What are the chances? - Probability

“What are the chances of that happening?” Theory of Probability

Weather forecast predicts 80% chance of rain. You know based on experience that going slightly over speed increases your chance of passing through more green lights. You buy a ticket for the Euro Millions because you figure someone has to win. A restaurant manager thinks about the probability of how many customers will come in in order to prepare accordingly.

17th Century Gambling Chevalier de Mere gambled frequently to increase his wealth. He bet on a roll of a die that at least one six would appear in four rolls. Tired of this approach he decided to change his bet to make it more challenging. He bet that he would get a total of 12, a double 6, on 24 rolls of two dice. THE OLD METHOD WAS MOST PROFITABLE!

He asked his friend Blaise Pascal why this was the case. Pascal worked it out and found that the probablity of winning was only 49.1% with the new method compared to 51.8% using the old approach! Pascal wrote a letter to Pierre De Fermat, who wrote back. They exchanged their mathematical principles and problems and are credited with the founding of probability theory. Correspondence leads to Theory

Probability is really about dealing with the unknown in a systematic way, by scoping out the most likely scenarios, or having a backup plan in case those most likely scenarios don’t happen. Life is a sequence of unpredctable events, but probability can be used to help predict the likelihood of certain events occuring.

Careers using the “Chance Theory” Actuarial Science Atmospheric Science Bioinformatics Biostatistics Ecological/Environmental Statistics Educational Testing and Measurement environmental Health Sciences Epidemiology Government Financial Engineering/Financial Mathematics/Mathematical Finance/Quantitative Finance Industrial Statistics Medicine Meteorology/Atmospheric Science Nurses/Doctors Pharmaceutical Research Public Health Public Policy Risk Analysis Risk Management and Insurance Social Statistics

First you gotta learn the rules and terms!

Problem: A spinner has four equal sectors colored yellow, blue, green and red. What are the chances on landing on blue after spinning the spinner? What are the chances of landing on red?

Terms Defintion Example An EXPERIMENT is a situation involving chance or probability that leads to results called outcomes. What color would we land on? A TRIAL is the act of doing an experiment in P! Spinning the spinner. The set or list of all possible outcomes in a trial is called the SAMPLE SPACE. Possible outcomes are Green, Blue, Red and Yellow. An OUTCOME is one of the possible results of a trial. Green. An EVENT is the occurence of one or more specific outcomes One event in this experiment is landing on blue.

Getting the rules! The P! of an outcome is the percentage of times the outcome is expected to happen. Every P! is a number (a percentage) between 0% and 100%. [Note that statisicians often express percentages as proportions-no. 0 and 1] If an outcome has P! of 0% it can NEVER happen, no matter what. If an outcome has a P! of 100% it will ALWAYS happen. Most P! are neither 0/100% but fall somewhere in between. The sum of all the Probabilities of all possible outcomes is 1 (or 100%).

Probability Scale In words RANGE IN NUMBER RANGE IN PERCENTAGE 0 - 1 0% - 100%

Want to have a guess? 0 1 0.5 A B C D E 5 EVENTS (A,B,C,D AND E ) ARE SHOWN ON A P! SCALE. COPY AND COMPLETE THE FOLLOWING TABLE: Probability Event Fifty-fifty C Certain Very unlikely Impossible Very likely

Back to the Spinner AFTER SPINNING THE SPINNER, WHAT IS THE PROBABILITY OF LANDING ON EACH COLOR? RED? BLUE? GREEN? ORANGE? 1/4!!!

In order to measure the P! of an event mathematicians have developed a method to do this!

Probability of an Event THE P! OF AN EVEN A OCCURING P(A) = THE NUMBER OF WAYS EVENT A CAN OCCUR THE TOTAL NUMBER OF POSSIBLE OUTSOMES

Enter P(not A) Complement of an Event

Probability of an Event Not Happening If A is any event, then ‘not A’ is the event that A does not happen. Clearly A and ‘not A’ cannot occur at the same time. Either A or ‘not A’ must occur.

Thus we have the following relationship between the probabilities of A and ‘not A’: P(A) + P(NOT A) = 1 OR P(NOT A) = 1 - P(A)

A spinner has 4 equal sectors colored yellow, blue, green and red. What is the probability of landing on a sector that is not red after spinning this spinner? Let’s understand! Sample Space: {yellow, blue, green, red} Probability: The probability of each outcome in this experiment is one fourth. The probability of landing on a sector that is not red is the same as the probability of landing on all the other colors except red. P(not red) = 1/4 + 1/4 + 1/4 = 3/4

Using the rule! P(not red) = 1 - P(red) P(not red) = 1 - 1/4 = 3/4

You try please! A single card is chosen at random from a standard deck of 52 playing cards. What is the probability of choosing a card that is not a club? P(not club) = 1 = P(club) 1 - 13/52 = 39/52 or 3/4

Notes: P(not A) can be also written as P(A’) or P(A). It is important NOT to count an outcome twice in an event when calculating probabilities. In Q’s on probability, objects that are identical are treated as different objects.

