Lesson Objective Understand the Basic Terminology used in probability Understand the use of the symbols ∩ and U and their rough translations in to everyday English What is Probability?

A trial is an experiment that has a number of possible outcomes Probability is about measuring the likelihood that an event happens as part of a trial A trial is an experiment that has a number of possible outcomes The possibility space is a list of all the possible outcomes connected with a trial An event is a subset of the possibility space consisting of the outcomes we are interested in If all the outcomes of a trial are equally likely then: P(event happens) = Number of ways an event can happen Number of items in the possibility space

In a survey of 200 people, 75 said they had shopped at Tesco in the last week, whilst 123 said that they had shopped at Sainsburys. 17 had not visited either supermarket. If you picked a shopper at random from the 200 in the survey, what is the probability: That they had not visited either Sainsburys or Tescos? That they had visited Tescos? That they had not visited Sainsburys? That they has shopped in at least one of Tescos or Sainsburys?

  A B A B A B A B

Venn Diagrams are Great For Probability Eg Suppose there are 100 pupils in a school sixth form, 50 study maths, 29 studying biology and 13 studying both Draw a Venn Diagram and calculate: P(B’) b) P(M ∩ B) c) P(M ∩ B’) d) P(M U B) e) P(M U B’) f) P(M’ ∩ B) g) What can you say about P(B) and P(B’) – is this generally true?

Venn Diagrams are Great For Probability Eg Suppose there are 100 pupils in a school sixth form, 50 study maths, 29 studying biology and 13 studying both M B 37 13 16 34

Lesson Objective Consolidate the use of the symbols ∩ and U and their rough translations in to everyday English Begin to consider rules for calculating P(A U B) A B   5 2 6 4  

I roll 2 fair six sided die. One is Red, the other is Blue. The word total refers to the sum of the score on the two dice. Let R = Score on the Red Die is Even B = Score on the Blue Die is less than 3 T = Total score is 7 W = Total score is Even Find: P(R) i) P(T U W) P(R ∩ B) j) P(W ∩ B) P(R U B) k) P(B’ U R) P(T ∩ R) l) P(T’ ∩ W) P(B ) m) P(T’ U W’) P(B ∩ T) n) P(B U T’) P(R ∩ W) o) P( (R U B) ∩ T) P(R U W) p) P( (W ∩ R) U T)

I roll 2 fair six sided die. One is Red, the other is Blue. The word total refers to the sum of the score on the two dice. Let R = Score on the Red Die is Even B = Score on the Blue Die is less than 3 T = Total score is 7 W = Total score is Even Find: P(R) i) P(T U W) P(R ∩ B) j) P(W ∩ B) P(R U B) k) P(B’ U R) P(T ∩ R) l) P(T’ ∩ W) P(B ) m) P(T’ U W’) P(B ∩ T) n) P(B U T’) P(R ∩ W) o) P( (R U B) ∩ T) P(R U W) p) P( (W ∩ R) U T) 24/36 6/36 30/36 18/36 1 32/36 4/36 15/36 ½ 6/36 24/36 3/36 1/3 2/36 9/36 27/36

 

I roll a red die and a black die. Find each probability and decide if the events are mutually exclusive: a) P(A total of 4 U a total of 6) = 8/36 ME b) P(A total of 8 U a total that is prime) = 20/36 ME c) P(A total that is even U a total more than 9) = 20/36 d) P(A 4 on the red die U a 6 on the black die) = 11/36 e) P(A total that is even U a total that is prime) = 32/36 f) P(A total that is more than 7 U a 5 on the black die) = 17/36 g) P(A total less than 5 U a total more than 10) =9/36 ME

Lesson Objective Understand what it means for two events to be mutually exclusive Begin to consider rules for calculating P(A U B) and understand how events being mutually exclusive affects these rules

If two events are Mutually Exclusive what does this mean? Give me an example of two events that are mutually exclusive

If two events are Mutually Exclusive what does this mean? Give me an example of two events that are mutually exclusive A B A B Mutually exclusive means that the two events do not overlap. They cannot happen at the same time or, in maths notation: P(A ∩ B) = 0

At GCSE we often say that P(A OR B) = P(A) + P(B) What we mean by this is that P(A U B) = P(A) + P(B) iff A and B are mutually exclusive What we should really say is that: P(A OR B) = P(A) + P(B) – P(overlap) Or using proper mathematical notation:

