PROBABILITY

~the heart of Statistics (probability distributions, hypothesis testing, & so on)

~probability vs. possibility (almost anything is possible as time passes)(can you give me an impossible event?)

~experiment: (procedure) the activity we are involved with (rolling dice, taking a survey, betting a horse, & so on)

~event: an outcome (getting a 7 with a pair of dice, people who fly in planes from a sample, horse A wins, & so on)

~sample space: all possible outcomes or events.

~The dice triangle: 1 way of rolling a 2, 2 ways for a 3, 3 ways for a 4, & so on. Note that a 7 can be rolled 6 ways. (most likely roll)
                                7
                            6       8
                        5               9
                    4                      10                
              3                                11
          2                                          12
          1  2   3  4  5  6   5  4   3   2    1 ----> 36 possible outcomes in the sample space.

~Definition:  The probability of an event A = P(A) = k/n, where k= the number of ways event A can occur  n= the total number of ways of all possible events in the sample space. (the bottom number in the probability includes the number in the top).

~Note:  Venn diagrams are very popular for visualizing the events in a given sample space.

~Types of probability:

~1)  a priori: mathematical, known in advance, doesn’t change (dice, cards, coins, roulette, & so on)
~2)  statistical or experimental: based on trials or collected data (horse racing, weather, & so on). Not fixed.

~Ex. of 2):  A sample of 855 randomly selected adults is taken. 710 said they have flown in an airplane. Find the Probability that a randomly selected adult has flown in an airplane.
Let F=event of flying. So, P(F)=710/855 = .83 or 83%  (3 ways to express probability)

~Ex. of 1):   Find the probability of rolling an "hard 8" with a pair of dice. P(4 & 4)=1/36

~Ex:  Assuming having a male or female born is equally likely, find the probability that a couple has 2 girls & one boy when having 3 children. List the sample space for this experiment first. Sample space: GGB,GBG,BGG,GGG,BBB,BBG, BGB, GBB.  P(2G&1B)= 3/8 = .375 or 37.5% chance

~Note:  If P(A)=0, then event A can not occur.
            If P(A)=1, then event A must occur.
            If P(A)>.5, then event A is likely to occur.
            If P(A)<.5, then event A is not likely to occur.
            If P(A)=.5, then it is equally likely that A does or doesn’t occur
           (this is know as a 50-50 chance)

~Generally in most all problems,  0 < P(A) < 1  (answers >1 or <0 make no sense)

~Note:  the Complement of A, designated as A bar or A' would be all points (events) in the sample space outside of A. (all outcomes for which A does not occur).

~Note: P(A) + P(A') = 1 so, P(A) = 1 - P(A'), this is the "backdoor" approach to probability & must be used in many problems.
 
~Odds: Come from probabilities, so to get odds, you must get the probability first.

~Ex:  P(A)=1/6 , A is not likely to occur , odds are against A occurring , 5 to 1 against or 5:1 against
 P(B)=4/7  , B is likely to occur , odds are 4 to 3 or 4:3 in favor
P(L)=½ ,  50-50 chance  , odds are even  , 1 to 1 or 1:1.

~many textbooks confuse students, i.e., they state 3 to 7 in favor?, should read 7 to 3 against)(the bigger number should always be placed first)

~at many betting parlors, you will see odds like 2-7, which means that you must bet $7 to win $2. Also, more common, +120, means you need to bet 100 to win 120, or -170, bet 170 to win 100

~Note: the 2 numbers for the odds always add to the bottom number of the probability.

~reduce odds, if you can ( i.e., 66 to 11 is reduced to 6 to 1, 4 to 2 is reduced to 2 to 1, & so on)

~Note: know how to go from odds back to the probability. ( i.e., 3 to 2 against A, P(A)=2/5)

~Note: Payoff odds (involved in games of chance) are usually not the same as the real odds. The game presenter (casino) is usually in business to make money, so odds will favor the "house". So, in the long run, the "house" will always win. But, in the short run, anything is possible. "luck" is usually used for short-term success or even a run of successes.

~To compute the payoff odds use: NET PROFIT to AMOUNT OF BET

~Note: A game is considered fair, if the payoff odds are the same as the real odds. (no body wins in the long run)

~Ex:  In roulette there are 38 numbers on the wheel. Usually, 18 red, 18 black, 0 & 00 green. Placing a bet on one position gives you a P(win) = 1/38. So, the real odds are 37 to 1 against you. However, the casino will pay 35 to 1. So, for a $1 bet on a number & it comes out, the casino will pay you $35. Your net gain is $35 (you keep the $1 you bet). There are many other bets you can make that will increase your odds of winning but will decrease the payoff. The worst odds in a casino is Keno (stay away). The best odds are from Blackjack, Craps (dice table), & Baccarat. You get the biggest "bang" for your buck at Slots (most popular casino game & produces the most revenue), although considered one of the worst games for the odds. (see my discussion on casino slot machines under the statistics link of selective topics)

~Many professional gamblers play Poker, since you don’t play against the house, but rather against other players. Many can make a living, if they are skilled in reading other players body language.