The phrase ‘ drawn at random ’ means each object is equally likely to be picked. ‘ Unbiased ’ means ‘ fair ’. ‘ Biased ’ means ‘ unfair ’ in some way.

Conditional P! With this you are normally given some prior knowledge or some extra condition about the outcome. This usually reduces the size of the sample space. EG. In a class - 21boys&15 girls. 3boys&5girls wear glasses. A PUPIL PICKED AT RANDOM FROM THE CLASS IS WEARING GLASSES. WHAT IS THE P! THAT IT IS A BOY? WE ARE CERTAIN THAT THE PUPIL PICKED WEARS CLASSES. THERE ARE 8 PUPILS THAT WEAR GLASSES AND 3 OF THOSE ARE BOYS P(WHEN A PUPIL WHO WEARS GLASSES IS PICKED, THE PUPIL IS A BOY) = 3/8

Combining two events There are many situations where we have to consider two outcomes. In these situations, all the possible outcomes, the sample space can be represented on a diagram - often called a two-way table!

Example Two fair six-sided dices, one red and one blue, are thrown. What is the P! of getting two equal scores or of the scores adding to 10? Solution: 36 POSSIBLE OUTCOMES P(TWO EQUAL SCORES OR A TOTAL OF 10) = 8/36 = 2/9 NOTE: (5,5) IS NOT COUNTED TWICE! 6 X X 5 X 4 X X 3 X 2 X 1 X 1 2 3 4 5 6

Relative Frequency Experimental Probability

Some P! cannot be calculated by just looking at the situation!

For example you cannot work out the P! of winning a football match by assuming that win, draw or loose are equally likely! But we can look at previous results in similar matches and use these results to estimate the P! of winning!

The Blues and Naoimh Martin Gaelic Teams are playing a match tonight and you want to know what is the P! that the Blues will win? They have played each other 50 times before. The Blues won 35 of those games and there was also 5 draws! So we can say so far the Blues have won 35/50 games or 7/10! Example 1

The fraction isn’t the P! of the Blues winning but an estimate! We say that the relative frequency og the Blues winning is 7/10.

Matthew decides to see what the P! is that buttered toast lands buttered side down when dropped. He drops 50 pieces of buttered toast. 30 pieces land buttered side down. His relative frequency is 30/50=3/5. Therefore he would estimate that the P! of landing buttered side down is 3/5. Example 2

Definition: Relative Frequency is a good estimate of how likely an event is to occur, provided that the number of trials is sufficiently large.

Experiment - Formula The relative frequency of an event in an experiment is given by: P(E) = RELATIVE FREQUENCY OF AN EVENT= NO. OF SUCCESSFUL TRIALS NO. OF TRIALS

How many times you expect a particular outcome to happen in an experiment. The expected number of outcomes is calculated as follows: EXPECTED NO. OF OUTCOMES = (RELATIVE FREQUENCY) X (NO. OF TRIALS) OR EXPECTED NO. OF OUTCOMES = P(EVENT) X (NO. OF TRIALS)

Example Sarah throws her fair six sided die a total of 1,200 times. Find the expected number of times the number 3 would appear. Solution: IF THE DIE IS FAIR, THEN THE P! OF A SCORE OF 3 WOULD BE 1/6! THUS THE EXPECTED NO OF 3’S = P(EVENT) X (NO. OF TRIALS) = 1/6 X 1200 =200.

Example 2 A spinner numbered 1-5 is biased. The P! that the spinner will land on each of the numbers 1 to 5 is given in the P! distribution table below. (I) Write down the value of B. (ii) If the spinner is spun 200 times, how many fives would expect? Number 1 2 3 4 5 Probability 0.25 0.2 0.25 0.15 B

Solution: (i) Since one of the no. 1-5 must appear, the sum of all the P! must add to 1! Expected no. of 5’s. Therefore 0.25 + 0.2 + 0.25 + 0.15 + B = 1 0.85 + B =1 thus B = 0.15 = P(5) X (no. of trials) = 0.15 X 200 = 30

Combined Events If A and B are two different events of the same experiment, then the P! that the two events, A or B, can happen is given by: It is often called the ‘ or ’ rule! It is important to remember that P(A or B) means A occurs, or B occurs, or both occur. By subtracting p(A and B), the possibility of double counting is removed. P(A OR B) = P(A) + P(B) - P(A AND B) REMOVES DOUBLE COUNTING

Mutually Exclusive Events Events A and B are said to be mutually exclusive events if they cannot occur at the same time. Consider the following event of drawing a single card from a deck of 52 cards. Let A be the even a king is drawn and B the event a Queen is drawn. The single card drawn cannot be a King and a Queen. The events A and B are said to be mutually exclusive events.

If A and B are mutually exclusive events, then P(A ∩ B) = 0. There is no overlap of A and B. For mutually exclusive events, P(A ∪ B) = P(A) + P(B).

