1.  What is probability?  Simply put, it’s the chance of an event occurring
 
 

Definitions: Probability experiment -  A process that leads to well-defineed results, or outcomes. Outcome -    The result of a single trial of a probability experiment Sample space -   The set of all possible outcoomes of a probability experiment Event -    One or more outcomes of a probability experiment Simple event - one outcome Compound event - two or more outcomes

 

Probability - 3 types Classical, Empirical, and Subjective

 

Type 1:  Classical - Uses sample spaces to determine the numerrical probability that an event will occur  a.  Assumes that all outcomes in the sample space are equally probable to occur  b.  Don’t actually have to perform the experiment to determine the probability - you can do this mathematically! Examples: Coin flip. Either heads or tails will come up. Since there are only two outcomes, there is a 1 in 2 chance of getting either heads or tails. Both heads and tails are just as equally likely to happen on a given toss of the coin. Roll of the dice. A single die, when rolled, will only have six possible outcomes, 1 through 6. 1,2,3,4,5, and 6 are equally likely to take place on a single roll (the chances of getting a "1" is 1/6, the chances of getting a "3" is 1/6, the chances of getting a "3" is 1/6, etc.) Formula for Classical Probability:   P(E) = n(E)/n(S)  where:           n(E) = number of outcomes in event “E”           n(S) = total # of outcomes in sample space “S”  **Notes: 1.  Probabilities are generally expressed as percentages, decimals, or fractions 2.  Decimals should rounded to 2 or 3 places 3.  Fractions should be reduced as far as possible  (8/16 = 1/2, e.g.) 4.  Probabilities will always be a number between 0 and 1, inclusive when using fractions or decimals 5.  The sum of the probabilities of all outcomes in a sample space is always one I realize that the terms "n(E) = number of outcomes in event E" and " n(S) = total # of outcomes in sample space S” sound quite strange. Look at it this way: Pretend  there is a box (see below) with 40 colored marbles (13 red, 12 blue, 9 orange, 6 green).  Now, I want to know the probability of getting a red ball when I pick one at random. 1. The n(S) ( the total # of outcomes in sample space “S”) is 40. 2. The n(E) ( number of outcomes in event “E”) is 13. 3. The probability of getting a red ball is P(E) = n(E)/n(S), which is  (13/40) or 0.325 or 32.5%.   O O O O               O O                        O O    O   O     O        OO                                 O                  O          O   O   O  O O O O       O O     O                                                                    O O O O O O                  O O O O                                                 O O O O O <<<<<----Here's the Box! Now, I want to know the probability of getting either a red ball or a blue ball when I pick one at random. 1. The n(S) ( the total # of outcomes in sample space “S”) is 40. 2. The n(E) ( number of outcomes in event “E”) is 25. 3. The probability of getting EITHER a red ball or a blue ball  is  P(E) = n(E)/n(S),  which is  (25/40) or 0.625 or 62.5%.

 

Complementary Events The complement of an event "E" is the set of outcomes in the sample space that are not included in the outcomes of the event E, and are denoted as E.  Rule for Complementary Events P(E) = 1 - P(E) -or-     P(E) = 1- P(E) ** aka :  If the probability of an event or its complement is known, then the other can be found by subtracting the known probability from 1

 
 

Type 2:  Empirical Probability - Relies on actual experience or experiments to determine the likelihood (probability) of outcomes What if someone weighted the dice (CHEATER!), and you found out about it? You watch the dice for a while, and find out that "2" is the preferred number? You watch the die for a while (say 100 throws) and "2" come up over half the time (you keep count). This is found out by experience - EMPIRICISM - or Empirical probabilty! Formula for Empirical probability:    P(E) = f/n  where:  f = frequency for the chosen class n = total frequencies in the distribution **Note:  There's another side of this, though.  While the probability of getting a ‘heads’ when flipping a coin once is exactly 1/2, when it’s tossed 100 times it may not come up heads exactly 50 times (due largely to chance variation).  It's possible that through simple good fortune, heads may come up 55 times out of 100 - this might trick someone into thinking that there is a preference for heads.  However, as the number of trials increases, the empirical probability of getting a ‘heads’ will approach 1/2, if the coin is balanced correctly.  This phenomenon is known as The Law of Large Numbers and will hold true for any type of gambling game.

 

Type 3 - Subjective Probability - a probability value based on an educated guess or estimate, employing opinion and inexact information Examples: Doctor says there’s a 30% chance a man has cancer Seismologist says there’s a 40% chance that Mount St. Helens will erupt again this decade TV weatherman says there’s a 70% chance of rain tomorrow No equations! This is an educated guess, based on expertise, qualitative experience (not empirical), etc.

