Probability How Likely In the real world events

Probability How Likely? In the real world events can not be predicted with total certainty. The best we can do is say how likely they are to happen, using the idea of probability. Tossing a Coin When a coin is tossed, there are two possible outcomes: • heads (H) or • tails (T) We say that the probability of the coin landing H is ½ Similarly, the probability of the coin landing T is 1/2.

Throwing Dice When a single die is thrown, there are six possible outcomes: 1, 2, 3, 4, 5, 6 The probability of throwing any one of these numbers is 1/6. Probability of an event happening = Number of ways it can happen Total number of outcomes

Example: the chances of rolling a "4" with a die Number of ways it can happen: 1 (there is only 1 face with a "4" on it) Total number of outcomes: 6 (there are 6 faces altogether) So the probability = 1/6

Probability Line You can show probability on a Probability Line: The probability is always between 0 and 1 Probability does not tell us exactly what will happen, it is just a guide.

Probability Definitions A trial is a single act by which an observation is noted e. g. a roll of a dice. An experiment is the process by which an observation is noted and consists of one or more trials. When a statistical experiment is conducted, there a number of possible outcomes. The possible outcomes are called a sample space and are often denoted by the letter S. An event is a subset of a sample space. The impossible event (or empty set) is one that contains no outcomes. It is often denoted by the Greek letter Ф (phi).

Note • The definition P(A) = Number of outcomes that yield event A Total number of possible outcomes makes sense only if the number of possible outcomes is finite. We will only consider these cases. • If an event can never occur, its probability is 0. An event that always happens has a probability of 1 • The value of a probability must always lie between 0 and 1 - a result outside this range indicates an error has occurred • A probability may be expressed as a decimal or fraction • All outcomes are considered equally likely

Here we can record data and predict outcomes based on past data – Tennis player A has beaten player B in 8 out of the last 10 matches – we could say there is an 80% chance of them winning the next match

Subjective probabilities differ from person to person. Because the probability is subjective, it contains a high degree of personal bias. An example of subjective probability could be asking a sports journalist , before the football season starts, the chances of St George winning the grand final. While there is no absolute mathematical proof behind the answer to the example, he might still reply in actual percentage terms, such as they have a 25% chance of winning the grand final – it is an educated guess, an estimate without any real data to support it.

Probability of throwing a head or a tail is Probability of a Head + probability of a tail = 1/2 +1/2 = 1 that is we can be absolutely certain you will throw a head or a tail But what is the probability of a head and then a tail? when you have the word "AND" - that is two events must occur it is harder and we multiply the probabilities 1/2 x 1/2 = 1/4 Prove this using a Tree Diagram …. . One more example - the probability of finding someone who is 6 foot four AND has blue eyes is much more remote than finding someone who is 6 foot four OR has blue eyes. The less chance something has of occurring the closer the probability is to zero

I have $100, 000 to invest - project B costs $80, 000, project A costs $90, 000 - if I run with Project A - it will exclude project B and vice versa

More probability examples can be reviewed in the following video check this out if you would like further examples…. . https: //www. khanacademy. org/math/probability/independent-dependentprobability/addition_rule_probability/v/addition-rule-for-probability

Max, Susan and Karen earn over $50, 000. What is the probability of selecting someone from the group at random who is Female or earns more than $50, 000 ? P(F) = 2 / 3 , P (>$50, 000) = 3/3 2/3 + (OR) 3/3 = 5/3 – (the overlap) 2/3 (two out of the three belong to both groups) = 3/3 or 1 That is there is a 100% chance of selecting someone at random who is female or earns over $50, 000

YOUTUBE INDEPENDENT EVENTS http: //www. khanacademy. org/math/probability/v/independent -events-1

Is it possible to toss a coin and get 14 heads in a row? If the last 40 tosses have been heads what is the chance that the next toss will also be a head?

http: //www. mathsonline. com. au/r. html? lesson=5218

Suppose you have a box with 3 blue marbles, 2 red marbles, and 4 yellow marbles. You are going to pull out one marble, record its colour, put it back in the box and draw another marble. What is the probability of pulling out a red marble followed by a blue marble? The multiplication rule says we need to find P(red) P(blue). P(red) = 2/9 P(blue) = 3/9 P(red and then a blue) = (2/9) x (3/9) = 6/81 = 2/27 The events in this example were independent. Once the first marble was pulled out and its colour recorded, it was returned to the box. Therefore, the probability for the second marble was not effected by what happened on the first marble.

Consider the same box of marbles as in the previous example. However in this case, we are going to pull out the first marble, leave it out, and then pull out another marble. What is the probability of pulling out a red marble followed by a blue marble? We can still use the multiplication rule which says we need to find P(red) P(blue). But be aware that in this case when we go to pull out the second marble, there will only be 8 marbles left in the bag. P(red) = 2/9 P(blue) = 3/8 P(red and then blue) = (2/9) x (3/8) = 6/72 = 1/12 The events in this example were dependent. When the first marble was pulled out and kept out, it effected the probability of the second event. This is what is meant by dependent events.

