Math Basics

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Math Basics

Postby Giacomo on May 5th, 2008, 5:44 am 

Ratio

A ratio is usually a quotient of two numbers, like 2/3.

When you calculate a ratio, the two things must have the same units.

Ratios tell how one number is related to another number.

A ratio may be written as A:B or A/B or by the phrase "A to B".

A ratio of 1:5 says that the second number is five times as large as the first.

The following steps will allow determination of a number when one number and the ratio between the numbers is given.

Example: Determine the value of B if A=6 and the ratio of A:B = 2:5

. Determine how many times the number A is divisible by the corresponding portion of the ratio. (6/2=3)

. Multiply this number by the portion of the ratio representing B (3*5=15)

. Therefore if the ratio of A:B is 2:5 and A=6 then B=15

How to Determine a Ratio

Ratios represent how one quantity is related to another quantity.

A ratio may be written as A:B or A/B or by the phrase "A to B".

A ratio of 1:5 says that the second quantity is five times as large as the first.

The following steps will allow a ratio to be determination if two numbers are known.

Example: Determine the ratio of 24 to 40.

. Divide both terms of the ratio by the greatest common factor (24/8 = 3, 40/8=5)

. State the ratio. (The ratio of 24 to 40 is 3:5)


Unit Rates

A rate is a ratio that is used to compare different kinds of quantities.

A unit rate describes how many units of the first type of quantity corresponds to one unit of the second type of quantity.

Some common unit rates are miles (or kilometers) per hour, cost per item, earnings per week, etc. In each case the first quantity is related to 1 unit of the second quantity.
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Postby Giacomo on May 5th, 2008, 5:46 am 

Odds:

The odds in favour of an event or a proposition are the quantity p / (1 − p), where p is the probability of the event or proposition.

In other words, an event with m to n odds would have probability n/(m + n).

E.g. If you choose a random day of the week, say Sunday, the probability of choosing Sunday is p = 1/7, the odds would be

(1/7) / (1 - 1/7) = (1/7) / (6/7) = 1/6

Odds are not quoted in this format to the general public to avoid confusion with probabilities.

So, odds is expressed in the form

6 to 1, or 6-1, or 6/1

al read as six-to-one.

If the odds are 5:3 that a certain horse will win. This terminology is related to ratios.

In particular, 5:3 is associated with the ratio

5/(5+3) = 5/8.

That means, on average, the horse will win 5 out of every 8 races.
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Postby Giacomo on May 20th, 2008, 4:35 am 

Probability Review



Independent Events

Suppose you choose a random number from 1 to 3. Then the probability of choosing the number 1 is 1/3. If you were to do this twice in a row, then the probability of choosing 1 both times would be 1/3 times 1/3, that is, 1/9. In general, if you have two unrelated (also called independent) events, then the probability of them both occurring is the product of the probabilities of each of the individual events.


Probabilities Sum to One

Each time we run a trial, with probability one we get some outcome. Therefore, the sum of the probabilities of all possible outcomes must be one. This allows you to calculate the probability of an event by finding the probability that the event does not occur and subtracting this probability from one. In many situations this is much easier than calculating the probability directly. For example, in the previous situation, if you want to find the probability of choosing any pair except for two 1's, then, since you know that the probability of actually getting two 1's is 1/9, the probability of not getting the pair of 1's must be 1 - [probability of getting the pair of 1's] = 1 - 1/9 = 8/9.


Factorials

5 people can stand in a line in 5! = 5 · 4 · 3 · 2 · 1 = 120 different ways (the first person in line can be any of the five people, the second person any of the remaining four, the third person any of the remaining three, and so on...). The symbol 5! is read as "five factorial". For n people the answer is n! = n · (n - 1) · · · 2 · 1 ways.


Permutations

A club with 23 members can elect a president, vice-president, secretary, and treasurer in 23 · 22 · 21 · 20 ways. In general, for n people and m offices there are nPm = n · (n-1) · (n-2) · · · (n-m+1) = n!/(n-m)! different ways.


Combinations

From a group of size 23 you can pick a committee of size four in (23 · 22 · 21 · 20)/4! ways (if we make a list of the permutations in the previous problem, each committee appears 4! times. When choosing the committee, none of the 4 positions are distinguished, so the order in which we choose the four committee member does not matter). Picking m things out of n can be done in Cn,m = n!/[m!(n-m)!] different ways. This is often read as "n choose m". The numbers Cn,m are also called binomial coefficients.


