Today I'm going to discuss a very known concept Probability of Quantitative aptitude with you all. I hope you are well versed with this topic. I will try to teach you very easy method to solve problems related to this topic.

For example: I toss a coin. There will be two events: Head and Tail i.e. either head will occur or tail will occur. Therefore, both the events will have half probability.

Probability concept includes some terms. Firstly, go through these terms, then i will start discussing the basic concept.

n(S)= number of ways of choosing `2` balls out of `12`= C`(12,2)`

`C(12,2) = (12!)/ (2! 10!)` = `66`

n(E)= number of ways of drawing same color balls i.e. either

`n(E) = C(8,2) + C(4,2) = 34`

Probability = `34/66` = `17/33`

i) card drawn is black

ii) card drawn is a queen

iii) card drawn is black and queen

iv) card drawn is either black or queen

Solution: n(S)= `52`

i) n(E) = n(black) = `26`

P(black) = `26/52` = `1/2`

ii) n(E) = n(queen) = `4`

P(queen) = `4/52` = `1/13`

iii) n(black and queen) = `2` ( two queens are of black color)

P(black and queen)= `2/52` = `1/26`

iv) n(black or queen) = `26` (black cards including `2` queen) + `2` (rest two queens)= `28`

P(black or queen) = `28/52` = `7/13`

**Probability**means measuring the uncertainty. In other words, probability means how likely an event is to occur.For example: I toss a coin. There will be two events: Head and Tail i.e. either head will occur or tail will occur. Therefore, both the events will have half probability.

Probability concept includes some terms. Firstly, go through these terms, then i will start discussing the basic concept.

## Some basic terms

### Random experiment

Experiment which does not give same results on performing repeatedly. The only factor is there should be homogeneous conditions.

### Sample Space

While performing any experiment,

**all possible outcomes**gives you a sample space. It is denoted by alphabet S.
For example: Tossing a coin. Sample space i.e. S = {H,T}

Throwing a dice. Sample space will be S = {1,2,3,4,5,6}

### Event

Any occurrence of an experiment is known as Event. Or I can say a subset of sample space. It is denoted as E.

For example: Occurrence of head, Occurrence of any odd number while throwing dice and so on.

## Probability

If S is a sample space and E is occurrence of any event, then probability of occurrence of any event will be as follows:

P(E) = `(n(E))/(n(S))`

In simple words, Probability = Favorable Outcomes/ Total number of possible outcomes

**Example1:**An unbiased is thrown. What will be the probability of occurrence of a number multiple of `3`?

**Solution:**When a dice is thrown, numbers occurs are `1,2,3,4,5,6`. Therefore, sample space i.e. total number of possible outcomes will be:

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

Event = Occurrence of a number multiple of `3` i.e. `3` and `6`. Therefore, favorable outcomes are:

`E = {3,6}`

Therefore, Probability = Favorable outcomes/ Total number of possible outcomes

`P = 2/ 6 = 1/3`

## Properties of Probability

Before start solving the questions, lets have a look on properties of probability. You should know about the basic properties of probability as it will help you to solve questions.

- P(S) = `1` i.e. Probability of total number of outcomes (Sample Space) is `1`
- 0 `le` P(E)`le` 1 i.e. Probability of Sample space lies between `0` and `1`.its never less than zero.
- `barP(A) = 1- P(A)` i.e. Probability of not occurring of an event = 1- Probability of occurring an event.
- If in any question, "and' is used for events, then this means we have to multiply the probabilities if events are independent i.e. P(A and B) = P(A) `times` P(B)
- Similarly, if or is used then both the probabilities will be added i.e. P(A or B) = P(A) + P(B)

**Example2:**A bag contains**8 red balls**and**4 blue balls**. Two balls are drawn at random. Find the probability of getting both balls of same color.**Solution:**Total number of balls = `8 + 4= 12`n(S)= number of ways of choosing `2` balls out of `12`= C`(12,2)`

`C(12,2) = (12!)/ (2! 10!)` = `66`

n(E)= number of ways of drawing same color balls i.e. either

**2 out of 8 red balls**OR**2 out of 4 blue balls**`n(E) = C(8,2) + C(4,2) = 34`

Probability = `34/66` = `17/33`

**Example3:**One card is drawn from a pack of `52` cards. Find the probability thati) card drawn is black

ii) card drawn is a queen

iii) card drawn is black and queen

iv) card drawn is either black or queen

Solution: n(S)= `52`

i) n(E) = n(black) = `26`

P(black) = `26/52` = `1/2`

ii) n(E) = n(queen) = `4`

P(queen) = `4/52` = `1/13`

iii) n(black and queen) = `2` ( two queens are of black color)

P(black and queen)= `2/52` = `1/26`

iv) n(black or queen) = `26` (black cards including `2` queen) + `2` (rest two queens)= `28`

P(black or queen) = `28/52` = `7/13`