**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

### Sample Space

**all possible outcomes**gives you a sample space. It is denoted by alphabet S.

### Event

## Probability

**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:

## Properties of Probability

- 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 that

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`

can you please post about, how to solve when three dice are tossed at the moment

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