 
 

The Addition Rules for Probability   Many problems involve finding the probability of two or more events occurring.  There are different ways of determining probability, based on whether the events are mutually exclusive. ** Two events are considered mutually exclusive if they cannot occur at the same time (i.e., they have no outcomes in common) Addition Rule 1- When two events A and B are mutually exclusive, the probability that either A or B will occur is given by:    P(A or B) = P(A) + P(B) ** But what if the events are NOT mutually exclusive? Addition Rule 2 - If events A and B are NOT mutually exclusive, then the probability that either A or B will occur is given by:    P(A or B) = P(A) + P(B) - P(A and B)   ** Note - the addition rules can be extended for 3 or more events in the following ways: Addition Rule 1 (mutually exclusive)    P(A or B or C) = P(A) + P(B) + P(C) Addition Rule 2 (not mutually exclusive)    P(A or B or C) =             P(A) + P(B) + P(C)           - P(A and B)          - P(A and C)           - P(B and C)          + P(A and B and C) Addition Rule 2 (not mutually exclusive) -------OUCH!!!!!! WOW! What are you saying? WHAT IS THIS?!?!?!?!? OK, this is going to be a long explanation - get a cup of coffee..... START OF Explanation  for Addition Rule 2 (not mutually exclusive) Look  at this graphic... This figure shows the Black Box "D", which contains the numbers - 1 through 25.  Since Box "D" contains all the numbers - 1 through 25 - it also includes the numbers in Circle "A", Circle "B" and Circle "C".  Notice:   Circle "A"  has "1", "2", "3", "4", "5", "6", "7" Circle "B"  has "1", "3", "8", "9", "10", "11", "12", "13" Circle "C"  has "2", "3", "4", "8", "9", "14", "15", "16"   Circles "A" and "B" have "1" and "3" in common.  Circles "A" and "C" have "2", "3", and "4" in common.  Circles "B" and "C" have "3", "8", and "9" in common.  All circles have "3" in common. Here the question: What is the probabilty of getting a number that is in "A", "B", or "C" (assuming a random pick)?   Probabilty of getting a number from Circle "A" 7/25 Probabilty of getting a number from Circle "B" 8/25 Probabilty of getting a number from Circle "C" 8/25 TOTAL of "A", "B", or "C": 23/25 But WAIT!  The numbers  "17", "18", "19", "20", "21", "22", "23", "24" , and "25" are NOT IN "A", "B", or "C"! That's 9 out of 25 total numbers - 9/25, or 0.36 (or 36%). Now get this: These numbers represent the COMPLEMENT of our intended goal (picking a number that IS IN "A", "B", or "C"!  So, the actual  probabilty of getting a number that is in "A", "B", or "C" (assuming a random pick) is : 1-0.36 = 0.64 -or- 100%-36%=64% So, we know the REAL ANSWER is 64%. What happened? Well, look at the graphic again. We counted the probabilty of getting a number from Circle "A".  Then, we counted the probabilty of getting a number from Circle "B".  PROBLEM: when we counted "A", we already included "1" and "3". When we counted "B", we included  "1" and "3" again  - in ERROR! Then, we counted the probabilty of getting a number from Circle "C".  PROBLEM: when we counted "B", we already included "3" and "8", and "9". When we counted "C", we included  "3", "8" and "9" again  - in ERROR! HOW DO WE FIX THIS? Simple: SUBTRACT ALL THE DOUBLE  COUNTED ITEMS AT LEAST ONCE! For the point when we counted "A" and "B", we counted  "1" and "3" twice.  So, let's subtract out the probability of getting a number that is in BOTH "A" and "B". This is:          - P(A and B) This is from the equation for Addition Rule 2 (not mutually exclusive) - above.  For the point when we counted "A" and "C", we counted  "2", "3", and "4" twice.  So, let's subtract out the probability of getting a number that is in BOTH "A" and "C". This is:           - P(A and C)  For the point when we counted "B" and "C", we counted  "3", "8", and "9" twice.  So, let's subtract out the probability of getting a number that is in BOTH "A" and "C". This is:           - P(B and C)  BUT WAIT!!! We added in the number "3" a total of three times, then subtracted it three times! so we did not actually include "3" yet! The problem here is that "3" is in circles "A", "B", AND "C"! So, then, let's ADD "3" back in by adding in the probability of getting a number that is in "A" and "B" AND "C". This is: + P(A and B and C) So, here's the calculation: P(A or B or C) =             P(A) + P(B) + P(C)           - P(A and B)          - P(A and C)           - P(B and C)          + P(A and B and C)     English way of saying it: Equation term Numerical result  Probabilty of getting a number from Circle "A" P(A) +7/25 Probabilty of getting a number from Circle "B" +P(B) +8/25 Probabilty of getting a number from Circle "C" +P(C) +8/25 Subtract out the probability of getting a number that is in BOTH "A" and "B" - P(A and B) -2/25 Subtract out the probability of getting a number that is in BOTH "A" and "C" - P(A and C)  -3/25 Subtract out the probability of getting a number that is in BOTH "B" and "C" - P(B and C)  -3/25 Add in the probability of getting a number that is in "A", "B" &  "C" + P(A and B and C) +1/25 TOTAL of "A", "B", or "C": P(A or B or C) =  16/25  (It works!) END OF Explanation  for Addition Rule 2 (not mutually exclusive) _____________________________________

I hope that made sense!

:-)