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Probability, in common terms, is the chance that something is likely to happen. Probability theory is the formal mathematical study of the principles and rules that help us understand how probability works. You can find the probability problems and strategies that the VMT community has worked on in the main Probability page.

1 Introduction to Probability
2 Basic Terminology of Probability Theory
3 Probability of an event
4 Basic Probability Theorems
5 Conditional Probability
6 Independent Events
7 A Review of Concepts on Permutations and Combinations
8 Probability in Our Lives
9 Some Sample Questions and Answers on Probability from Aks Dr. Math Archives
10 Other Resources
11 References

[edit] Introduction to Probability

The study of probability helps us figure out the likelihood of something happening. For instance, when you roll a pair of dice, you might ask how likely you are to roll a seven. In math, we call the "something happening" an "event." The probability of the occurrence of an event can be expressed as a fraction or a decimal from 0 to 1. Events that are unlikely will have a probability near 0, and events that are likely to happen have probabilities near 1. In any probability problem, it is very important to identify all the different outcomes that could occur. For instance, in the question about the dice, you must figure out all the different ways the dice could land, and all the different ways you could roll a seven. See more details about basic conepts at Ask Dr. Math

[edit] Basic Terminology of Probability Theory

Experiment: a process whose outcome is not known in advance with certainty, but all possible outcomes are known.

e.g. toss a coin (you know that the outcome will be either heads or tails in advance)

e.g. draw a card from a deck of 52 cards

e.g. roll a dice

Outcomes: all possible results of an experiment

e.g. for the experiment "toss a coin twice" there are 4 outcomes {H,H}, {H,T}, {T,H}, and {T,T}

Sample Space: is the set of all possible outcomes of a given experiment. It is often denoted by S.

e.g. for the experiment roll a dice S = {Heads, Tails}

Events: are collections of outcomes i.e. subsets of a sample space S

e.g. the event that "He drew an ace from a deck of cards" consists of 4 outcomes, namely {ace of spades, ace of clubs, ace of diamonds, ace of hearts}, which is a subset of S including all 52 possibilities.

Disjoint Events: Two events E_1 and E_2 are said to be disjoint (or mutually exclusive) if they have no common outcomes (i.e. <math>{E_1 \cup E_2}</math> is empty).

e.g. Consider the experiment that a coin is tossed 3 times. Then he sample space S will look like:

S = {HHH, THH, HTH, HHT, HTT, THT, TTH, TTT}

Consider the following events

: a head is obtained on the second toss i.e. {HHH, THH, HHT, THT}

: a tail is obtained on the third toss i.e. {HHT, HTT, THT, TTT}

: no heads are obtained i.e. {TTT}

and

are disjoint events since they do not have a common outcome, yet E_3

and

are not disjoint since TTT is a common outcome for both events.

[edit] Probability of an event

The probability of an event (E) is a number assigned to it in a sample space S. It is denoted by Pr(E) or P(E). This assignment is done according to the following axioms (by Kolmogorov, 1920s):

Axiom 1. For all events E in S, Pr(E) >= 0

Axiom 2. Pr(S) = 1

Axiom 3. For every infinite sequence of pairwise disjoint events E1, E2,… the probability that one or the other will occur is the sum of their individual probabilities. In other words,

Pr (union of E_1, E_2, ...) = Pr(E_1) + Pr(E_2) + ...

If we assume "an equally likely" model of probability where each possible outcome of an experiment has the same chance of occurrence, then P(E) can be defined as

Pr(E) = |E| / |S|

where |E| denotes the number of outcomes in E, and |S| denotes the number of outcomes in S

e.g. If a card is randomly drawn from a deck of 52 cards, then the probability of the event E that "an ace is observed" is Pr(E) = 4 / 52

[edit] Basic Probability Theorems

Theorem 1: Probability of an event with an empty outcome set equals 0

Theorem 2: For any event E in S, Pr(E) = 1 - Pr(E^c), where E^c denotes the complement of event E with respect to the sample space S

Theorem 3: If event A is a subset of event B, then Pr(A) <= Pr(B)

Theorem 4: For any event E in S, 0 <= Pr(E) <= 1

Theorem 5: For any two events A and B in S, Pr (A union B) = Pr(A) + Pr(B) - Pr(A.B) where A.B denotes the intersection of A and B.

[edit] Conditional Probability

Sometimes the probability of an event needs to be updated when a certain related event is observed. The updated probability of the event A after we learn that event B has occured is called the conditional probability of A given B.

This is denoted by Pr(A|B) i.e. probability that event A will occur given event B occurred.

