Probability in maths is a very important and one of the scariest chapters. Permutation and Combination, the set theory, and Probability fall in the same domain. These all chapters are interconnected.

**Definition**

Probability is the mathematical representation of the possibility (chance) of the occurrence of an event.**Example:** (1) Chances of getting head after flipping a coin.

(2) The likelihood of India’s winning the Cricket world cup.

**Importance of Probability in Real-Life**

(1) Before investing in a business, the estimation of the probability of success is required. (Probability in Maths)

(2) Life insurance companies decide the amount of premium according to life expectancy. Like premium of a person starting insurance at age of 28 is far lower than that person of age 60 because its a very high chance to die soon for the person of age 60.

(3) In the share market, the price fluctuations of a particular share purely depend on the probability of its future success or failure.

Concepts of Permutation and Combination: Click Here

**The terminology of Probability**

**Experiments:** An activity, the outcomes of which are not known. The few outcomes of an experiment could be favorable.

Example: One thousand failed Experiments by Thomas Alva Edison during the invention of the Light Bulb. (Probability in Maths)

**Random Experiments:** If all possible outcomes of an experiment are known but the result of a particular execution is not clear.

Example: In the toss of a coin either Head or Tail would come but which one will come is not clear.

**Trial:** The numerous attempts, in a particular process.

Example: In the numerous tosses (Trials) of a coin, some results head, and others are tail.

**Outcome: **The possible result of an event.

**Event:** A trial, with a clearly defined outcome.

Example: Getting 3 after rolling a Dice.

**Equally likely Events: **The events with the same possibility of occurrence.

Example: Getting a head or tail after tossing a coin, Getting an even or odd number after randomly picking a natural number.

**Sample Space:** The set of possible outcomes of all trials.

Example: The Sample Space, after rolling a Dice= {1,2,3,4,5,6}

**Impossible Events:** The possibility (Probability) of occurrence is Zero.

Example: Getting 7 after rolling a Standard Dice.

**Certain Event:** Its probability is equal to One because, in each trial, we get only a favorable outcome.

Example: Getting a number that is less than 7 after rolling a Standard Dice.

**Elementary Events: **An event with only one outcome. The sum of probabilities of all the possible elementary events in an experiment is One. (Probability in Maths)

Example: In a random experiment of tossing a coin there are two elementary events:

(a) Event of getting Head

(b) Event of getting Tail

Getting more than 4 after rolling a Dice is not an elementary event because it has two possible outcomes which are 5 & 6.

**Complementary Event:** Two events are complementary if the occurrence of one is the surety of not occurring of the other. Such events can’t happen simultaneously.

Example: Getting Head and Tail at the same time after tossing a coin is impossible. (Probability in Maths)

**Odds: **The ratio of the probability of an event and the probability of its complementary event or the ratio of the number of favorable outcomes and unfavorable outcomes.

**The formula of Probability in Maths**

**Que:** What is the probability of getting a number greater than 2 in a throw of normal dice?**Solution: **Possible outcomes, n(T)= {1,2,3,4,5,6} ⇨6 events

Required outcomes, n(R)={3,4,5,6} ⇨4 events

Probability= \frac{n(R)}{n(T)} = \frac{2}{3}

**The use of Conjunctions (AND & OR)**

**Que 1:** The probabilities of hitting a target by A and B are 1/3 & 1/2 respectively. If one shot is taken by both of them then Find the probability of hitting the target by **A & B**?**Solution:** P(A)=1/3 and P(B)=1/2

The required probability= P(A)×P(B)= \frac{1}{3}\times \frac{1}{2}=\frac{1}{6}

**Que 2:** In the above problem, find the probability that either** A or B** hits the target?**Solution:** The required probability= P(A)+P(B)= \frac{1}{3}+\frac{1}{2}=\frac{5}{6}.

**Mutually exclusive events**

The two events which can’t occur at the same time are called mutually exclusive events. Example: Getting head and tail at the same time after flipping a coin is impossible. (Probability in Maths)

If the probability of occurrence of an event= P(A)

Then the probability of its exclusive event= P(A’)

Here, P(A’)= 1- P(A)

**Example:** If two dice are thrown then find the probability [P(A)] of getting a sum more than 3? (Probability in Maths)

Here the minimum possible sum is 1+1=2 and maximum=6+6=12

The Required Sample Space={4,5,6,7,8,9,10,11,12}

First, we would calculate the probability [P(A’)] of getting a sum less than 4.

The total number of outcomes= 6×6=36

The number of require outcomes= {(1,1),(1,2),(2,1)}=3

P(A’)=\frac{3}{36}=\frac{1}{12}.

Now, P(A)= 1-P(A’)= 1-\frac{1}{12}=\frac{11}{12}.

