What are the chances? - Probability

“What are the chances of that happening?” ,[object Object]

[object Object],[object Object],[object Object],[object Object]

17th Century Gambling ,[object Object],[object Object],[object Object],[object Object],THE OLD METHOD WAS MOST PROFITABLE!

[object Object],[object Object],[object Object],[object Object],Correspondence leads to Theory

Probability ,[object Object],[object Object]

Careers using the “Chance Theory” ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]

First you gotta learn the rules and terms!

Problem: ,[object Object],[object Object],[object Object]

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! ,[object Object],[object Object],[object Object],[object Object]

Probability Scale ,[object Object],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) ,[object Object]

Probability of an Event Not Happening ,[object Object],[object Object],[object Object]

[object Object],P(A) + P(NOT A) = 1 OR P(NOT A) = 1 - P(A)

[object Object],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! ,[object Object],[object Object]

You try please! ,[object Object],[object Object],[object Object],[object Object]

Notes: ,[object Object],[object Object],[object Object]

[object Object],[object Object],[object Object]

Conditional P! ,[object Object],[object Object],[object Object],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 ,[object Object]

Example ,[object Object],[object Object],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 ,[object Object]

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

[object Object],[object Object]

[object Object],[object Object],[object Object],Example 1

[object Object],[object Object],[object Object],[object Object],[object Object],Example 2

Experiment - Formula ,[object Object],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. ,[object Object],EXPECTED NO. OF OUTCOMES = (RELATIVE FREQUENCY) X (NO. OF TRIALS) OR EXPECTED NO. OF OUTCOMES = P(EVENT) X (NO. OF TRIALS)

Example ,[object Object],[object Object],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 ,[object Object],[object Object],[object Object],Number 1 2 3 4 5 Probability 0.25 0.2 0.25 0.15 B

Solution: ,[object Object],[object Object],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 ,[object Object],[object Object],[object Object],P(A OR B) = P(A) + P(B) - P(A AND B) REMOVES DOUBLE COUNTING

Mutually Exclusive Events ,[object Object],[object Object]

[object Object],[object Object],[object Object],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 ,[object Object],[object Object],[object Object]

Solution: ,[object Object],[object Object],[object Object],= 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! ∕

Solution: ,[object Object],[object Object],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 ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],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 ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],33/50 20/50 = 2/5 5/50 = 1/10 27/50 Boys Girls Aged 17 18 12 Aged 18 15 5

Conditional Probability ,[object Object],[object Object],[object Object]

[object Object],[object Object],[object Object],[object Object],[object Object],# (A ∩ B) P(A ∩ B) #B P(B) =

[object Object],[object Object],Q 4. Pg: 79 {GGG, GGB, BGG, GBG, BBG, GBB, BGB, BBB} 2/8 = 0.25

Q. 9 PG: 80 ,[object Object],[object Object],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(FP(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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