Probabilty

Probability

Probability is the branch of mathematics that deals with measuring or determining the

likelihood that an event or experiment will have a particular outcome. Probability is based on

the study of permutations and combinations and is also necessary for statistics.17th-century

French mathematicians Blaise Pascal and Pierre de Fermat is usually given credit to the

development of probability, but mathematicians as early as Gerolamo Cardano had made

important contributions to its development. Mathematical probability began when people tried to

answer certain questions that was in games of chance, such as how many times a pair of dice

must be thrown before the chance that a six will appear is 50-50. Or, in another example, if two

players of equal ability, in a match to be won by the first to win ten games, is the other player

suspend from play when one player has won five games, and the other seven, how should the

stakes be divided?

Permutations and combinations are the arrangement of objects. The difference between

permutations and combinations is that combinations pays no attention to the order of

arrangement and permutations includes the order of arrangements of objects.

Permutations is the idea of permuting n number of objects. For example, when n = 3

Probability

Probability is the branch of mathematics that deals with measuring or determining the

likelihood that an event or experiment will have a particular outcome. Probability is based on

the study of permutations and combinations and is also necessary for statistics.17th-century

French mathematicians Blaise Pascal and Pierre de Fermat is usually given credit to the

development of probability, but mathematicians as early as Gerolamo Cardano had made

important contributions to its development. Mathematical probability began when people tried to

answer certain questions that was in games of chance, such as how many times a pair of dice

must be thrown before the chance that a six will appear is 50-50. Or, in another example, if two

players of equal ability, in a match to be won by the first to win ten games, is the other player

suspend from play when one player has won five games, and the other seven, how should the

stakes be divided?

Permutations and combinations are the arrangement of objects. The difference between

permutations and combinations is that combinations pays no attention to the order of

arrangement and permutations includes the order of arrangements of objects.

Permutations is the idea of permuting n number of objects. For example, when n = 3

and the objects are x, y, and z, the permutations or the number of arrangements are xyz, xzy,

yzx, yxz, zyx, and zxy. That means that there is 6 ways that x, y, and z can be arrange. Another

way of finding out the answer is using factorial. Here there are 6 permutations, or 3 · 2 · 1 = 3!

The answer 3! is read as three factorial and that tells you all the positive integers numbers

between 1 and 3. The formula for the factorial is:

n ! = n · (n - 1) · … · 1 permutations

For example, if there are n teams in a league, and ties are not possible, then there are

n ! possible team rankings at the end of the season. A slightly more complicated problem

would be finding the number of possible rankings of the top r number of teams at the end of a

season in a league of n teams. Here the formula is

nPr = n · (n - 1) · … · (n - r 1) = n !/(n - r) !

so that the number of possible outcomes for the first four teams of an eight-team league is

8P4 = 8 · 7 · 6 · 5 = 840.

Now what if we weren’t interested in the order in which the top four teams finished,

but interested in only about the number of the possible combinations of teams that could be in

the top four positions in the league at the end of the season. This is what finding a four-object

combinations out of an eight-object set or 8C4. In general, an r-combination of n objects(n is

greater than r) is the number of distinct groupings of r elements pulled from a set of n

elements. The formula for this number, written (nCr) or (nPr)/r !. For example, the

2-combinations of the three elements a, b, and c are ab, ac, and bc or can be written as 3C2 =

3. The general formula for (nCr) is:

n !/[r !(n - r) !] (This expression can also me written as (nr))

yzx, yxz, zyx, and zxy. That means that there is 6 ways that x, y, and z can be arrange. Another

way of finding out the answer is using factorial. Here there are 6 permutations, or 3 · 2 · 1 = 3!

The answer 3! is read as three factorial and that tells you all the positive integers numbers

between 1 and 3. The formula for the factorial is:

n ! = n · (n - 1) · … · 1 permutations

For example, if there are n teams in a league, and ties are not possible, then there are

n ! possible team rankings at the end of the season. A slightly more complicated problem

would be finding the number of possible rankings of the top r number of teams at the end of a

season in a league of n teams. Here the formula is

nPr = n · (n - 1) · … · (n - r 1) = n !/(n - r) !

so that the number of possible outcomes for the first four teams of an eight-team league is

8P4 = 8 · 7 · 6 · 5 = 840.

