Suppose an event E have the right to happen in r ways out of a full of n possible equally likely methods.

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Then the probcapacity of event of the event (referred to as its success) is dedetailed by

`P(E)=r/n`

The probcapability of **non-occurrence** of the event (called its failure) is dedetailed by

`P(barE)=(n-r)/n=1-r/n`

Notice the bar above the *E*, indicating the event does **not** take place.

Hence,

`P(barE)+P(E)=1`

In words, this means that the amount of the probabilities in any type of experiment is `1`.

## Definition of Probcapacity making use of Sample Spaces

When an experiment is perdeveloped, we erected a sample room of all feasible outcomes.

In a sample of N equally likely outcomes we assign a chance (or weight) of `1/N` to each outcome.

We define the **probability** of an occasion for such a sample as follows:

The probcapability of an occasion E is identified as the variety of outcomes favourable to E separated by the complete variety of equally most likely outcomes in the sample space S of the experiment.

That is:

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

where

`n(E)` is the variety of outcomes favourable to E and

`n(S)` is the total variety of equally most likely outcomes in the sample space S of the experiment.

## Properties of Probability

(a) 0 ≤ P(event) ≤ 1

In words, this implies that the probcapacity of an event should be a number between `0` and also `1` (inclusive).

(b) P(difficult event) = 0

In words: The probcapability of an impossible event is `0`.

(c) P(certain event) = 1

In words: The probcapability of an absolutely specific event is `1`.

### Example 1

What is the probability of...

(a) Getting an ace if I select a card at random from a standard load of `52` playing cards.

Answer

In a traditional fill of 52 playing cards, we have:

♥ 2 3 4 5 6 7 8 9 10 J Q K A ♦ 2 3 4 5 6 7 8 9 10 J Q K A♣ 2 3 4 5 6 7 8 9 10 J Q K A ♠ 2 3 4 5 6 7 8 9 10 J Q K A

Tbelow are 4 aces in a normal pack. So the probability of obtaining an ace is:

`P("ace")=4/52 = 1/13`

(b) Getting a `5` if I roll a die.

Answer

Image source

A die has actually 6 numbers.

There is only one 5 on a die, so the probcapability of obtaining a 5 is provided by:

`P(5)=1/6`

(c) Getting an also number if I roll a die.

Answer

Even numbers are `2, 4, 6`. So

`P("even")=3/6=1/2`

(d) Having one Tuesday in this week?

Answer

Each week has a Tuesday, so probcapability = `1`.

### Example 2

There are `15` balls numbered `1` to `15`, in a bag. If a perchild selects one at random, what is the probcapacity that the number published on the round will be a prime number higher than `5`?

Answer

The primes in between `5` and `15` are: `7, 11, 13`.

So the probability `=3/15=1/5`

### Example 3

The names of four directors of a company will be put in a hat and a 2-member delegation will be selected at random to represent the agency at an global meeting. Let A, B, C and D signify the directors of the firm. What is the probcapacity that

(a) A is selected? (b) A or B is selected? (c) A is not selected?

Answer

The feasible outcomes are: AB, AC, AD, BC, BD, CD.

**Explacountry 1: **The probcapability is `3/6=1/2` considering that as soon as we select A, we must choose one of the remaining 3 directors to go through A. There are `C_2^4=6` feasible combicountries.

**Explanation 2:** Probability that A is schosen is `C_1^1 times C_1^3/C_2^4 = 3/6 = 1/2`

**Explacountry 1:** The probcapability of gaining A or B first is `2/4=1/2`.

Now to think about the probcapability of selecting A or B as the second director. In this case, the first director hregarding be C or D via probcapacity `2/4` (2 certain directors out of 4 possible).

Then the probability of the second being A or B is `2/3` (2 certain directors out of the staying 3 directors).

We have to multiply the 2 probabilities.

So the probability of getting A or B for the second director is `2/4 xx 2/3 = 1/3`

The complete is: `1/2 + 1/3 = 5/6`

Explacountry 2: Probcapability that A or B is schosen is

`fracC_1^1 times C_1^3 + C_1^1 times C_1^2C_2^4` `=frac3+26` `=5/6`

**Explanation 3:** If A or B is chosen, then we cannot have the case C **and** D is preferred. So the probcapacity of A or B is provided by:

`P("A or B") = 1-P("C and D")` `=1-1/6` `=5/6`

Part (c)Probcapability that A is not selected is `1-1/2=1/2`

### Extension

Consider the case if we are picking 2 directors from 5. The probabilities would certainly currently be:

(a) Probcapacity that A is schosen is

`fracC_1^1 times C_1^4C_2^5=4/10=2/5`

(b) Probcapability that A or B is selected is

`fracC_1^1 times C_1^4 + C_1^1 times C_1^3C_2^5` `=frac4+310` `=7/10`

(c) Probcapability that A is not selected is `1-2/5=3/5`.

♥ 2 3 4 5 6 7 8 9 10 J Q K A ♦ 2 3 4 5 6 7 8 9 10 J Q K A♣ 2 3 4 5 6 7 8 9 10 J Q K A ♠ 2 3 4 5 6 7 8 9 10 J Q K A