The steps involved in continuous improvement of any process are: collection of data, its analysis, drawing inferences, taking appropriate actions, again collecting the data and this cycle continues. We now focus on drawing inferences i.e, making meaningful conclusions. In order to draw inference(s), we use the language of probability.

**Probability is a measure of the likeliness that an event will occur**.

It is used to quantify an attitude of mind towards some proposition of whose truth we are not certain. The proposition of interest is usually of the form "Will a specific event occur ?" The attitude of mind is of the form "How certain are we that the event will occur ?" This certainty can be described in terms of a numerical measure called its **probability**.

** The probability number lies between 0 and 1 where 0 indicates impossibility and 1 indicates certainty**.

Thus the higher the probability of an event, the more certain we are that the event will occur.

A simple example would be the toss of a fair coin. Since the 2 outcomes are deemed equiprobable, the probability of "heads" equals the probability of "tails". Each probability is 1/2 or equivalently a 50% chance of either "heads" or "tails".

## Experiment (Random phenomenon):

An experiment is an activity in which we know what outcomes could happen. But we don't know which particular outcome will happen !

**Example:** If we toss a coin, we know that we will get head or tail, but we don't know which particular outcome will happen.

Similarly, if we roll a six-sided die, we know that we will get a 1 or 2 or 3 or 4 or 5 or 6. Again we don't know which particular outcome will happen.

A single attempt or realization of an experiment is known as 'trial' and one of the possible result of an experiment is known as an 'outcome'.

Ex: The possible outcome for tossing a single coin is: head or tail;

The possible outcome for rolling a six-sided die is: 1 or 2 or 3 or 4 or 5 or 6.

## Sample space:

The set of all possible outcomes of an experiment is known as sample space. It is denoted by the capital letter 'S'.

Ex: Sample space for the roll of a single die, S = {1, 2, 3, 4, 5, 6};

Sample space for tossing a single coin, S = {H, T}

## Event:

An event is a collection of outcomes of an experiment i.e., an event is the subset of the sample space.

For example: Sample space for the roll of a single die, S = {1, 2, 3, 4, 5, 6}

Let event P = "face value of the die is an even number".

Then P = {2, 4, 6}

Let event Q = "face value of the die is an odd number".

Then Q = {1, 3, 5}

As seen, events P and Q are subsets of the sample space S.

## Occurrence of events:

In a random experiment, if E is the event of a sample space S and w is the outcome, then we say that

i) the event E has occurred if w E

ii) the event E has not occurred if w E

## Simple event:

Any event which consists of only a single outcome in the sample space is called simple event. It can also be known as **elementary event**. For example: in tossing a single coin experiment, obtaining head or tail, is an elementary event because it consists of only a single outcome in the sample space i.e., {H} or {T}.

## Compound event:

Any event which consists of more than one outcome in the sample space is called compound event. It can also be known as **mixed event**. For example: in rolling a six-sided die experiment, obtaining an even number is a compound event because it consists of three outcomes in the sample space i.e., {2, 4, 6}.

## Sure (or Certain) event:

In a random experiment, an event will be called a sure event or a certain event if it always occurs whenever an experiment is performed. In other words, let S be a sample space and if E S, then E is an event called sure event.

For example, if we roll a six sided die, the possible outcomes of the sample space will be 1, 2, 3, 4, 5, 6. Now we consider an event of "Getting an even or odd number" in any particular event. We find that this event is represented with the following elements in the set: 1, 2, 3, 4, 5, 6 – which is exactly same as the sample space. So it is called sure event.

## Equally likely events:

Events are said to be equally likely, if we have no reason to believe that one is more likely to occur than the other. For example: when a six sided die is thrown, all the six-faces {1, 2, 3, 4, 5, 6} are equally likely to come-up.

## Impossible event:

In a random experiment, let S be a sample space and if S, then is an event called an impossible event. It is also known as **null event**. For example: in rolling a six-sided die experiment, obtaining the face value of the die above 6 (or below 1) is an impossible event because sample space = {1, 2, 3, 4,5, 6}.

## Mutually exclusive (or Disjoint) events:

Two or more events are said to be mutually exclusive events if and only if they have no outcomes in common. For example: when a coin is tossed, the event of occurrence of a head and the event of occurrence of a tail are mutually exclusive events.