Now what if we weren’t interested in the order in which the top four teams finished,

but interested in only about the number of the possible combinations of teams that could be in

the top four positions in the league at the end of the season. This is what finding a four-object

combinations out of an eight-object set or 8C4. In general, an r-combination of n objects(n is

greater than r) is the number of distinct groupings of r elements pulled from a set of n

elements. The formula for this number, written (nCr) or (nPr)/r !. For example, the

2-combinations of the three elements a, b, and c are ab, ac, and bc or can be written as 3C2 =

3. The general formula for (nCr) is:

n !/[r !(n - r) !] (This expression can also me written as (nr))

If repetitions or a given element can be chosen more than once is permitted, then the

last example would also include aa, bb, and cc which adds up to 6. The general formula for the

number of r-combinations from an n-element set is

(n r - 1) !/[r !(n - 1) !]

For example, if a teacher must make a list containing three names from a class of 15,

and if the list can contain a name two or three times and order does not matter, then there are

(15 3 - 1) !/[3 !(15 - 1) !] = 680 possible lists. In the case of r-permutations with repetition

from an n-element set, the formula is nr. For example, to the six 2-permutations of a, b, c

without repetitions (ab, ba, ac, ca, bc, and cb) are added the three with repetitions (aa, bb, and

cc), for a total of 9, which is equal to 32. Thus, if two prizes are to be awarded among three

people, and it is possible that one person could receive both prizes, then nine possible

outcomes exist.

Finally, suppose there are n1 objects of one type, n2 of another type, on to n3 objects of

some third type. Let n = n1 n2 … n2. In how many ways can these objects be arranged

and also keeping order? The answer is n !/(n1 !n2 ! … n3 !), One example is how many letters

of the word banana can be arranged? 60 letters because 6 !/(3 !2 !1 !) = 60. This is also the

coefficient of x3y2z1 in (x y z)6.

The most common use of probability is used in statistical analysis. For example, the

probability of throwing a 7 in one throw of two dice is 1/6, and this answer means that if two

dice are randomly thrown a very large number of times, about one-sixth of the throws will be 7s.

This method is most commonly used to statistically determine the probability of an outcome that

cannot be tested or is impossible to obtain. So, if long-range statistics show that out of every 100

last example would also include aa, bb, and cc which adds up to 6. The general formula for the

number of r-combinations from an n-element set is

(n r - 1) !/[r !(n - 1) !]

For example, if a teacher must make a list containing three names from a class of 15,

and if the list can contain a name two or three times and order does not matter, then there are

(15 3 - 1) !/[3 !(15 - 1) !] = 680 possible lists. In the case of r-permutations with repetition

from an n-element set, the formula is nr. For example, to the six 2-permutations of a, b, c

without repetitions (ab, ba, ac, ca, bc, and cb) are added the three with repetitions (aa, bb, and

cc), for a total of 9, which is equal to 32. Thus, if two prizes are to be awarded among three

people, and it is possible that one person could receive both prizes, then nine possible

outcomes exist.

Finally, suppose there are n1 objects of one type, n2 of another type, on to n3 objects of

some third type. Let n = n1 n2 … n2. In how many ways can these objects be arranged

and also keeping order? The answer is n !/(n1 !n2 ! … n3 !), One example is how many letters

of the word banana can be arranged? 60 letters because 6 !/(3 !2 !1 !) = 60. This is also the

coefficient of x3y2z1 in (x y z)6.

The most common use of probability is used in statistical analysis. For example, the

probability of throwing a 7 in one throw of two dice is 1/6, and this answer means that if two

dice are randomly thrown a very large number of times, about one-sixth of the throws will be 7s.

This method is most commonly used to statistically determine the probability of an outcome that

cannot be tested or is impossible to obtain. So, if long-range statistics show that out of every 100