3) Given that , P(A’) = , P(B) = , P(A ∩ B) = . Find P(A U B) 1) In a group of 20 adults, 4 out of the 7 women and 2 out of the 13 men wear glasses. What is the probability that a person chosen at random from the group is a woman or someone who wears glasses?   2) In a street containing 20 houses, 3 households do not own a television set, 12 households have a black and white set, and 7 households have a colour and a black and white set. Find the probability that a household chosen at random owns a colour television set. 3) Given that , P(A’) = , P(B) = , P(A ∩ B) = . Find  P(A U B) 4) A card is selected at random from a regular set of 52 playing cards. Let Q be the event that the card is a queen and D be the event that the card is a diamond. Draw a Venn diagram to show these events. Find a) P(Q ∩ D) b) P(Q U D) c) P(Q’ U D) d) P(Q ∩ D’)  

3) Given that , P(A’) = , P(B) = , P(A ∩ B) = . Find P(A U B) 9/12 1) In a group of 20 adults, 4 out of the 7 women and 2 out of the 13 men wear glasses. What is the probability that a person chosen at random from the group is a woman or someone who wears glasses? 9/20   2) In a street containing 20 houses, 3 households do not own a television set, 12 households have a black and white set, and 7 households have a colour and a black and white set. Find the probability that a household chosen at random owns a colour television set. 12/20 3) Given that , P(A’) = , P(B) = , P(A ∩ B) = . Find  P(A U B) 9/12 4) A card is selected at random from a regular set of 52 playing cards. Let Q be the event that the card is a queen and D be the event that the card is a diamond. Draw a Venn diagram to show these events. Find a) P(Q ∩ D) 1/52 b) P(Q U D) 16/52 c) P(Q’ U D) 49/52 d) P(Q ∩ D’) 3/52  

In a class of 23 students 15 are girls and the rest are boys. 7 of the girls complete their homework on time and 4 of the boys complete their homework on time. If a randomly selected student is chosen from the class what is the probability that they are either a boy or have completed their homework on time?

In a class of 23 students 15 are girls and the rest are boys. 7 of the girls complete their homework on time and 4 of the boys complete their homework on time. If a randomly selected student is chosen from the class what is the probability that they are either a boy or have completed their homework on time?   B H 4 7 4 8

Lesson Objective Derive a rule to find the P(A ∩ B) Understand the importance of independence and be able to recognise when two events are independent.

In maths it is often said that P(A ∩ B) = P(A) × P(B) Consider the following situations and decide if this rule is true: I roll a fair Blue die and a fair Red die R = Score a Red 6 M = Red Score more than 3 B = Score a Blue 6 N = Blue Score more than 4 Find P(R) P(B) P(N) P(M) P(R ∩ B) P(M ∩ N) P(B ∩ N) P(B ∩ M) P(R ∩ M) P(R ∩ N) Is it true that: P(R ∩ B) = P(R) × P(B) P(M ∩ N) = P(M) × P(N) P(B ∩ N) = P(B) × P(N) P(B ∩ M) = P(B) × P(M) P(R ∩ M) = P(R) × P(M) P(R ∩ N) = P(R) × P(N)

In maths it is often said that P(A ∩ B) = P(A) × P(B) Consider the following situations and decide if this rule is true: I roll a fair Blue die and a fair Red die R = Score a Red 6 M = Red Score more than 3 B = Score a Blue 6 N = Blue Score more than 4 Find P(R) 1/6 P(B) 1/6 P(M) 3/6 P(N) 2/6 P(R ∩ B) 1/36 P(M ∩ N) 6/36 P(B ∩ N) 6/36 P(B ∩ M) 3/36 P(R ∩ M) 6/36 P(R ∩ N) 2/36 Is it true that: P(R ∩ B) = P(R) × P(B) yes P(M ∩ N) = P(M) × P(N) yes P(B ∩ N) = P(B) × P(N) no P(B ∩ M) = P(B) × P(M) yes P(R ∩ M) = P(R) × P(M) no P(R ∩ N) = P(R) × P(N) yes Why does the rule work sometimes and not others? What stops it from working all the time?