## Complementary event:

In a random experiment, let S be the sample space and E be an event. The complement of an event E with respect to S is the set of all the elements of S which are not in E. The complement of E is denoted by **E'** or **E ^{c}**.

We know that there are different ways of combining two or more sets using the operations such as union, intersection, complement and difference. In a similar manner, two or more events can be combined.

In the a sample space S of an experiment, let A and B be two related events.

**(i) Complementary event**

For every event, there corresponds another complementary event which is "not that event". This shall be made clear through an example.

Consider the experiment of tossing of three coins. If H represents the outcome of a head and T represents the outcome of a tail, there will altogether be 8 combined outcomes. Thus the sample space of the experiment,

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

Now consider an event A of "minimum two heads shall appear".

Clearly, A = {HHH, HHT, HTH, THH}

The complement of the event 'not A' is represented by A' and is given by

A' = S – A

= {HTT, THT, TTH, TTT}

**(ii) Event "A or B"**

The event "A or B" is the event "event A or event B or both events". It is represented by A ∪ B as in sets.

A ∪ B = {ω: ω ∈ A or ω ∈ B}.

In continuation of the earlier experiment of "tossing 3 coins", let B be the event of "three tails appear". Then

B = {TTT}

∴ A ∪ B = {HHH, HHT, HTH, THH, TTT}

**(iii) Event "A and B"**

A ∩ B = {ω: ω ∈ A and ω ∈ B}

In the above example, A and B have nothing in common.

∴ A ∩ B = ϕ (or null set)

**(iv) Event "A but not B"**

From set theory we know that A – B = A ∩ B'

In the same example,

A = {HHH, HHT, HTH, THH}, B = {TTT}

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

A – B = A ∩ B' = {HHH, HHT, HTH, THH} = A (because A and B are mutually exclusive events).

Note that B – A is another event different from A – B.

It is defined, mathematically, as the relative frequency of an outcome. In other words, it is the fraction of times that the outcome would occur if the experiment were repeated indefinitely. For any event A, probability is given by:

where s = number of ways an outcome can succeed,

f = number of ways an outcome can fail

(s + f) is the total number of outcomes in the sample space S.

When s = 0, P(A) = 0 and when f = 0, P(A) = s/s = 1.

∴ The probability of any event 'A' ranges from 0 to 1, both inclusive.

i.e, **0 ≤ P(A) ≤ 1 **

This is an algebraic result from the definition of probability. When success is guaranteed f = 0, s = 1 and if failure is guaranteed f = 1, s = 0. The sum of the probabilities of all possible outcomes in a sample space is equal to one i.e., if the sample space is composed of 'n' possible outcomes,

then

**Example:**There are 4 white marbles and 5 blue marbles in a bag. If a marble is drawn from a bag, what is the probability that it is a white color marble?**Solution:**In the experiment of drawing a marble, let the event E = obtain a white color marble. The sample space contains 9 marbles. Here, s = 4 because there are 4 ways for outcome to be considered a success (drawing a white marble). And f = 5 because there are 5 ways for outcome to be considered a failure (not drawing a white marble).

Therefore, required probability is:

P(E) = .

## Probability of combined events:

**P(A or B): **The probability that either an event A or an event B occurs.

Using set notation, this can be written as P(A B).

A B is spoken as: A union B.

**P(A and B): **The probability that both an event A and an event B occur.

Using set notation, this can be written as P(A B).

A B is spoken as: A intersection B.

## Complement of an event:

Set of all the outcomes in the sample space that are not included in the outcomes of event A is known as complement of an event A.

The complement of an event A can be represented by either A^{c} or A or A'.

It is read as "A complement" or as "not A".

The probability of complement of an event A is equal to the '1' minus the probability of an event A

P(A) = 1 – P(A)

## Probability of mutually exclusive events:

Two or more events are said to be mutually exclusive events, if and only if, they have no outcomes in common. If A and B are two mutually exclusive events, then

A B = Φ P(A B) = 0.

**Example: **In the two-dice rolling experiment, let event A = "at least one face shows a 2" and event B = "sum of the two dice is 9".

If one of the dice shows a 2, the other can show a maximum of 6. The sum then would be 8 at the maximum.

So there is no way to get a sum of 9 if one die shows a 2.

That is, events A and B cannot both occur. In such a case, A and B are said to be mutually exclusive events.

