**Event**: the outcome from an experiment.

**Experiment**: the process which results in an outcome.

**Random Experiment**: an experiment where there is uncertainty over which outcome will occur.

Tossing a coin is a random experiment. One event is 'head' and another 'tail'.

Sample Space U: the total of all possible events. tossing a coin has a sample set of 2: [H, T]

n(U) = 2

The probability of an event $P(E)$ is the relative abundance of the event in the sample space n(E), divided by the total spample space:

For example, when two die are thrown, there are 6 combinations which add up to 7, out of 36 possible combinations of the two die. The probability of throwing a 7 from two die is:

$P(E)={n(E)}/{n(U)} = 6/{36} = 1/6$

Probability has the following laws:

- The sum of all possible events = 1
- P(E) = 1 - P'(E)
- 0 ≤ P(A) ≤ 1
- P(A) = n(A)/n(U), where U is the total number elements in the sample space (U)

The probability of an outcome = 1 - the probability of it not occurring

A probability lies between zero and one.

Events are mutually exclusive if the cannot occur at the same time. This means there is no overlap of their probability set.

For example, the chance of getting a head is exactly 50%, and a tail 50%. There is no outcome in a series of trials where a head and a tail occur simultaneously.

The probability of an outcome = 1 - the probability of it not occurring: P(E) = 1 - P'(E).

P'(E) is the 'complement' to the event E, or all outcomes where E does not occur.

e.g., the probability of drawing an ace from a deck is $4/{52} = 1/{13}$. The probability of not drawing an ace is P'(Ace) = 1 - P(Ace) = 1 - $1/{13}$ = ${12}/{13}$.

A dice can have any number of sides (including 2, which is the same as tossing a coin).

Player A throws first. If he gets the highest number on the die, he scores a point. If he does not, Player B gets to throw. If B throws the highest value on the die, B wins, otherwise A throws again, and so on till one of them throws the highest value.

Player A obviously has the advantage, since he can win without B even throwing. But how much of an advantage is this, and how does changing the number of sides on the die change player's A's chances of winning?

The greater the number of games played the closer the distribution of scores is to the probabilities. Experiment with the number of games and see what effect it has.

The probability of two independent events occurring is the product of the two probabilities.

$P(E_1$ and $E_2) = P(E_1) × P(E_2)$

For example, the probability of getting a six and a six on two consecutive throws of the same dice is:

$P(6$ and $6) = P(6) × P(6) = 1/6 × 1/6 = 1/{36}$

However, if the two events not independent the probability will be different.

For example,if a bag contains 6 red balls, and 6 white balls, the probability of drawing two red balls on consecutive draws, where the first ball is replaced is:

$P($red and red$) = 6/{12} × 6/{12} = {1}/{4}$

However, if the first ball is not replaced before the second draw, the probability of drawing two red balls is now:

$P($red and red$) = 6/{12} × 5/{11} = {30}/{132}$

This is lower than the independent events of two balls being drawn with replacement.

Another way to calculate probability is by relative frequency. This is best done by creating a table:

Students of a school conduct a survey of colours of cars passing their school. They recorded the number of cars of a particular colour that passed in 1 hour as frequency:

Colour | Frequency |
---|---|

white | 22 |

green | 12 |

yellow | 14 |

brown | 5 |

black | 8 |

red | 16 |

silver | 32 |

1. Calculate the probability that the next car will be white.

2. What is the probability that the next car will not be silver?

$P(A) = {n(A)}/{n(U)}$

U is the sample space, or the total frequency of all colours of cars: 109

White: P(white) = n(white)/n(U)

= 22/109 = 0.202 = 20.2%

Not silver: P(silver)' = n(silver)'/n(U)

= (109 - 32)/109 = 0.706 = 70.6%

Content © Andrew Bone. All rights reserved. Created : February 11, 2014

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