ssessment Standard: 11.4.2 a) Correctly identify dependent and independent events (e.g.: from two way contingency tables or Venn - Diagrams) and therefore appreciate when it is appropriate to calculate the probability of two independent events: P( and B) = P(). P(B) b) Use Tree and Venn - Diagrams to solve probability problems (where events are not necessarily independent) Over the next two weeks we will give you a brief review of PROBBILITY from Grade 11. Remember that about 50% - 60% of the Examination is set on Descriptive Statistics and. lot of the was done in Grade 11. Collect your Paper 3 Lessons every week!! Guys, both NSC and IEB examinations candidates have the option of writing Paper 3 at the end of the year! Paper 3 covers additional mathematics material and is out of 100 marks. Maths Paper 3 will really set you apart in the job market, and make studying technical subjects at tertiary level easier. We have hooked you up with these lessons - written by IEB Maths Paper 3 examiner Heather Frankiskos. Though the lessons apply to both IEB and NSC candidates,where there are differences, we will point them out! The lesson this week applies to candidates from both examining bodies. Give it a go! So recall the following: PROBBILITY is the likelihood of something happening or being true. PROBBILITY is assigned a value between 0 (the impossible) and 1 (absolutely certain) even chance 0 1 2 1 The PROBBILITIES of the possible outcomes in a sample space must sum up to 1. Let s see if you can do these basic probabilities without any fancy techniques: 1. regular die is rolled. Find: a) P(1) - (the probability of getting a 1) = 1_ 6 b) P (7) - (the probability of getting a 7) = 0 Page 1
c) P (3 or 4) - (the probability of getting a 3 or 4 ) = 1 6 + 1_ 6 = 1_ 3 d) P (not a 2) = 1-1_ 6 = 5_ 6 e) P (even) = 3_ 6 = 1_ 2 2. card is drawn from a normal pack of cards (No jokers included) Find: a) P (ce) = 4 = 1 13 b) P (Heart) = 13 = 1_ 4 c) P (ce or a Heart) = 16 = 4 13 d) P (ce and a Heart) = 1 e) P (Red) = Remember that if is an event, then is called the complement of, and means not. lso P() + P() = 1 So P(Heart) = 4 P (Heart) = P (not a heart) = 48 and 4 + 48 = 1 Two events which have no outcomes in common are called MUTULLY EXCLUSIVE events. For example: 1) P(Heart) and P(Diamond) are mutually exclusive ) P (roll a 1) and P (roll a 3) are mutually exclusive NB: For mutually exclusive events P( or B) = P() + P (B) 1) P (Heart or Diamond) = P(Heart) + P(Diamond) = 13 + 13 = 26 = Page
11) P (roll a 1 or roll a 3) = P (roll a1) + P (roll a 3) = 1_ 6 + 1_ 6 = 2/6 = 1_ 3 But what would happen if we wanted P (King or a Heart)? P (King) = 4 P (Heart) = 13 Now can you see that the card King of Hearts (which is a heart and a king has been counted into both categories (so counted in twice) so P (King or Heart) = P (King) + P (Heart) - P (King of Hearts) = 4 + 13-1 = 16 = 4 13 So if events are NOT mutually exclusive then P ( or B) = P () + P (B) - P ( and B) We have to subtract off the overlap or intersection. So, if we roll a single die what is the probability of getting a prime number or an even number? P (Prime) = 3_ 6 = 1_ 2 { 2 ; 3; 5} P (Even) = 3_ 6 = 1_ 2 { 2 ; 4; 6} P (Prime or Even) = 1_ 2 + 1 2 1_ 6 = 5_ 6 Clearly the only number not allowed is the number 1. We can also see this in a Venn-Diagram Page 3