Binomial Distribution

The binomial distribution tells us that the probability of getting m successes in n trials when the success probability is p is

Cn,m pm (1 - p)n-m

The first factor gives the number of ways to pick the m trials on which success happens. The second gives the probability of any particular outcome with m successes and n-m failures.


Expected value

In a situation where you have a set of outcomes each with a given probability of occurring and each with some numerical "payoff" value, you might be interested in what average payoff you will earn. This average payoff, also called the expected value, is calculated by taking the sum over all possible outcomes of the probability of each outcome times the payoff for that outcome. For example, suppose that you are to choose one door among three possible choices. Behind one door there are 6 dollars, behind another one there are 12 dollars and behind the third door there is nothing, but you don't know what is behind each door. Then the expected payoff you will earn is the sum over all possible doors of probability of choosing each door (in this case 1/3 for each door) times the payoff for that door:


6 * 1/3 + 12 * 1/3 + 0 * 1/3 = 6 dollars
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Postby Giacomo on May 24th, 2008, 1:26 am 

Permutations with repetitions

When the order matters, and an object can be chosen more than once, the number of permutations is

nr

where n is the number of objects from which you can choose and r is the number to be chosen.


Example:
--------

If you have the letters A, B, C, and D and you wish to discover the number of ways to arrange them in three letter patterns (trigrams)

order matters (e.g., A-B is different from B-A, both are included as possibilities)
an object can be chosen more than once (A-A possible)

You find that there are 43 or 64 ways.


Permutations without repetitions

When the order matters and each object can be chosen only once, then the number of permutations is

(n)r = n!/(n-r)!

where n is the number of objects from which you can choose, r is the number to be chosen and "!" is the standard symbol meaning factorial.


Example :
---------

If you have five people and are going to choose three out of these, you will have 5!/(5 − 3)! = 60 permutations.

If n = r (meaning the number of chosen elements is equal to the number of elements to choose from; five people and pick all five)

then the formula becomes

n!/(n-n)! = n!/0! = n!, where 0! = 1

For example, if you have the same five people and you want to find out how many ways you may arrange them, it would be 5! or 5 × 4 × 3 × 2 × 1 = 120 ways. The reason for this is that you can choose from 5 for the initial slot, then you are left with only 4 to choose from for the second slot etc. Multiplying them together gives the total of 120.


Combinations without repetitions

When the order does not matter and each object can be chosen only once, the number of combinations is the binomial coefficient:

Cn,k = n! / k!(n-k)!

where n is the number of objects from which you can choose and k is the number to be chosen.

Example :
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If you have ten numbers and wish to choose 5 you would have 10!/(5!(10 − 5)!) = 252 ways to choose. The binomial coefficient is also used to calculate the number of permutations in a lottery.


Combinations with repetitions

When the order does not matter and an object can be chosen more than once, then the number of combinations is

(n+k-1)! / k!(n-1)!

where n is the number of objects from which you can choose and k is the number to be chosen.

Example :
---------

If you have ten types of donuts (n) on a menu to choose from and you want three donuts (k) there are (10 + 3 − 1)! / 3!(10 − 1)! = 220 ways to choose
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Postby Giacomo on June 30th, 2008, 7:25 pm 

PROBABILITY -- PART ONE



Experiment

An experiment is an act for which the outcome is uncertain.

Examples of experiments are rolling a die, tossing a coin, surveying a group of people on their favorite soft drink or beer, etc.



Sample Space

A sample space S for an experiment is the set of all possible outcomes of the experiment such that each outcome corresponds to exactly one element in S. The elements of S are called sample points. If there is a finite number of sample points, that number is denoted n(S), and S is said to be a finite sample space.

For example, if our experiment is rolling a single die, the sample space would be S = {1, 2, 3, 4, 5, 6}.
If our experiment is tossing a single coin, our sample space would be S = {Heads, Tails}.
If our experiment is surveying a group of people on their favorite soft drink, our sample space would be all of the soft drinks or beer on the survey.



Event

Any subset E of a sample space for an experiment is called an event for that experiment.

For example, if our experiment is rolling a single die, an event E could be rolling an even number, thus E = {2, 4, 6}.