If we know that event B has happened, then the sample space for A has been reduced to those outcomes that are also included in B. This suggests the following definition for the conditional probability of A given B:

Pr(A|B) = Pr(A.B) / Pr(B), provided Pr(B)>0

e.g. Suppose that a dice is rolled, and let A be the event that 6 is observed and let B be the event that an even number is observed. Pr(A) = 1/6 and Pr(B) = 3/6. Now. if we were informed that event B has occurred, what would be the probability that A will occur under this new condition?

Pr(A|B) = Pr(A.B)/Pr(B) = (1/6) / (3/6) = 1/3

Intuitively this result makes sense since when B occurs we know that the outcome has to be either 2, 4, or 6. A will occur only if 6 is observed, so under the new condition the probability will be 1/3.

Hint: Conditional probability can be a useful tool to calculate the probability of the intersection of two events. i.e.

Pr(A.B) = Pr(A|B).Pr(B) = Pr(B|A).Pr(B) (since A.B = B.A)

[edit] Independent Events

Two events A and B are called independent if Pr(A.B) = Pr(A)*Pr(B)

e.g. Suppose a card is drawn from a deck and let A be the event that an ace is observed and let B be the event that a spade is observed. Are these events independent?

Pr(A) = 4/52 since there are 4 aces in a deck of 52

Pr(B) = 13/52 since there are 13 cards with spades in a deck of 52

Pr(A.B) = 1/52 since there is a single "ace of spades" in a deck of 52

Pr(A)*Pr(B) = 4/52 * 13/52 = 1/52 = Pr(A.B), thus A and B are independent by definition

This definition can be intuitively justified by using the formula for the conditional probability. If A and B are independent events, then knowing that B happened won't effect the occurrence of A. In other words, Pr(A|B) = Pr(A). If we plug this to the formula for conditional probability, then we will get:

Pr(A.B) = Pr(A|B).Pr(B) = Pr(A).Pr(B)

[edit] A Review of Concepts on Permutations and Combinations

Permutations

Suppose we want to find the number of ways to arrange the three letters in the word CAT in different two-letter groups where CA is different from AC and there are no repeated letters.

Because order matters, we're finding the number of permutations of size 2 that can be taken from a set of size 3. This is often written 3_P_2. We can list them as:

     CA   CT   AC   AT   TC   TA

Now let's suppose we have 10 letters and want to make groupings of 4 letters. It's harder to list all those permutations. To find the number of four-letter permutations that we can make from 10 letters without repeated letters (10_P_4), we'd like to have a formula because there are 5040 such permutations and we don't want to write them all out!

For four-letter permutations, there are 10 possibilities for the first letter, 9 for the second, 8 for the third, and 7 for the last letter. We can find the total number of different four-letter permutations by multiplying 10 x 9 x 8 x 7 = 5040. This is part of a factorial (see note).

To arrive at 10 x 9 x 8 x 7, we need to divide 10 factorial (10 because there are ten objects) by (10-4) factorial (subtracting from the total number of objects from which we're choosing the number of objects in each permutation). You can see below that we can divide the numerator by 6 x 5 x 4 x 3 x 2 x 1:

          10!     10!    10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1
10_P_4 = ------- = ---- = --------------------------------------
       (10 - 4)!   6!                     6 x 5 x 4 x 3 x 2 x 1
                        = 10 x 9 x 8 x 7 = 5040

From this we can see that the more general formula for finding the number of permutations of size k taken from n objects is:

          n! 
n_P_k = --------  
       (n - k)!

For our CAT example, we have:

         3!     3 x 2 x 1
3_P_2 = ---- = ----------- = 6
         1!         1

We can use any one of the three letters in CAT as the first member of a permutation. There are three choices for the first letter: C, A, or T. After we've chosen one of these, only two choices remain for the second letter. To find the number of permutations we multiply: 3 x 2 = 6.

Combinations

When we want to find the number of combinations of size 2 without repeated letters that can be made from the three letters in the word CAT, order doesn't matter; AT is the same as TA. We can write out the three combinations of size two that can be taken from this set of size three:

     CA   CT   AT

We say '3 choose 2' and write 3_C_2. But now let's imagine that we have 10 letters from which we wish to choose 4. To calculate 10_C_4, which is 210, we don't want to have to write all the combinations out!