Maths Science 250 130 80 30 10 random pupil is selected from 250 grade 11s. Find: a) that he/she takes Maths P (Maths) = 210 250 = 0,84 b) that he /she takes Maths or Science P (Maths) + P (Science) - P (Maths and Science) = 210 250 + 110 250-80 250 = 240 250 Must subtract off P (M S) - the intersection = 0,96 Sometimes we are interested not in one outcome, but in two or three or more of them. For example, we may toss a coin twice, or select two or 3 cards from a pack or take 3 beads from a bag containing different colours Drawing up a TREE DIGRM is usually useful to assist with these kinds of problems Example 1 coin is tossed and H or T is recorded. Find the probability of getting two heads and a tail (this means in any order) Page 4
Toss 1 Toss 2 Toss 3 P(H) (H; H; H) 1 8 P(Head) P(H) P(T) P(T) P(H) 3 3 (H; H; T) 1 8 (H; T; H) 1 8 P(T) (H; T; T) 1 8 P(H) 3 (T; H; H) 1 8 P(Tail) P(H) P(T) (T; H; T) 1 8 P(T) P(H) (T; T; H) 1 8 \ P (two heads and a tail) = 1 8 + 1 8 + 1 8 = 3 8 Example 2 Draw two cards from a normal pack with replacement 1) Find probability of two Kings 2) Find probability of one King P(T) (T;T; T) 1 8 Draw 1 Draw 2 P (King) P (King) 4 P (Not a King) 4 48 (K;K) = _ 169 1 (K; not K) = _ 169 12 P (not a King) P (King) 48 4 (not K;K) = 169 12 P (Not a King) 48 (not K; not K) = 169 144 Page 5
1) P (K;K) = 1 169 2) P (one King) = 12 169 + 12 169 = 24 169 = 0,142 Can you see in the previous two examples when we came to toss the coin the second time or draw the second card - there was no effect of the first result on the second? When this happens the two events are said to be INDEPENDENT For two INDEPENDENT events P ( and B) = P (). P (B) So these are INDEPENDENT EVENTS: 1) Toss a coin twice 2) Toss a die four (or however many) times 3) Toss a coin, roll a die 4) Pick cards from a pack WITH replacement 5) Choose sweets from a bag WITH replacement Question 1 P () = 0,2 P (B) = 0,5 P ( B) (which means P( or B)) = 0,6 a) re events and B mutually exclusive? Motivate b) re events and B independent? Motivate a) P ( B) P () + P (B); so not mutually exclusive b) P ( B) = P() + P(B) - P ( and B) 0,6 = 0,2 + 0,5 - P ( and B) P ( and B) = 0,7-0,6 P ( and B) = 0,1 but P (). P (B) = 0,2 x 0,5 =0,1 \ are independent Page 6
Example 2 weather forecaster classifies all days as either wet or dry. He claims that the probability that 1 September will be wet is 0,4. If any particular day in September is wet, the probability that the next day is wet is 0,5; otherwise the probability that the next day is wet is 0,3. Find the probability that: 1 Sep 2 Sep P (wet) 0,5 P (wet) P (dry) 0,4 P (dry) 0,6 0,5 0,7 Key (1-0,4) = P (dry) = 0,6 (1-0,3) = P (second day dry) = 0,7 P (wet) 0,3 a) The first two days in September are wet b) September 2nd is wet a) 0,4 x 0,5 = 0,2 [(W;W)] b) 0,4 x 0,5 + 0,6 x 0,3 [(W;W) or (D; W)] = 0,38 and P (W,W) + P (W,D) + P (D, W) + P (D,D) = 0,2 + 0,2 + 0,42 + 0,18 = 1 Example 3 P () = 0,25 and P (B) = 0,5 and P ( and B) = 0,15 a) re events and B independent? b) re events and B mutually exclusive? Page 7
a) P() x P (B) = 0,25 x 0,5 = 0,125 P ( and B) = 0,15 so 0,125 0,15 NOT independent b) no since P ( or B) = P() + P (B) - P ( and B) Example 4 P () + P (B) nother way that we use to decide on Independence is by using a CONTINGENCY TBLE 4a) Let us look at this table which gives information about Males and Females and whether they prefer Red, Blue or Yellow Red Blue Yellow Total Male 20 40 50 110 Female 50 20 20 90 Total 70 60 70 200 This is called a two-way 2x3 Contingency Table. It has 2 rows and 3 columns. Can you see that we could get the following from the table? 1) P (Male) = 110 200= 0,55 2) P (Blue) = 60 200 = 0,3 3) P (Blue and Male) = 40 200 = 0,2 4) P (Blue or Male) = 60 200 + 110 200-40 200 = 130 200 = 0,65 They are NOT mutually exclusive 5) re the events being Male and preferring Blue independent or not? P (Blue and Male) = 0,2 P (Blue) x P (Male) = 60 200 x 110 200 = 33 200 = 0,165 They are NOT independent 6) re the events being Female and preferring Red independent or not? P (Female and Red) = 50 200 = 0,25 P (Female) x P (Red) = 90 200 x 70 200 = 63 400 = 0,1575 They are NOT independent Page 8