If our experiment is tossing a single coin, an event E could be tossing a Tail, where E = {Tails}.

If our experiment is surveying a group of people on their favorite soft drink, an event E could be picking a diet soft drink.

If beer, an event E could be picking one brand of beer.

Giacomo wrote:Empirical Probability

Finding the probability of an empirical event is specifically based on direct observations or experiences.

For example, a survey may have been taken by a group of people. If the data collected is used to find the probability of an event tied to the survey, it would be an empirical probability. Or if a scientist did research on a topic and recorded the outcome and the data from this is used to find the probability of an event tied to the research, it would also be an empirical probability.


Empirical Probability Formula :

P(E) = (number of times event E occurs) / (total number of observed occurences)

P(E) represents the probability that an event, E, will occur.

The numerator of this probability is the number of times or ways that specific event occurs.

The denominator of this probability is the overall number of ways that the experiment itself could occur.

Equiprobable Space


A sample space S is called an equiprobable space if and only if all the simple events are equally likely to occur.

Some quick examples of this are:


A toss of a fair coin. It is equally likely for a head to show up as it is for a tail.

Select a name at random from a hat. Since it is at random, each name is equally likely to be picked.

Throwing a well balanced die. Each number on the die has the same amount of chance of coming up.

Theoretical Probability


Theoretical probability is finding the probability of events that come from an equiprobable sample space or, in other words, a sample space of known equally likely outcomes.

For example, finding various probabilities dealing with the roll of a die, a toss of a coin, or a picking of a name from a hat.

Theoretical Probability Formula :

P(E) = (number of outcomes in E) / (number of outcomes in S) = n(E)/n(S)

If E is an event of sample space S, where n(E) is the number of equally likely outcomes of event E and n(S) is the number of equally outcomes of sample space S, then the probability of event E occurring can be found using the Theoretical Probability Formula above.

Mutually Exclusive

In general, events E and F are said to be mutually exclusive if and only if they have no elements in common.

For example, if the sample space is rolling a die, where S = {1, 2, 3, 4, 5, 6},
and E is the event of rolling an even number, E = {2, 4, 6}
and F is the event of rolling an odd number, F = {1, 3, 5},

E and F are mutually exclusive, because they have NO elements in common.

Now let's say that event G is rolling a number less than 4, G = {1, 2, 3}.

Question : Would event G and E be mutually exclusive?

If you say no, you are correct, they have one element, the number 2, in common. G and F would not be mutually exclusive either.

Properties of Probability


Property #1: 0 =< P(E) =< 1 (=< means equal or less)

The probability can be equal to 0 or less or equal to 1, it will never exceed 1, since the bottom number of the probability is the total number - which is the highest number.


Property #2: P(not E) = 1 - P(E)

N.B. I will use the symbol "~" for not. So, ~ E means not E. We can now write:

P(~ E) = 1 - P(E)


Property #3: P(E) = 1 - P(~ E)

This is just like Property 2 in reverse.


Property #4: "Or" probabilities with mutually exclusive events

P(E OR F) = P(E) + P(F)

We could use the symbol "\/" for OR. We could rewrite it as follows:

P(E \/ F) = P(E) + P(F)

Since we are dealing with sets that are mutually exclusive, this means they have no elements in common. So we can just add the two probabilities together without running a risk of having something counted twice.


Property #5: "Or" probabilities with events that are NOT mutually exclusive

P(E OR F) = P(E) + P(F) - P(E AND F)

We could use the symbol "/" for AND. We could then rewrite it as follows:

P(E \/ F) = P(E) + P(F) - P(E /\ F)


Since we are dealing with events that are NOT mutually exclusive, we run the risk of elements being counted twice if we just add them together as in Property 4 above.

You need to subtract the intersection to get rid of the elements that were counted twice. In other words, you may have some elements in common, so if we add the number of elements in E to the number of elements of F, we may be adding some elements twice, so to avoid this we need to subtract the number of elements in the intersection of the two events - which would be all the elements that are in both sets.

Independent Events

Two events are independent of each other if the outcome of one event does not affect the outcome of the other event.

E and F are Independent Events if an only if

P(E AND F) = P(E)*P(F)

P(E /\ F) = P(E)*P(F)
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Postby Giacomo on April 24th, 2009, 4:30 pm 

Concept mentioned by PeSla in PCF: The Fano Plane
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