Since we already know that 10_P_4 = 5040, we can use this information to find 10_C_4. Let's think about how we got that answer of 5040. We found all the possible combinations of 4 that can be taken from 10 (10_C_4). Then we found all the ways that four letters in those groups of size 4 can be arranged: 4 x 3 x 2 x 1 = 4! = 24. Thus the total number of permutations of size 4 taken from a set of size 10 is equal to 4! times the total number of combinations of size 4 taken from a set of size 10: 10_P_4 = 4! x 10_C_4.

When we divide both sides of this equation by 4! we see that the total number of combinations of size 4 taken from a set of size 10 is equal to the number of permutations of size 4 taken from a set of size 10 divided by 4!. This makes it possible to write a formula for finding 10_C_4:

              10_P_4      10!         10!
    10_C_4 = -------- = ------- = ----------
                4!      4! x 6!    4!(10-4)!

              10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1
           =  --------------------------------------
               4 x 3 x 2 x 1  (6 x 5 x 4 x 3 x 2 x 1)

             10 x 9 x 8 x 7    5040
           = -------------- = ------ = 210
              4 x 3 x 2 x 1     24

More generally, the formula for finding the number of combinations of k objects you can choose from a set of n objects is:

           n!
n_C_k = ----------
       k!(n - k)!

For our CAT example, we do the following:

         3!      3 x 2 x 1     6 
3_C_2 = ------ = ----------- = --- = 3
       2!(1!)    2 x 1 (1)     2

Pascal's TriangleRead more about Pascal's Triangle from Ask Dr. Math

We can also use Pascal's Triangle to find combinations:

  Row 0                   1
  Row 1                 1   1
  Row 2               1   2   1
  Row 3             1   3   3   1
  Row 4           1   4   6   4   1
  Row 5         1   5  10   10  5   1
  Row 6       1   6  15  20   15  6   1

Pascal's Triangle continues on forever - it can have an infinite number of rows. Each number is the sum of the two numbers just above it. For the 1 at the beginning of each row, we imagine that Pascal's triangle is surrounded by zeros: to get the first 1 in any row except row 0, add a zero from the upper left to the 1 above and to the right. To get the 3 in row 4, add the 1 left and above to the 2 right and above.

To find the number of combinations of two objects that can be taken from a set of three objects, all we need to do is look at the second entry in row 3 (remember that the 1 at the top of the triangle is always counted as row zero and that a 1 on the lefthand side of the triangle is always counted as entry zero for that row).

Looking at the triangle, we see that the second entry in row 3 is 3, which is the same answer we got when we wrote down all the two-letter combinations for the letters in the word CAT.

  Row 0                   1
  Row 1                 1   1
  Row 2               1   2   1
  Row 3             1   3   3   1

Suppose we want to find 10_C_4? To use Pascal's Triangle we would need to write out 10 rows of the triangle. This is a good time to use a formula.

More generally, to find n_C_k ("n choose k"), just choose entry k in row n of Pascal's Triangle.

One of the hardest parts about doing problems that use permutations and combinations is deciding which formula to use.

[edit] Probability in Our Lives

A basic understanding of probability makes it possible to understand everything from batting averages to the weather report or your chances of being struck by lightning! Probability is an important topic in mathematics because the probability of certain events happening - or not happening - can be important to us in the real world.

Weather forecasting

Suppose you want to go on a picnic this afternoon, and the weather report says that
the chance of rain is 70%? Do you ever wonder where that 70% came from?

Flipping coins

If you want to know the probability of a coin landing heads, heads is the favorable 
outcome. There is only one way for a coin to land heads, so the numerator of the 
probability fraction is 1.

Batting averages

Let's say your favorite baseball player is batting 300. What does this mean?

The lottery

What if you want to know the probability of winning the lottery? Combination and 
permutation formulas are very useful for solving probability problems.

[edit] Some Sample Questions and Answers on Probability from Aks Dr. Math Archives

Drawing Marbles

A jar contains 2 red, 3 blue, and 4 green marbles. Niki draws one marble from the jar,
and then Tom draws a marble. What is the probability that Niki will draw a green 
marble and Tom will draw a blue marble?

Three Heads in a Row

How many flips of a coin on the average will it take to hit/get three 
heads (or tails) in a row?

Tossing a Coin and Rolling a Die

If you toss a coin and roll a die, what is the probability of obtaining: 
a) heads and a five b) heads or a five c) tails or a two?

Probability of Being Born on Monday

What is the probability that of any seven persons exactly one was born on a Monday?

A Triangle in a Circle

Suppose you randomly place 2 points on the circumference of a circle. 
What is the probability that a 3rd point placed randomly on the circle's 
circumference will form a triangle that will contain the center of the circle?

View More Questions/Answers from Ask Dr. Math Archives

Middle School Probability

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