P(R) 1/6 P(B) 1/6 P(M) 3/6 P(N) 2/6 R = Score a Red 6 M = Red Score more than 3 B = Score a Blue 6 N = Blue Score more than 4 P(R) 1/6 P(B) 1/6 P(M) 3/6 P(N) 2/6 P(B ∩ N) = P(B) × P(N) no P(B ∩ M) = P(B) × P(M) yes

The problem is independence! The rule P(A ∩ B) = P(A) × P(B) only works if the events A and B are events that are independent P(Blue 6) P(Blue score more than 4) These events not independent because if I know that the score is a Blue 6 then the Blue score must be bigger than 4 If knowing the result of one outcome alters the probability of the second outcome then we say that the items are not independent

Independent or not? I flip a fair coin twice. A = Head on 1st flip B = Head on 2nd flip A draw has 3 blue and 3 yellow socks. I draw two socks from the a draw at the same time A = 1st sock yellow B = 2nd sock yellow I pick 2 different students from this class. A = 1st student a girl B = 2nd student a Boy I look at the weather on the 1st day and last days of a 10 day holiday A = It snows on day 1st day B = It snows on 2nd day I pick a teacher and student from a school A = Teacher’s name is David B = Student’s name is David

In fact the rule should be: P(A ∩ B) = P(A) × P(B׀A) This is called a conditional probability and means: The probability that ‘B happens given that we know A has already happened’ If A and B are independent then P(B׀A) = P(B) And the rule reverts back to P(A ∩ B) = P(A) × P(B)

8 kittens in a bag. 5 Black, 3 White. Randomly pick a kitten from a bag. Look at it and replace it. Shake the bag, then take another kitten. What is the probability that the two kittens are different colours? 8 kittens in a bag. 5 Black, 3 White. Randomly pick a kitten from a bag. Throw the kitten away then take another kitten from the bag. What is the probability that the two kittens are different colours?

8 kittens in a bag. 5 Black, 3 White. Randomly pick a kitten from a bag. Look at it and replace it. Shake the bag, then take another kitten. What is the probability that the two kittens are different colours? 8 kittens in a bag. 5 Black, 3 White. Randomly pick a kitten from a bag. Throw the kitten away then take another kitten from the bag. What is the probability that the two kittens are different colours?

P(A U B) = P(A) + P(B) – P(A ∩ B) Lesson Objective Be able to calculate conditional probabilities by counting Begin to use conditional probabilities with our rules two probability rules: P(A U B) = P(A) + P(B) – P(A ∩ B) P(A ∩ B) = P(A) x P(B ׀A) to solve problems. Eg I roll a fair six sided die E = Get an even P = Get a prime B = Get a number below 3 Find a) P(E|P) b) P(E|B) c) P(P|B)

Consider rolling two fair die – one red and one blue: R = 6 on the Red die B = 6 on the Blue die T = Total score of 12 E = Total score is Even Find: a) P(R ׀ B) b) P(B ׀ R) c) P(R ׀ T) d) P(T ׀ R) e) P(R ׀ E) f) P(E ׀ R) g) P(B ׀ T) h) P(T ׀ B) i) P(T ׀ E) j) P(E ׀ T) k) P(R’ ׀ B) l) P(B’ ׀ R) m)P(T’ ׀ E) b) P(E’ ׀ T) a) Is it true that P(A׀ B) = P(B ׀ A) ? b) What do you notice about P(A׀ B) and P(A’ ׀ B )?

Consider rolling two fair die – one red and one blue: R = 6 on the Red die B = 6 on the Blue die T = Total score of 12 E = Total score is Even Find: a) P(R ׀ B) b) P(B ׀ R) c) P(R ׀ T) d) P(T ׀ R) e) P(R ׀ E) f) P(E ׀ R) g) P(B ׀ T) h) P(T ׀ B) i) P(T ׀ E) j) P(E ׀ T) k) P(R’ ׀ B) l) P(B’ ׀ R) m)P(T’ ׀ E) b) P(E’ ׀ T) a) Is it true that P(A׀ B) = P(B ׀ A) ? b) What do you notice about P(A׀ B) and P(A’ ׀ B )? 1/2 1/6 1 1/6 3/18 3/6 1 1/6 1/18 1 5/6 5/6 17/18 No, not in general They always add up to 1

Use puzzle pieces on the next two pages with this table to complete the puzzle:

These are not the same so dependent P(A ∩ B) = P(A) x P(B|A) = 0.4 x 0.6 = 0.24 0.16 0.06 0.24 1 - 0.4 -0.06 = 0.54 P(B ∩ A’) =0.06 P(Choc ∩ Crisps) = 0.75 P(Choc) x P(Crisps) = 0.9 x 0.8 = 0.72 These are not the same so dependent Choc Crisps 0.15 0.75 0.05 P(Choc ∩ Crisps’) = 0.15 P( (Choc U crisps)’) = 1 – 0.15 – 0.75 -0.05 = 0.05 A B P(A ∩ B) = P(B) x P(A|B) = 0.5 x 0.1 = 0.05 0.1 0.05 0.45 No P(A ∩ B) = 0.05 is not the same as P(A)xP(B)=0.075 0.4 No – she should eat more takeaways because P(A|B) = 0.1 and P(A|B’) = 0.2 P(A) = 0.15

Lesson Objective Be able to use tree diagrams to solve probability questions Focus in particular on questions that include a ‘reversed conditional probability’ part (Bayes theorem) Consider the following probability question that may (or may not be) answered with a tree diagram: The probability that it is sunny on any given day is 1/3. If it is sunny the probability that I go swimming is 3/4 , if it is not sunny the probability that I go swimming is only 1/5. Draw a tree diagram to illustrate this information What is the probability that I go swimming? If you know that I went swimming what is the probability that it was sunny on that particular day?

Consider the following probability question that may (or may not be) answered with a tree diagram: The probability that it is sunny on any given day is 1/3. If it is sunny the probability that I go swimming is 3/4 , if it is not sunny the probability that I go swimming is only 1/5. Draw a tree diagram to illustrate this information What is the probability that I go swimming? If you know that I went swimming what is the probability that it was sunny on that particular day?

Note: To answer these ‘backwards’ conditional probability questions it is important to notice that: P(A ∩ B) = P(B ∩ A) So that P(A) × P(B ׀A) = P(A ∩ B) = P(B ∩ A) = P(B) × P(A ׀B) Meaning that P(A ׀B) = P(A ∩ B) P(B) (Essentially the number of times that A happens at the same time as B as a proportion of the number of times B happens)

3. A student regularly has to take two train journeys on a Sunday 3. A student regularly has to take two train journeys on a Sunday. Over a long period of time she has worked out the probability that the first train is late is 0.4. If the first train is late, the probability that the second train is late is 0.5. If the first train is not late, the probability that the second train is late is 0.3. (i) Draw a tree diagram to show the possible outcomes for her next 2 journeys on a Sunday. (ii) Using the tree diagram calculate the probability: (A) that exactly one journey is on time, (B) that both trains are on time, given that the second journey is on time. 5. In the build up to the Olympics a high jumper measured her success at a particular height. She has a maximum of 3 attempts at this height; once she has jumped successfully she does not jump that height again. On 60% of occasions she clears it at the first attempt. When she attempts the height for the second time, she is successful in 75% of the attempts. When she attempts the height for the third time, she is successful in only 30% of the attempts. (i) Draw a tree diagram to show the possible outcomes. (ii) From the tree diagram calculate the probability (A) that she successfully clears the height in her three attempts, (B) that, given she clears the height, that she does so successfully at the first attempt.

Lesson Objective Understand the seriousness of misunderstanding the difference between P(A ׀B) and P(B ׀A). Recognise that people do this all the time in every day life and it is very easy to slip into Imagine that there is a rare, but deadly, disease that only affects 0.1% of the population. The test for the disease is very reliable, being accurate 95% of the time. You decide to test for the disease and the result comes back positive. How worried should you really be?

Imagine that there is a rare, but deadly, disease that only affects 0 Imagine that there is a rare, but deadly, disease that only affects 0.1% of the population. The test for the disease is very reliable, being accurate 95% of the time. You decide to test for the disease and the result comes back positive. How worried should you really be? Have Dis and +ve = 0.001 x 0.95 = 0.00095 Test positive given that you have disease 0.95 Have disease 0.001 Test negative given that you have disease 0.05 Have Dis and -ve = 0.001 x 0.05 = 0.00005 Test positive given that you do not have disease 0.05 Don’t have Dis and +ve = 0.999 x 0.05 = 0.04995 Don’t have disease 0.999 P(have disease ׀ +ve) = P(+ve ∩have disease) P(+ve) Test negative given that you do not have disease 0.95 Don’t have Dis and -ve = 0.999 x 0.95 = 0.94905 = 0.00095 ÷ (0.00095+0.04995) = 0.0187 ie around 2% chance !!!