A and B are two events such that P(A ∪ B) = 9/10, P(A) = 7/10 and P(A ∩ B) = 3/20 Find: (i) P(B) (ii) P(B’) (iii) P[(A ∪ B)’] Solution: Example 1 (I) P(A ∪ B) = P(A) + P(B) - P(A ∩ B) 9/10 = 7/10 + P(B) - 3/20 P(B) = 9/10 - 7/10 + 3/20 = 7/20 (II) P(B’) = 1 - P(B) = 1 - 7/20 = 13/20 (III) P[(A ∪ B)’] = 1 - P(A ∪ B) = 1 - 9/10 = 1/10 A B 3/20 1/10 U 11/20 2/5 (7/10 - 3/20 = 11/20) (7/20 - 3/20 = 2/5)

Example 2 An unbiased 20-sided die, numbered 1 to 20, is thrown. (i) What is the P! of obtaining a no. divisible by 4 or 5? (ii) Are these events mutually exclusive?

Solution: There are 20 possible outcomes: (i) No ÷ by 4 are 4, 8, 12, 16 or 20 ∴ P(no÷4) = 5/20 No ÷ by 5 are 5, 10, 15 or 20 ∴ P(no÷5) = 4/20 No ÷ by 4 and 5 is 20 ∴ P( no divisible by 4 and 5 ) = 1/20 P(no÷4/5) = P(no÷4) + P(no÷5) - P(no÷4 and 5) = 5/20 + 4/20 - 1/20 = 8/20 = 4/5 THE NO 20 IS COMMON TO BOTH EVENTS, AND IF THE PROBABILITIES WERE SIMPLY ADDED, THEN THE NO 20 WOULD HAVE BEEN COUNTED TWICE. (II) P(NUMBER DIVISIBLE BY 4 AND 5) = 0 ∴ THE EVENTS ARE Not MUTUALLY EXCLUSIVE! ∕

Example 3 A bag contains five red, three blue and yellow discs. The red discs are numbered 1, 2, 3, 4 and 5; the blue are numbered 6, 7, and 8; and the yellow are 9 and 10. A single disc is drawn at random from the bag. What is the P! that the disc is blue and even? Are these mutually exclusive?

Solution: Let B represent that a blue disc is chosen and E represent that a disc with an even no is chose. P(B or E) = P(B) + P(E) - P(B and E) 1 2 3 4 5 6 7 8 9 10 = 3/10 + 5/10 - 2/10 = 6/10 = 3/5 P(B AND E) = 2/10 = 0 ∴ THE EVENTS ARE NOT MUTUALLY EXCLUSIVE! ∕

Q. 8 Pg: 73 Active Math Given the Venn diagram, write down: P(E) = P(F) = P(E ∩ F) = P(E ∪ F) = Verify that P(E ∪ F) = P(E)+P(F)-P(E ∩ F) s S E F 0.1 0.3 0.5 0.1 0.4 0.6 0.1 0.9 0.9 = 0.4 + 0.6 - 0.1

Q. 18 Pg: 74 The data shows the number of boys and girls aged either 17 or 18 on a school trip, in which only these 50 students took part. A students is chosen at random from the party. Find the P! that the student: (i) Is a boy: (ii) is aged 18: (iii) is a girl of 18 years: (iv) is a girl aged 17 or a boy aged 18: 33/50 20/50 = 2/5 5/50 = 1/10 27/50 Boys Girls Aged 17 18 12 Aged 18 15 5

Conditional Probability In a class there are 15 male students, 5 of whom wear glasses and 10 female students, 3 of whom wear glasses. We will let M = {male students}, F = {female students} and G = {students who wear glasses}. A student is picked at random from the class. What is the P! that the student is female, given that the student wears glasses?

So we write this as: P(F|G) [The P! of F, given G] 8 students wear glasses. 3 of these are girls. Hence P(F|G) = 3/8 In general the P(A|B) = # (A ∩ B) P(A ∩ B) #B P(B) =

A family has three children. Complete the outcome space: {GGG, GGB, ...}, where GGB means the first two are girls and the third is a boy. Find the P! that all 3 children are girls, given that the family has at least two girls. Q 4. Pg: 79 {GGG, GGB, BGG, GBG, BBG, GBB, BGB, BBB} 2/8 = 0.25

Q. 9 PG: 80 E and F are two events such that P(E) = 2/5, P(F) = 1/2, and P(E|F) = 1/9. Find: (i) P(E ∩ F) (ii) P(F|E) (iii) P(E ∪ F) P(E|F) = P(E ∩ F) \ P(F) 1/9 = P(E ∩ F) \ 1/2 ∴ P(E ∩ F) = 1/18 P(F|E) = P(F ∩ E) \ P(E) P(F\ P(F|E) = 1/18 \ 2/5 P(F|E) = 5/36 P(E ∪ F) = P(E) + P(F) - P(E ∩ F) 2/5 + 1/2 - 1/18 38/45

What are the chances? - Probability

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