YOUTUBE DEPENDENT EVENTS http: //www. khanacademy. org/math/probability/v/probabilityof-dependent-events

Or just focus on this event 2/3 chance of a number less than 4 given that we know it is odd

YOUTUBE PROBABILITY WITH CARDS http: //www. khanacademy. org/math/probability/v/probabilitywith-playing-cards-and-venn-diagrams

Is it correct to say the probability of them not moving is 0. 3?

Draw a tree diagram for this - example next slide 6 possible combinations are HHTT, HTHT, HTTH, TTHH, THTH, THHT

H T H H T T H H T T T H T H T

Or using the multiplication rule ½ x ½ because the first has to be green AND the second has to be green AND the third has to be green

This looks like the same question but it is slightly different…

G G G R R G R G R R Given that 2 green counters have already been drawn we are now just looking at the probability that the third green counter will also be green. Here we are only looking at the top last branch of the tree - the options are one in two GGG or GGR

A different way of solving the same question …. ½ x ½ prob of first green AND 2 nd green ½ x ½ Prob 1 st AND 2 nd AND 3 rd Because they are independent events we don’t need to do this we just say each event is ½ of drawing a green counter

70 / 120 =. 58 32 / 102 =. 31

A table is a better idea - less confusing Rich Not Rich Banker 3 2 Non Banker 1 4 Total 4 6 For (a) just focus on the rich column 3 out of four chances you are a banker given that you are rich For (b) just focus on the banker row - 3 of the five bankers are rich - 3/5 = 6 /10 =. 6

Combinations / Permutations There are three runners in a race A, B and C. List all the possible opportunities that first and second place can be awarded to each runner? This is a permutation question where order does matter – A, B is not the same as B, A that is A first B second is not the same as B first and A second On the calculator - 3, 2 nd F , n. Pr , 2 =6 Proof AB, BA, BC, CB, AC, CA I need to select two names from a hat containing three names, how many different combinations of two names could I select? Now order doesn’t matter – B, A is the same as A, B On the calculator - 3, 2 nd F , n. Cr 2 =3 AB, BC, AC

This question doesn’t exist in Ed 3

3 defective parts, 3 acceptable parts I randomly select 2 out of the six parts what are all the different combinations of defective parts Work down the column, to find combinations e. g. D 1, D 2, down to D 1, D 3, Down to D 1, A 4 etc Using the Financial Calculator to find the number of combinations 6, 2 nd F , n. Cr 2 = 15

Answer…. 107 = 10 000 different numbers If digits were not repeated I think it would be a permutation question. . with the calculator… 10 2 nd F, n. Pr , 7 = 604, 800. . . Test this on the next slide

Any assumption should be able to be tested by using smaller, simpler examples. So imagine the question said the company had a 2 digit serial number and that numbers could be repeated the answer would be 102 = 100 – examples 00, 01, 02, 03, 04, 05, 06, 07, 08, 09, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, ………. 96, 97, 98, 99, 100 Now check the permutation solution ……. with the calculator… 10 2 nd F, n. Pr , 2 = 90 …… the reason for the difference is that with permutations you are saying imagine you have 10 numbers in a hat how many combinations of two numbers can you draw from the hat. Now you can’t pull out 00, 11, 22, 33, 44, 55, 66, 77, 88, 99 – because there is only one of each number in the hat.

(a) 6 x 3 x 5 = 90 (b) 7 x 6 x 3 x 5 = 630

Using the Financial Calculator to find the number of combinations 52, 2 nd F , n. Cr 2 = 1326

This is a permutation question. . with the calculator… 10 2 nd F, n. Pr , 5 = 30, 240

Here order does matter - this is a permutation question. . so with the calculator… 7 2 nd F, n. Pr , 3 = 210 One last comment on Permutations and Combinations – there will always be more permutations than combinations because when order matters that is AB is not BA so with Permutations we have possibilities yet with combinations AB is treated the same as BA we only have one combination. So in the above example A is on the left, B is in the middle and C is on the right is one permutation and is different than B on the left, A is in the middle and C on the right etc. If the question said how many combinations of photos are possible if each photo has three people in it and there are seven people to choose from then the answer would be 7 2 nd F, n. Cr , 3 = 35 Check out the table on the next slide for all the possible combinations

Seven people A, B, C, D, E, F, G, Combinations of three people in a photo 1 ABC 11 ADF 21 BDF 31 CFG 2 ABD 12 ADG 22 BDG 32 DEF 3 ABE 13 AEF 23 BEF 33 DEG 4 ABF 14 AEG 24 BEG 34 DFG 5 ABG 15 AFG 25 BFG 35 EFG 6 ACD 16 BCD 26 CDE 7 ACE 17 BCE 27 CDF 8 ACF 18 BCF 28 CDG 9 ACG 19 BCG 29 CEF 10 ADE 20 BDE 30 CEG Each one of these photo combinations can be arranged 6 ways. 35 combinations x 6 = 210 permutations – for example the people in photo 1 can be arranged ABC, ACB, BCA, BAC, CAB or CBA

You Tube Video – Conditional Probability Bayers Theorem http: //www. youtube. com/watch? v=Zxm 4 Xxvzohk

Suggested Questions from Textbook…… Select a range of questions from the Problems in this chapter – enough so that you feel comfortable with this topic