**Mutually inclusive events**

If two events have some overlap with each other then these are called mutually inclusive events.

Example: If you choose a card from a standard deck then what is the probability of being it a card of Diamond or Queen?

In this case, the first event is ‘choosing a card of Diamond’ and the second event is ‘choosing a card of Queen’. (Probability in Maths)

The probability of getting a card of Diamond, P(A)= \frac{13}{52}.

Note: There are 13 cards of diamond in a deck of 52 cards.

The probability of getting a card of Queen, P(B)= \frac{4}{52}.

Note: There are 4 Queens in a deck of 52 cards.

The probability of getting a Queen of Diamond, P(A⋂B)= \frac{1}{52}.

Note: There is a single Queen of Diamond in a deck of 52 cards.

Then, the required probability, P(A⋃B)= P(A)+ P(B)- P(A⋂B)

P(A⋃B)=\frac{13}{52}+\frac{4}{52}-\frac{1}{52}=\frac{4}{13}.

**The Standard Deck of Playing Cards**

A standard deck of playing cards consists of 52 cards which are divided into four suits of 13 cards each. These suits are: (Probability in Maths)

1. Heart (Red Color)

2. Club (Black Color)

3. Diamond (Red Color)

4. Spade (Black Color)

**Face Cards or Court Cards: **The King, Queen, and Jack are the face cards because they depict a person. There are 12 face cards in a deck.

**Honour Cards:** The Ace, King, Queen, and Jack are called Honour Cards. There are 16 honour cards in a deck.

**Number Cards:** The cards on which numerals (From 2 to 10) are imprinted. A total of 36 number cards are there in a deck.

**Conditional Probability**

The probability of occurrence of event A if event B has already occurred, is called Conditional Probability. It is represented by P(A|B).

Example: If you draw a Red card from a standard deck of 52 cards then find the probability that it’s four? (Probability in Maths)

P(Four|Red)=\frac{2}{26}=\frac{1}{13}.

Note: There are 26 Red cards in a deck out of which only 2 are four.

**Formula:** P(A⋂B)= The joint probability of events A & B.

P(A|B)= Probability of occurring event A if B already occurred.

P(B|A)= Probability of occurring event B if A already occurred.

P(A|B)=\frac{P(A\cap B)}{P(B)}.

In the above question, There are two events:

Event A: Choosing card number 4 from 52 cards

Event B (Already occurred): Choosing the Red card from 52 cards.

P(A⋂B)=\frac{2}{52}.

P(B)= \frac{26}{52}.

P(A|B)= \frac{1}{13}.

**Key Points:**

(1) \frac{P(A|B)}{P(B|A)}=\frac{P(A)}{P(B)}.

(2) For independent events, P(A|B)=P(A) & P(B|A)=P(B)

(3) For mutually exclusive events, P(A|B) or P(B|A)=0

**Que 1:** A bag contains 2 blue and 3 red marbles. From which two marbles are drawn one by one without replacement. Find the probability of getting two blue marbles?

**Solution:** Here we can use Tree Method. (Probability in Maths)

**The first marble is drawn:Case 1:** Probability of drawing Blue marble = 2/5

**The second marble is drawn:**

Case i: Probability of drawing Blue marble = 1/4

Case ii: Probability of drawing Red marble = 3/4

**Case 2:** Probability of drawing Red marble = 3/5**The second marble is drawn:**Case i: Probability of drawing Blue marble = 2/4

Case ii: Probability of drawing Red marble = 2/4

Here, the Probability of Both marbles to be blue= Probability of the first blue × Probability of the second blue

= \frac{2}{5}\times \frac{1}{4}.

=\frac{1}{10}.

**The solution by using the formula:** (Probability in Maths)

Event A: First marble is blue

P(A)=2/5

P(B|A)= Probability of second blue marble if the first one is already blue

=1/4

P(A⋂B)=P(A)×P(B|A)

=\frac{2}{5}\times \frac{1}{4}.

=\frac{1}{10}.

**Que 2:** Two dice are thrown simultaneously and the sum of the numbers obtained is found to be 7. Find the probability that the number 3 appeared at least once?

**Solution:** The number of total outcomes= 6×6= 36

Event A: The sum of the digits is 7

A= {(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)}⇨ 6 outcomes

P(A)=\frac{6}{36}.

Event B: At least one number is 3

B={(3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (1,3), (2,3), (4,3), (5,3), (6,3)}⇨ 11 outcomes

Event A & B both: {(3,4), (4,3)}⇨ 2 outcomes

P(A∩B)=\frac{2}{36}.

P(B|A)=\frac{P(A\cap B)}{P(A)}=\frac{1}{3}.

More concepts on Probability in Maths are coming soon…