~Probability Rules:

Addition Rules:

1)  P(A or B) = P(A) + P(B), if A and B are mutually exclusive (disjoint)(Events A & B cannot happen at the same time)

    Ex:  P (getting a heart or club when one card is selected) = P(heart) + P(club) = 13/52 + 13/52 = ½

2)  P(A or B) = P(A) + P(B) - P(A and B), if they are not mutually exclusive.

    Ex:  P (getting a heart or a King when one card is selected) = P(H) + P(K) - P(H & K) = 13/52 + 4/52 - 1/52 =16/52 = 4/13

~Important Note: Use the backdoor approacch when dealing with multiple events coming from different sample spaces & repeated events from the same sample space.

~Note: the negative of (A OR B) is (not A AND not B).

~Example:  P( head on a coin OR a roll < 12 on a pair of dice). Since the sample spaces are different, you must use backdoor approach. This gives 1 - P(no Head and a roll of 12)=1- (.5)(1/36)= .986

~Example:  P(when tossing a coin 3 times, getting a head on the 1st toss OR a head on the 2nd toss OR a head on the 3rd toss). You cannot use the addition rule directly. The backdoor must be used. We get, 1-P(tail & tail & tail) = 1-(1/2)(1/2)(1/2)=1-1/8 = 7/8 or .875

~Multiplication Rules:

3)  P(A and B) = P(A) P(B), if A and B are independent  (one event has no effect on the other)

    Ex:  P (head on a coin toss and rolling a 6 on a die) = P(H) P(6) = (½)(1/6) = 1/12

4)  P(A and B) = P(A) P(B|A), if A and B are dependent. (A occurring effects B’s probability) (This is conditional probability)

     See  Conditional Probability

    P(B|A) reads "probability of B given that A has occurred"

    Ex:  P (getting a jack| the card is red) =  2/26 = 1/13  (since there are 2 jacks in the red suits)

~Note:  Experiments are done (selecting data) with replacement (so data point can be selected again) or without replacement (same data point can not be selected)(see link under topics of interest). Usually with large populations, we don’t make a distinction. However, will relatively small sample spaces, a distinction must be made, since probabilities change (rule 4 takes hold).

~Note: Read my article on replacement & without replacement under Topics of Interest. 

~Note:  probability tree diagrams are often helpful in setting up many multistage events.

~Note: multiply probabilities along one path, but add results for different paths  (outcomes).

~Note:  Probabilities are written along each stem or limb. The probabilities along each stem horizontally must add to 1. Probabilities are multiplied along consecutive stems to get the desired result. If you have 2 different paths to the end result, the probabilities are added for each completely different path.

~Example where the backdoor approach saves much time:  A coin is tossed 5 times. What is the probability of getting at least 2 heads?

~Translating, P(2 or more heads) = P( 2 or 3 or 4 or 5 heads). Adding probabilities does not work. Here, you must use the basic definition of probabiliy or the backdoor approach. Using the basic definition, we need to figure how many ways we can get 2 or more heads in 5 tosses. Here Pascal's triangle is useful. Those numbers for n=5 in Pascal's triangle are: 1 5 10 10 5 1. So, for 2 or more, we add 10+10+5+1=26 ways. Then we place that number over the total number of possible outcomes 2⁵ = 32.

~Using the backdoor approach, P(at least 2H) = 1 - P(complement of at least 2H) = 1- P (0 or 1) = 1 - (1+5)/32 = 1-6/32 = 1 - 3/16 = 13/16 (much easier) (see link on Bernoulli’s backdoor)

~Conditional probability: For dependent events (over lapping events).

~Symbols: P(B|A) reads "probability of B given A has occurred"

~Note: So, assume event A has occurred, then compute the probability of B based on that assumption

~So, P(B|A) will differ from P(B) alone since event A has affected the outcome for event B. So, P(A and B)) = P(A) P(B|A) instead of just P(A) P(B), since P(B) and P(B|A) are not the same.

~Note: Since we are after the probability of B, we write the left side with B first,: P(B and A) = P(A) P(B|A). Solving this for P(B|A), we get  P(B|A) = [P(B and A)]/P(A) (a popular form) (see Bayes’ thm link)

~Note: This is important, since the denominator, P(A) must correspond to the event A in the numerator (written 2nd).

