Lesson 5 - Probability

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.

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:

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)
_____________________________________

:-)

 Multiplication Rules and Conditional Probability Used to find the probability of two or more events that occur in sequence. Independent events - Two events A and B are said to be independent if the fact that A occurs does not affect the probability of B occurring. Example: Two die are rolled separately (and one does not hit the other). That fact that the first was a "1" does not affect the outcome of the other die - it could be 1,2,3,4,5, or 6. Multiplication Rule 1 - When two events are independent, the probability of BOTH of them occurring is given by:    P(A and B) = P(A)  P(B) or , this is the also stated as:    P(A and B) = P(A)  x  P(B) In English: The probability of both "A" and "B" happening at the same time is equal to the probability of "A" occurring, TIMES the probability of "B" occuring.   We can also extend this rule to 3 or more events: P(A and B and C and ... and K) = P(A) P(B) P(C) ... P(K) But what if the events under consideration arent independent? Dependent events - When the outcome or occurrence of the first event affects the outcome or occurrence of the second event in such a way that the probability is changed, then the events are said to be dependent. **This brings up the concept of Conditional Probability   The probability of an event B occurring given that an event A has already occurred is called conditional probability and has the notation:  P(B|A):  the probability that B occurs given that A already occurred Huh? OK, let's play a card game. Pick a card - any card. First one to get an ACE Wins! 1.  There are 52 cards in the deck (NO JOKERS!) 2. Your chances of picking an ACE is 4/52, or 1/13. 3. You pick, and get a Deuce. YOU DO NOT PUT THE DEUCE BACK IN THE DECK! 4. On your next pick, your chances of getting an ACE is 4/51.        NOTE: Your chances of getting an ACE are a little better now!  5. You pick, and get another Deuce. YOU DO NOT PUT THE DEUCE BACK IN THE DECK! 6. On your next pick, your chances of getting an ACE is 4/50, or 2/25.        NOTE: Your chances of getting an ACE are again a little better now!  So, the previous picking  of a card AFFECTS your chances of SUCCESS! This is "Conditional Probability". Multiplication Rule 2 - When two events are dependent, the probability of both occurring is given by: P(A and B) = P(A) P(B|A) For 3 or more events: P(A and B and C) = P(A) P(B|A) P(C|B) Remember we found that multiplication rule 2 was given by:    P(A and B) = P(A) P(B|A) We can use this formula and simple math to find the conditional probability!  Divide through both sides by P(A) to get:    P(A and B)/P(A) = P(B|A) or rewritten, this becomes:    P(B|A) = P(A and B)/P(A)  the Formula for Conditional Probability I realize that "P(B|A)" looks odd.  We say this as "The probabilty of 'B' taking place , given that  'A' just took place".

 Probabilities involving the phrase at least When dealing with problems of this nature (see examples 5-35, 36 & 37), it is usually easier to find the probability when the event in question doesnt happen at all.  This is then the complementary to at least, and we know from before that: 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 So simply determine the probability for the event not occurring, then subtract that probability from 1 to get the probability of the at least case.

 Probability and Counting Techniques We often need to use a combination of techniques to determine probabilities in certain situations.  Using the counting rules described in Chapter 4, plus the probability rules in this chapter, we can find probabilities for more complex problems.

 HOMEWORK: No reading assignment. Go to Lesson 6.