What are the chances of getting 4, 5 or 6 in the first throw and 1, 2, 3 or 4 in the second throw if the dice are thrown twice.

Solution:

Let P(A) = probability of getting 4, 5 or 6 in the first throw=3/6=1/2

and P(B)= probability of getting 1, 2, 3 or 4 in the second throw= 4/6=2/3

be the probability of the events then

Example: A book having total 100 number of pages, if any one of the page is selected arbitrary.  What is the possible chance that the sum of all the digits of the page number of the selected page is 11.

Solution:  The number of Favorable ways to get 11 will be (2, 9), (9, 2), (3, 8), (8, 3), (4, 7), (7, 4), (5, 6), (6, 5)

Hence required probability = 8/100=2/25

Example: A bucket contains 10 white, 6 red, 4 black & 7 blue marbles. 5 marbles are pull out at random. What is the probability that 2 of them are red colour and one is black colour ?

Solution:

Total no. of marbles= 10 + 6 + 4 + 7 =27

5 marbles can be drawn from these 27 marbles= 27 choose 5 ways

= 27C5=27!/

=(27*26*25*24*23)/(5*4*3*2)=80730

Total no. of exhaustive events = 80730

2 red marbles can be drawn from 6 red marbles= 6 ways

= 6C2=6!/

=(6*5)/2=15

1 black marbles can be pull out from 4 black marbles= 4 choose 1 ways= 4C1=4

∴ No. of favorable cases = 15 × 4 = 60

Hence required probability = Number of favourable casesTotal number of exhaustive cases

## Conclusion:

The probability theory is very interesting and applicable in our daily day to day life so probability theory and examples looks familiar to us, this is actually a complete theory which is used now a days in numerous technologies and applications,  This article was just a glimpse of the concept of probability the consecutive articles will deal with the detail concept and results of Probability, for more study, please refer below book:

Ref: Schaum’s Outlines of Probability and Statistics.