4b) Research with regard to the effects of a new headache tablet involving 100 males and 80 females showed that 60 males and 50 females responded positively to the tablet. Can we conclude that the success of the tablet is independent of gender? Male Female Total Positive 60 50 110 Not Positive 40 30 70 Total 100 80 180 P (Male) = 100 180 P (Female) = 80 180 P (Positive) = 110 180 P (not Positive) = _ 70 180 P (Male and Positive) = 60 180 = 0,3 P (Male) x P (Positive) = _ 100 180 x 110 180 = 0,34 Mathematically these are NOT independent (OR) P (Female and Positive) = 50 180 = 0,27 P (Female) x P (Positive) = 80 180 x 110 180 = 0,2716 There is a slight difference so they are still NOT independent So now we have seen the multiplication law for INDEPENDENT events. But what happens if we were to draw two cards from a pack and NOT put the first one back before we took the second one? (This is called WITHOUT replacement) It is very important to realise that because we do not put the first card back, we are now going to draw the second card from a smaller sample and the next outcome is influenced by (depends on) what happened in the first. So if we draw the first card P (King) = 4 Having removed one King and not replaced it means that there are now only 3 Kings in the remaining 51 cards so the probability of the second card also being a King is 3 51 So the of two Kings in two successive draws of a pack of cards WITHOUT replacement is P (King and King) = 4 x 3 51 = 1 221 Page 9
So what if we drew two cards and wanted the probability of one King.? The tree diagram would look as follows: 1st Draw 2nd Draw P (King) P (not King) P (King) 4 P (not King) P (King) 48 3 51 48 51 51 4 So the of one King means P (not King) 47 51 = P (King followed by not King) or P (not King followed by King) = 4/ x 48/51 + 48/ x 4/51 = 32 221 Please note: P (K,K) = 4 x 3 51 = 1 221 P (K, not K) = 4 x 48 51 = 16 221 P (not K; K) = 48 x 4 51 = 16 221 P (not K; not K) = 48 x 47 51 = 188 221 1 now 221 + 16 221 + 16 221 + 188 221 =221 221 = 1 method that we can use to help us organise the relationships between events so that we can make conclusions about probabilities, mutually exclusive and / or independence is called a VENN-DIGRM. The best way to get to understand how useful they are is with an example or two: Page 10
Example 1 Given P() = 0,7; P (B) = 0,4 and P ( B) = 0,28 Find out: a) P ( B) b) whether and B are independent Best way to solve this is draw a picture: B x 0,28 y now x + 0,28 = 0,7 x = 0,42 and y + 0,28 = 0,4 y = 0,12 a) Here P ( B) = P ( or B) = 0,42 + 0,28 + 0,12 = 0,82 so P ( B) = 1-0,82 = 0,18 b) P () x P (B) = 0,7 x 0,4 = 0,28 P ( B) = 0,28 so they are independent events Sometimes we get given the picture organiser and we are asked to interpret it. Page 11
S R C 12 108 54 36 6 18 134 97 J Example 2 survey was conducted to determine the preferences for three different music types Rock (R), Classic (C), Jazz (J) 1) How many people were surveyed? 2) How many people preferred Rock (R) only? 3) What percentage of people surveyed like all three types? 4) What percentage liked Rock (R) or Classic (C)? 5) 134 do not like music at all. Is this statement true or false? Motivate your answer. 6) Find P (R C ) 1) 465 2) 108 3) 36 465 = 0,0774 = 7,74% 4) 108 + 12+ 36 + 6 + 54 + 18 46 =_ 234 465 = 0,503 = 50,3% 5) False. They may well like other kinds of music - they just don t like these three. 6) 97+134 465 = 77 155 = 0,4968 Page 12