Suppose an event E can happen in r methods out of a full of n possible equally most likely ways.

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Then the probability of event of the event (called the success) is denoted by

`P(E)=r/n`

The probability of non-occurrence that the occasion (called that failure) is denoted by

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

Notice the bar above the E, denote the event does not occur.

Thus,

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

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

## Definition the Probability making use of Sample Spaces

When one experiment is performed, we collection up a sample an are of all feasible outcomes.

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

We specify the probability of an event for such a sample as follows:

The probability of an occasion E is defined as the variety of outcomes favourable to E separated by the total number 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 number of outcomes favourable to E and

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

## Properties the Probability

(a) 0 ≤ P(event) ≤ 1

In words, this method that the probability of an event must be a number in between `0` and also `1` (inclusive).

(b) P(impossible event) = 0

In words: The probability of an impossible occasion is `0`.

(c) P(certain event) = 1

In words: The probability of one absolutely certain event is `1`.

### Example 1

What is the probability of...

(a) getting an ace if I pick a map at random from a standard load of `52` playing cards.

In a standard load 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

There are 4 aces in a normal pack. For this reason the probability of acquiring an ace is:

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

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

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A die has actually 6 numbers.

There is just one 5 top top a die, for this reason the probability of gaining a 5 is provided by:

`P(5)=1/6`

(c) getting an also number if I role a die.

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

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

(d) having one Tuesday in this week?

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

### Example 2

There are `15` balls numbered `1` come `15`, in a bag. If a human being selects one at random, what is the probability that the number printed on the ball will it is in a prime number higher than `5`?

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

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

### Example 3

The name of 4 directors of a company will be put in a hat and also a 2-member delegation will certainly be selected at arbitrarily to represent the agency at an international meeting. Permit A, B, C and also D represent the directors of the company. What is the probability that

(a) A is selected? (b) A or B is selected? (c) A is no selected?

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

Part (a)

Explanation 1: The probability is `3/6=1/2` since when we choose A, us must choose one that the remaining 3 directors to go through A. There space `C_2^4=6` possible combinations.

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

Part (b)

Explanation 1: The probability of obtaining A or B very first is `2/4=1/2`.

Now to consider the probability of picking A or B as the second director. In this case, the first director needs to be C or D through probability `2/4` (2 specific directors the end of 4 possible).

Then the probability that the second being A or B is `2/3` (2 particular directors the end of the staying 3 directors).

We need to multiply the two probabilities.

So the probability of gaining A or B because that the second director is `2/4 xx 2/3 = 1/3`

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

Explanation 2: Probability that A or B is selected is

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

Explanation 3: If A or B is chosen, then us cannot have actually the situation C and D is chosen. So the probability the A or B is offered by:

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

Part (c)

Probability that A is not selected is `1-1/2=1/2`

### Extension

think about the instance if us are choosing 2 directors from 5. The probabilities would now be:

(a) Probability the A is selected is

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

(b) Probability the A or B is selected is

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

.

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(c) Probability that A is no 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