1. Page 79 Exercise 5A Homework - using GCSE notes for review prior to starting this unit

2. 100 in a sixth form 34 take Maths 42 take Physics 24 take Chemistry 18 take Biology How would you represent this data? Why does this come to more than 100? Intersections and Unions

3. If I tell you 24 Mathematicians also take Physics what is the probability I pick at random a student from this sixth form who only takes Maths? What is the probability a student takes Maths or Physics?

5. A card is selected at random from a normal set of 52 playing cards. Let Q be the event that the card is a Queen and D be the event the card is a Diamond. What do you think the following questions are asking? P(Q D) P(QUD) P(Q'UD) P(Q D') Try solving them.

6. How might you calculate P(A U B)?

7. It's important we know the names for these notations too A U B means A's "union" with B A or B happens or both A B means A's "intersection" with B A and B happens A' means A's complement Everything that isn't A happens P (A U B) = P(A) + P(B) - P(A B) Can be rearranged to... P(A B) = P(A) + P(B) - P (A U B)

8. When solving these style of questions a Venn Diagram is helpful Don't forget that drawing these on Venn diagrams makes it much simpler enter the values in the correct bits and shade in what is required. If it's an "Intersection" you use the values that have been shaded by BOTH If it's a "Union" you use ALL the values that have been shaded

9. Page 83 Exercise 5B Q1 - 6 Finish for homework Now check your answers and correct or seek help where necessary

10. Page 87 Exercise 5C We'll do Q4 together Now try Q1, 3, 6 and 7 Putting the information into a Venn diagram WILL help Complete for homework and check it. Seek help with those questions you don't understand BEFORE you hand it in.

12. A bag contains 6 red and 4 blue beads. A bead is picked out and retained and then a second bead is picked out. Find the probability that: a) both beads are red b) the beads are of different colours c) the second bead is red given that the first one is blue? Conditional Probability

13. Remember our sixth form from earlier 100 in a sixth form 34 take Maths 42 take Physics But only 24 of the mathematicians also study Physics If I went into the Maths classroom what is the probability I pull out a physicist? How does this relate to the figures we have?

14. Another piece of notation we will use shows conditional probability P(A / B) means the probability A happens given that B has already happened You have already met this when you learned about tree diagrams. We can use our new notation with tree diagrams and generate formulae which will be useful Don't forget formulae can be rearranged.

15. Of course this can be rearranged as shown below. Sometimes this is a more useful format so learn both It's useful to note that the event after the slash / is the event you divide by and also the event which happens first. It can often be useful to sketch a tree diagram to put the information on. However you may also need to consider a Venn diagram to calculate the P(A B) part

16. Page 90 Exercise 5D Q1, 2 and 3 warm you up to the phrase "given that" We'll try Q7 together Now try Q4 and 6 so you practice using the notation Tree Diagrams Page 94 Exercise 5E Q1,3 and 5 As always finish for homework

17. The Chevalier du Mere's Problem. A seventeenth-century French gambler had run out of takers for his bet that, when a fair cubical dice was thrown four times, at least one six would be scored. He therefore changed the game to throwing a pair of dice 24 times. What is the probability that, out of these 24 throws, at least one is a double six? (You may wish to consider his first problem with 4 dice to warm up) The Birthday Problem What is the probability that out of 23 randomly chosen people, at least two share a birthday? Assume that all 365 days of the year are equally likely and ignore leap years. (You may want to find the probablities that two people have different birthdays, that three people have different birthdays, and so on first)

18. A question might give you information about P(B/A) but then ask you P(A/B) which requires you to consider it from the perspective that B happened first. For any events A and B, write P(A and B) and P(B) in terms of P(A), P(A'), P(B/A) and P(B/A'). Now write P(A/B) in terms of P(A), P(A'), P(B/A) and P(B/A')

19. This is known as Bayes' Theorem: You may look the formula up on Google and use it to help in any questions where the order of events has been reversed

20. The Doctor's Dilemma It is known that among all patients displaying a certain set of symptoms the probability they have a particular rare disease is 0.001. A test for the disease has been developed. The test shows a positive result on 98% of the patients who have the disease and on 3% of patients who do not have the disease. The test is given to a patient displaying the symptoms and it records a posiitve result. Find the probability that the patient has the disease. Comment on your result. The Prosecutor's Fallacy. An accused prisoner is on trial. The defence lawyer asserts that in the absence of further evidence the probability that the prisoner is guilty is one in a million. The prosecuting lawyer produces a further piece of evidence and asserts that if the prisoner was guilty the probability that this evidence could be obtained is 999 out of 1000 and if he were not guilty it would only be 1 in 1000; in other words, P(evidence / guilty) = 0.999 and P(evidence / not guilty) = 0.001. Assuming that the court excepts the legality of the evidence and that both lawyers figures are correct what is the probability that the prisoner is guilty? Comment on your answer.

21. You should remember earlier we considered a particular sixth form. 100 in a sixth form 24 take Chemistry 18 take Biology If I told you the probability a student takes Chemistry or Biology is 0.42 what else does that tell you?

22. Because there couldn't possibly be anyone taking Biology and Chemistry we call them Mutually Exclusive - they can't both happen If two events A and B are Mutually Exclusive then P(A and B) = 0 This makes our equation from before: P(AUB) = P(A) + P(B) There is a similar result affecting the other formula we've been looking at in this chapter P(A) x P(B/A) = P(A and B) logically if P(A) x P(B) = P(A and B) then P(B/A) = P(B) which means B is unaffected by A this is known as being Independent of A

23. Page 100 Exercise 5F We'll do Q3 and 6 together Now try Q1, 2, 4 and 8 Don't forget to check your answers, ask and correct.

24. Page 104 has a mixed exercise of exam style questions for revision practice of your own.