According to this rule, probabilities of all the related events are added together to compute the probability of their joint occurrence.

If A and B are any two events associated with a random experiment, then addition rule for A and B is given as:

P(A or B) |
= | P(A) + P(B) – P(A and B) |

i.e, P(A ∪ B) |
= | P(A) + P(B) – P(A ∩ B) |

Proof by Venn diagram is given at the adjacent.

## Special case of the addition rule:

If A and B are two **mutually exclusive** events i.e, A and B are two disjoint sets,

A ∩ B = ϕ ⇒ P(A ∩ B) = 0

Then addition rule for A and B is simplified to:

**P(A or B) = P(A ∪ B) = P(A) + P(B)**

If A and B are two events in the sample space, it can be concluded from Venn diagrams that

i) P(B – A) = P(B) – P(A ∩ B)

ii) P(A – B) = P(A) – P(A ∩ B)

It is easy to understand probability problems involving rolling of dice or picking coloured balls from different boxes. But to solve problems based on "playing cards" you have to be thorough with its terminology or jargon.

A pack of cards or a **deck** contains 52 cards (excluding the Joker card). In probability problems, the Joker is not considered as a card unless mentioned specifically.

A pack of cards has four **suits**, namely **Spades, Hearts, Diamonds and Clubs.**

These are abbreviated as S, H, D and C respectively.

Spades and Clubs are called black suits as they are black in colour.

Hearts and Diamonds are called red suits as they are red in colour.

Thus there are 26 black cards and 26 red cards in a deck making a total of 52.

Each of the four suits contains 13 cards

→ 2, 3, 4, 5, 6, 7, 8, 9, 10, J, Q, K, A (Refer images above).

So the deck again consists of 4 × 13 = 52 cards.

Note: In a game of "bridge", Spades & Hearts are known as "major suits" and Diamonds & Clubs as "minor suits".

And the decreasing order of priority of the suits is S → H → D → C

You can see that, each suit contains 9 cards that are numbered: 2, 3, 4, 5, 6 7, 8, 9, 10.

Refer images below.

The remaining 4 cards (i.e, J, Q, K & A) are called **honour cards**.

J stands for Jack (or knave), Q → Queen, K → King, A → Ace.

So, in all there will be 16 honour cards in a deck.

The 4 honour cards in the spade suit are shown below.

2, 3, 4, .... J, Q, K, A are termed as "face values" which are 13 in number.

Since there are four suits, the number of cards, in a deck, with same face value is 4.

**Example:** If you consider a face value of 8, there is a eight of spades (8S), a eight of hearts (8H), a eight of diamonds (8D) and a eight of clubs (8C) in the deck i.e, 4 cards in all as shown below.

So is the case with any other face value card.

**Note: ** The decreasing order of priority of the face value cards is A, K, Q, J, 10, . . . . 3, 2.

In the probability problems involving cards two cases arise:

i. **With replacement** – a card once selected (or drawn or picked) is put back in to the deck. So the number of cards remain as 52 for the next event.

ii. **Without replacement** – a card once drawn is NOT put back in to the deck. So the number of cards reduces to 51 for the next event.

Some formulae to solve problems based on tossing of coins, rolling of dice, word formation, insertion of letters into envelopes are given next.

The number of favorable cases to get a sum 'k'

i) when two dice are rolled is given by

ii) when three dice are rolled is given by

iii) when 'r' dice are rolled

k = coefficient of x^{k} in (x^{1} + x^{2} + .... + x^{6})^{r}

Say a coin is tossed "m + n" times (m > n).

Then the probability of getting :

i) At least 'm' consecutive heads (or tails) =

ii) Exactly 'm' consecutive heads (or tails) =

Suppose 'n' coins are tossed simultaneously.

Then the probability of getting :

iii) At least one head (or tail) =

iv) Exactly 'r' heads (or tails) =

Let a word contain a total of 'T' letters.

Let it contain 'C' number of consonants and 'V' number of vowels.

Each consonant and vowel should not repeat i.e., it appears only once in the word (that should not repeat).

Now, let the letters of the word be arranged at random.

Then the probability that:

i) Relative positions of vowels and consonants do not change =

ii) Order of the vowels do not change =

iii) Order of the consonants do not change =

iv) Order of both vowels and consonants do not change =