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Probability 3 Exam Paper Assessment Standard: 11.4.2 a) Correctly identify dependent and independent events (e.g .: from two way contingency tables or Venn - Diagrams) Collect your Paper 3 Lessons every week!! and therefore appreciate when it is Guys, both NSC and IEB examinations appropriate to calculate the probability candidates have the option of writing Paper 3 at the of two independent events: end of the year! Paper 3 covers additional P(A and B) = P(A). P(B) mathematics material and is out of 100 marks. Maths Paper 3 will really set you apart in the job market, b) Use Tree and Venn - Diagrams to and make studying technical subjects at tertiary level solve probability problems easier. We have hooked you up with these lessons (where events are not necessarily - written by IEB Maths independent) Paper 3 examiner Heather Frankiskos. Though the lessons apply to both IEB and NSC candidates,where Over the next two weeks we will give you a there are differences, we will point them out! The brief review of PROBABILITY from Grade 11. lesson this week applies to candidates from both Remember that about 50% - 60% of the examining bodies. Give it a go! Examination is set on Descriptive Statistics and Probability. A lot of the Probability was done in Grade 11. So recall the following: • PROBABILITY is the likelihood of something happening or being true. • A PROBABILITY is assigned a value between 0 (the impossible) and 1 (absolutely certain) even chance 0 __ 2 • The PROBABILITIES 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. A regular die is rolled. Find: 1 a) P(1) - (the probability of getting a 1) = _ 6 b) P (7) - (the probability of getting a 7) = 0 Page 1 Probability 3 Exam Paper 1 1 1 c) P (3 or 4) - (the probability of getting a 3 or 4 ) =__ + _ = _ 6 6 3 1 5 d) P (not a ) = 1- _ = _ 6 6 3 1 e) P (even) = _ = _ 6 . A card is drawn from a normal pack of 5 cards (No jokers included) Find: 4 1 A A a) P (Ace) = __ = __ 5 13 13 1 b) P (Heart) = __ = _ 5 4 A A A 16 4 c) P (Ace or a Heart) = __ = __ 5 13 A A 1 __ d) P (Ace and a Heart) = 5 1 e) P (Red) = _ A A • Remember that if A is an event, then A’ is called the complement of A, and means ‘not A’. Also P(A) + P(A)’ = 1 4 So P(Heart) = __ 5 48 P (Heart)’ = P (not a heart) = __ 5 4 48 and __ + __ = 1 5 5 • Two events which have no outcomes in common are called MUTUALLY 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(A or B) = P(A) + P (B) 1) P (Heart or Diamond) = P(Heart) + P(Diamond) 13 13 = __ + __ 5 5 6 = __ 5 1 =_ Page Probability 3 Exam Paper 11) P (roll a 1 or roll a 3) = P (roll a1) + P (roll a 3) 1 1 =_+_ 6 6 = /6 1 =_ 3 But what would happen if we wanted P (King or a Heart)? 4 P (King) = __ 5 13 P (Heart) = __ 5 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 = __ + __ - __ 5 5 5 16 = __ 5 4 = __ 13 So if events are NOT mutually exclusive then P (A or B) = P (A) + P (B) - P (A 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? 3 1 P (Prime) = _ = _ { ; 3; 5} 6 3 1 P (Even) = _ = _ { ; 4; 6} 6 1 1 1 5 P (Prime or Even) = _ + __ − _ = _ 6 6 Clearly the only number not allowed is the number 1. We can also see this in a Venn-Diagram Page 3 Probability 3 Exam Paper 50 M a t hs Science 130 80 30 10 A random pupil is selected from 50 grade 11s. Find: a) Probability that he/she takes Maths 10 P (Maths) = ___ = 0,84 50 b) Probability that he /she takes Maths or Science P (Maths) + P (Science) - P (Maths and Science) 10 110 80 = ___ + ___ - ___ 50 50 50 40 = ___ 50 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 DIAGRAM is usually useful to assist with these kinds of problems Example 1 A 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 Probability 3 Exam Paper Toss 1 Toss 2 Toss 3 P(H) 1 _ 1 __ (H; H; H) → 8 P(H) 1 _ 1 1 _ 3 (H; H; T) → __ P(T) 8 1 _ 1 _ P(H) 1 _ 1 __ P(Head) 3 (H; T; H) → P(T) 8 1 _ 1 __ (H; T; T) → P(T) 8 P(H) 1 _ 1 __ 3 (T; H; H) → 8 P(Tail) 1 _ P(H) 1 _ 1 _ 1 __ (T; H; T) → P(T) 8 1 _ P(H) 1 _ __ 1 (T; T; H) → P(T) 8 1 _ 1 __ (T;T; T) → P(T) 8 3 \ P (two heads and a tail) = __ + __ + __ = __ 8 8 8 8 Example 2 Draw two cards from a normal pack with replacement 1) Find probability of two Kings ) Find probability of one King Draw 1 Draw 2 4 __ P (King) 52 (K;K) = ___ 69 4 __ 52 P (King) 48 __ 2 (K; not K) = ___ P (Not a King) 52 69 P (King) 4 __ 2 P (not a King) 52 (not K;K) = ___ 69 48 __ 52 48 __ 44 P (Not a King) 52 (not K; not K) = ___ 69 Page 5 Probability 3 Exam Paper 1 1) P (K;K) = ___ 169 1 1 4 ) P (one King) = ___ + ___ = ___ = 0,14 169 169 169 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 (A and B) = P (A). P (B) So these are INDEPENDENT EVENTS: 1) Toss a coin twice ) 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 (A) = 0, P (B) = 0,5 P (A∪B) (which means P(A or B)) = 0,6 a) Are events A and B mutually exclusive? Motivate b) Are events A and B independent? Motivate a) P (A∪B) ≠ P (A) + P (B); so not mutually exclusive b) P (A∪B) = P(A) + P(B) - P (A and B) 0,6 = 0, + 0,5 - P (A and B) P (A and B) = 0,7 - 0,6 P (A and B) = 0,1 but P (A). P (B) = 0, x 0,5 =0,1 \ are independent Page 6 Probability 3 Exam Paper Example 2 A 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: Sep 2 Sep P (wet) 0,5 0,4 Key P (wet) 0,5 P (dry) ( - 0,4) = P (dry) = 0,6 0,7 P (dry) ( - 0,3) = P (second day dry) 0,6 = 0,7 P (wet) 0,3 a) The first two days in September are wet b) September nd is wet a) 0,4 x 0,5 = 0, [(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, + 0, + 0,4 + 0,18 =1 Example 3 P (A) = 0,5 and P (B) = 0,5 and P (A and B) = 0,15 a) Are events A and B independent? b) Are events A and B mutually exclusive? Page 7 Probability 3 Exam Paper a) P(A) x P (B) = 0,5 x 0,5 = 0,15 P (A and B) = 0,15 so 0,15 ≠ 0,15 NOT independent b) no since P (A or B) = P(A) + P (B) − P (A and B) ≠ P (A) + P (B) Example 4 Another way that we use to decide on Independence is by using a CONTINGENCY TABLE 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 0 40 50 110 Female 50 0 0 90 Total 70 60 70 00 This is called a two-way x3 Contingency Table . It has rows and 3 columns. Can you see that we could get the following from the table? 110 1) P (Male) = ___ = 0,55 00 60 ) P (Blue) = ___ = 0,3 00 40 3) P (Blue and Male) = ___ = 0, 00 60 110 40 130 4) P (Blue or Male) = ___ + ___ - ___ = ___ = 0,65 00 00 00 00 They are NOT mutually exclusive 5) Are the events ‘being Male’ and ‘preferring Blue’ independent or not? P (Blue and Male) = 0, 60 110 33 P (Blue) x P (Male) = ___ x ___ = ___ = 0,165 00 00 00 They are NOT independent 6) Are the events ‘being Female’ and ‘preferring Red’ independent or not? 50 P (Female and Red) = ___ = 0,5 00 90 70 63 P (Female) x P (Red) = ___ x ___ = ___ = 0,1575 00 00 400 They are NOT independent Page 8 Probability 3 Exam Paper 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 80 180 100 P (Male) = ___ 180 80 P (Female) = ___ 180 110 P (Positive) = ___ 180 70 P (not Positive) = ___ 180 60 P (Male and Positive) = ___ = 0,3 180 100 110 P (Male) x P (Positive) = ___ x ___ = 0,34 180 180 Mathematically these are NOT independent (OR) 50 P (Female and Positive) = ___ = 0,7 180 80 110 P (Female) x P (Positive) = ___ x ___ = 0,716 180 180 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. 4 So if we draw the first card P (King) = __5 Having removed one King and not replaced it means that there are now only 3 Kings in the remaining 51 3 cards so the probability of the second card also being a King is __51 So the Probability of two Kings in two successive draws of a pack of cards WITHOUT replacement is 4 3 P (King and King) = __ x __ 5 51 1 = ___ 1 Page 9 Probability 3 Exam Paper So what if we drew two cards and wanted the probability of one King.? The tree diagram would look as follows: st Draw 2nd Draw P (King) 3 __ 5 4 __ 52 P (King) 48 __ P (not King) 5 P (King) 4 __ P (not King) 5 48 __ 52 47 __ P (not King) 5 So the Probability of one King means = P (King followed by not King) or P (not King followed by King) = 4/5 x 48/51 + 48/5 x 4/51 3 = ___ 1 4 3 1 Please note: P (K,K) = __ x __ = ___ 5 51 1 4 48 16 P (K, not K) = __ x __ = ___ 5 51 1 48 4 16 P (not K; K) = __ x __ = ___ 5 51 1 48 47 188 P (not K; not K) = __ x __ = ___ 5 51 1 1 16 16 188 1 ___ ___ ___ ___ ____ now 1 + 1 + 1 + 1 = 1 = 1 A 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-DIAGRAM. The best way to get to understand how useful they are is with an example or two: Page 10 Probability 3 Exam Paper Example 1 Given P(A) = 0,7; P (B) = 0,4 and P (A∩B) = 0,8 Find out: a) P (A∪B)’ b) whether A and B are independent Best way to solve this is draw a picture: A B x 0,8 y now x + 0,8 = 0,7 x = 0,4 and y + 0,8 = 0,4 y = 0,1 a) Here P (A∪B) = P (A or B) = 0,4 + 0,8 + 0,1 = 0,8 so P (A∪B)’ = 1-0,8 = 0,18 b) P (A) x P (B) = 0,7 x 0,4 = 0,8 P (A∩B) = 0,8 so they are independent events Sometimes we get given the picture organiser and we are asked to interpret it. Page 11 Probability 3 Exam Paper S R C 1 108 54 36 6 18 134 97 J Example 2 A survey was conducted to determine the preferences for three different music types Rock (R), Classic (C), Jazz (J) 1) How many people were surveyed? ) 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 ) 108 36 3) ___ = 0,0774 = 7,74% 465 108 + 1+ 36 + 6 + 54 + 18 34 4) ____________________ =___ = 0,503 = 50,3% 46 465 5) False. They may well like other kinds of music - they just don’t like these three. 97+134 77 6) ______ = ___ = 0,4968 465 155 Page 1

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probability theory, Conditional Probability, possible outcomes, sample space, the experiment, probability distribution, probability distributions, random variables, Probability Problems, Independent Events

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posted: | 2/13/2011 |

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