Lesson Objective: Final Probability Overview Practise Look at the variety of different probability questions

A school has 120 students in year 12. 94 students do English or Maths 57 students do maths and 49 do English What is the probability that a student chosen at random does neither Maths or English What is the probability that a student chosen at random does both English and Maths What is the probability that a student chosen at random does English but not Maths If two students are selected at random what is the probability that they both do Maths and English

Number of Xmas presents bought by sixth form students 1 2 3 Probability for Mrs Smith 0.6 0.2 0.1 Probability for Mr Banham 0.4 0.3 What is the probability that Mrs Smith receives at least one present? What is the probability that Mr Banham and Mrs Smith both receive no presents? What is the probability that Mr Banham receives more presents than Mrs Smith? Given that Mr Banham received more presents than Mrs Smith what is the probability that he received 3 presents.

There are two A-level maths classes in Year 12 There are two A-level maths classes in Year 12. Group X contains eleven girls and seven boys. Group Y contains six girls and eight boys. Two students are chosen at random. Find the probability that a) they are of the same gender, b) they are from the same group, c) they are of the same gender and from the same group, d) they are of the same gender, given that they are from the same group, e) they are from the same group, given that they are of the same gender. In a large Sixth Form, 40% of the students do Mathematics, 30% do Physics and 24% do both Mathematics and Physics. Illustrate this information using a Venn diagram. Hence or otherwise find the probability that a student, chosen at random: (i) does Physics but not Mathematics, (ii) does neither Mathematics nor Physics, (iii) does not do Physics, given that he or she does not do Mathematics.

A school has 120 students in year 12. 94 students do English or Maths 57 students do maths and 49 do English What is the probability that a student chosen at random does neither Maths or English What is the probability that a student chosen at random does both English and Maths What is the probability that a student chosen at random does English but not Maths If two students are selected at random what is the probability that they both do Maths and English English Maths 37 12 45 26 26/120 12/120 37/120 12/120 x 11/119 = 132/14280

Number of Xmas presents bought by sixth form students 1 2 3 Probability for Mrs Smith 0.6 0.2 0.1 Probability for Mr Banham 0.4 0.3 What is the probability that Mrs Smith receives at least one present? What is the probability that Mr Banham and Mrs Smith both receive no presents? What is the probability that Mr Banham receives more presents than Mrs Smith? Given that Mr Banham received more presents than Mrs Smith what is the probability that he received 3 presents. 0.2+0.1+0.1 = 0.4 0.6 x 0.4 = 0.24 Mrs S 0 and Mr B More 0.6 x 0.6 Mrs S 1 and Mr B More 0.2 x 0.3 Mrs S 2 and Mr B More 0.1 x 0.2 Total = 0.36 +0.06 +0.02 = 0.44 0.6 x 0.2 +0.2 x 0.2 + 0.1 x 0.2 0.44 = 0.18/0.44 = 9/22

For last question this may help: 0 1 2 3 0 0.24 0.08 0.04 0.04 B 1 018 0.06 0.03 0.03 2 0.06 0.02 0.01 0.01 3 0.12 0.04 0.01 0.02

Illustrate this information using a Venn diagram. There are two A-level maths classes in Year 12. Group X contains eleven girls and seven boys. Group Y contains six girls and eight boys. Two students are chosen at random. Find the probability that a) they are of the same gender, GG + BB = 17/32 x 16/31 + 15/32 x 14/31 b) they are from the same group, XX + YY = 18/32 x 17/31 + 14/32 x 13/31 c) they are of the same gender and from the same group, GXX BXX GYY BYY 11/18 x 10/17 + 7/18 x 6/17 + 6/14 x 5/13 + 8/14 x 7/13 = d) they are of the same gender, given that they are from the same group, answer c /answer b e) they are from the same group, given that they are of the same gender. answer c /answer a X Y G 11 6 17 B 7 8 15 18 14 32 In a large Sixth Form, 40% of the students do Mathematics, 30% do Physics and 24% do both Mathematics and Physics. Illustrate this information using a Venn diagram. Hence or otherwise find the probability that a student, chosen at random: (i) does Physics but not Mathematics, (ii) does neither Mathematics nor Physics, (iii) does not do Physics, given that he or she does not do Mathematics. 0.06 0.54 Maths Physics 0.54/0.6 = 0.9 0.16 0.24 0.06 0.54