~Note:  An alternate way to get P(B|A) is to count the # of ways both B & A can occur, then divide by  the # of ways A can occur alone. Using this method, probabilities are not used. However, in many problems, probabilities are given & the counts are not known, So the above formula must be used.

~Note: Just memorizing a formula without understanding is useless, so use common sense when ever possible.

~Ex: Let a sample space S={1,2,3,4,5,6,7,8,9}, Let A={odd digit}={1,3,5,7,9}, B={digit greater than 6}={7,8,9}. Then, If you randomly select one digit[TI-83 can do this, Math,PRB,5, (1,9,1)], the P(A)=5/9, P(B)=3/9=1/3, P(A|B)=2/3, P(B|A)=2/5

~Note: In this example, you can count the # of ways each event could occur, so you can use the alternate form. Not always the case.

~Ex: A candidate gets 60% of the vote, but only 50% of women voted  for A. Let A= a person votes for A and W=woman voter. Then, If we select a person at random, P(A)=.6 but P(A|W)=.5. This is known as the "gender gap" in politics. You can interpret P(A|W) as the ratio of women who voted for A divided by the women who voted.

~Counting possibilities is Sometimes very easy to do, however, there are many times it would be very impractical & extremely time consuming.

~the Fundamental principle of counting is used: i.e., If one act can be perfomed in n ways, and a 2nd act in m ways, and a 3rd act in r ways, & so on, then the number of ways they can occur in succession, will be the product of the # of ways. (many problems can be solved by just using this principle)

~Ex: If there are 3 different routes you can take from Poughkeepsie to Albany, and 4 different routes from Albany to Lake George,Then the number of completely different routes you can take from Poughkeepsie to Lake George would be (3)(4) = 12.

~Some are more complicated:

~Ex: How many ways can I choose 3 students from a class of 29 ? How many different ways can I rearrange the letters of the word MISSSISSIPPI? We need a way to figure these numbers.

~Note: Factorials. The symbol for factorial is !. So, 5!=(5)(4)(3)(2)(1).  8!=(8)(7)(6)(5)(4)(3)(2)(1), & so on. The TI-83 gives us these: insert #, MATH, PRB, 4, ENTER. Note: n!=n(n-1)(n-2)...1=n(n-1)!,  i.e., 12!=(12)(11!),  0!=1 by definition. (be careful with factorial arithmetic, it’s tricky)

~Permutations: An Arrangement of n different objects taking them r at a time.

~Ex: Take the letters A,B,C. How many permutations are there if we take the letters 2 at a time?   AB,AC,BA,BC,CA,CB, there are 6.

~Notation:   nPr = n!/(n-r)!,   nPn = n!/0!=n!,  The TI-83 will give these, just remember to enter the n first, then MATH, PRB, & find it. Let’s do 10 P 3 = 720 on your calculator.

~Finding distinct permutations: If objects are repeated and you want different arrangements, you must use the following formula: The distinct permutations of n objects, taken n at a time, with r alike, q alike, s alike, & so on , is  n!/[r!q!s1...]

~Ex: For the letters of the word MISSISSIPPI, we would have,  11!/[4!4!2!]=34,650 (try listing these, if you have too much fee time)

~Combinations: These deal with selection (what’s there) NOT arrangement. (how many committees can be formed) (how many golf 4somes can we have with a certain # of golfers) (how many ways we can select 6 numbers in a lottery) & so on.

~Notation: nCr =( nPr)/r! = n!/[r!(n-r)!].  nCn=n!/[n!0!]=1,  nC0 =1 by definition., 8C4 =8!/[4!4!]=(8)(7)(6)(5)(4!/[(4)(3)(2)(1)(4!)]=70 The TI-83 will give these too. Look for it in the same place, but remember, enter the number before you press that item. Note, If you don’t have a calculator, use the formulas. Combinations can also be found using PASCAL’S triangle (see link)

~Important note: When selection is made WITHOUT REPLACEMENT, combinations can be used to figure that number. However, selecting WITH REPLACEMENT requires a different technique since probabilities of selecting individual items remain the same.

(see Finding Probabilities With or Without Replacement).

~Example: When selecting 4 marbles from a bin containing 5 green & 7 blue marbles, the number of ways you can select 4 blue marbles (without replacement) is 7C4 = 35 (No repeated outcomes & permutations eliminated), but if you replace each one after selection, the number of ways of selecting 4 blue marbles is (7)(7)(7)(7) or 7⁴ = 2401 (repeated outcomes & all possible permutations included).

~Note: See link, under topics of interest, for 2-D probability.