S1: Chapter 5 Probability Dr J Frost (jfrost@tiffin.kingston.sch.uk) Last modified: 11th December 2013

Starter Draw a sample space (using a table) for throwing two dice and recording the product of the two values. What is the probability of the value being greater or equal to 24? ? Die 1 × 1 2 3 4 5 6 8 10 12 9 15 18 16 20 24 25 30 36 Die 2 ? 𝑃 𝑃𝑟𝑜𝑑𝑢𝑐𝑡≥24 = 6 36 = 1 6

Some fundamentals ? ? ? An experiment is: A repeatable process that gives rise to a number of outcomes. ? A sample space is: The set of possible outcomes of an experiment. e.g. The sample space 𝑆 of throwing two coins: 𝑆= 𝐻𝐻, 𝐻𝑇, 𝑇𝐻, 𝑇𝑇 ? ?

𝑃 𝑒𝑣𝑒𝑛𝑡 =0.3 Some fundamentals An event is a set of (one or more) outcomes. ? 𝑨 2 4 6 3 5 1 𝑩 𝑺 𝑆= the whole sample space 𝐴= even number on a die thrown 𝐵= prime number on a die thrown

Some fundamentals 𝑨 2 4 6 3 5 1 𝑩 𝑺 𝑆= the whole sample space 𝐴= even number on a die thrown 𝐵= prime number on a die thrown What does it mean in this context? What is the resulting set of outcomes? 𝐴′ ? Not A. i.e. Not rolling an even number. ? {1, 3, 5} 𝐴∪𝐵 ? A or B. i.e. Rolling an even or prime number. ? {2,3,4,5,6} 𝐴∩𝐵 ? A and B. i.e. Rolling a number which is even and prime. ? {2}

Some fundamentals 𝑨 2 4 6 3 5 1 𝑩 𝑺 𝑆= the whole sample space 𝐴= even number on a die thrown 𝐵= prime number on a die thrown What does it mean in this context? What is the resulting set of outcomes? 𝐴∩𝐵′ ? Rolling a number which is even and not prime. ? {4,6} (𝐴∪𝐵)′ ? Rolling a number which is not [even or prime]. ? {1} 𝐴∩𝐵 ′ ? Rolling a number which is not [even and prime]. ? {1,3,4,5,6}

What area is indicated? A C B S 𝐴∩𝐵 ?

What area is indicated? A C B S 𝐴∪𝐵 ?

What area is indicated? A C B S 𝐴∩𝐵∩𝐶 ?

What area is indicated? A C B S 𝐴∩ 𝐶 ′ ?

What area is indicated? A C B S 𝐴∩𝐵∩ 𝐶 ′ ?

A C B S 𝐴 ′ ∩ 𝐵 ′ ∩ 𝐶 ′ 𝐴∪𝐵∪𝐶 ′ What area is indicated? ? ? or alternatively… ? 𝐴∪𝐵∪𝐶 ′

What area is indicated? A C B S 𝐴 ′ ?

What area is indicated? A C B S 𝐴∩ 𝐵∩𝐶 ′ ?

Solving problems using Venn Diagrams A vet surveys 100 of her clients. She finds that 25 own dogs, 15 own dogs and cats, 11 own dogs and tropical fish, 53 own cats, 10 own cats and tropical fish, 7 own dogs, cats and tropical fish, 40 own tropical fish. Fill in this Venn Diagram, and hence answer the following questions: 𝑃 𝑜𝑤𝑛𝑠 𝑑𝑜𝑔 𝑜𝑛𝑙𝑦 𝑃 𝑑𝑜𝑒𝑠 𝑛𝑜𝑡 𝑜𝑤𝑛 𝑡𝑟𝑜𝑝𝑖𝑐𝑎𝑙 𝑓𝑖𝑠ℎ 𝑃(𝑑𝑜𝑒𝑠 𝑛𝑜𝑡 𝑜𝑤𝑛 𝑑𝑜𝑔𝑠, 𝑐𝑎𝑡𝑠, 𝑜𝑟 𝑡𝑟𝑜𝑝𝑖𝑐𝑎𝑙 𝑓𝑖𝑠ℎ) 𝑪 𝑺 ? 35 100 ? 11 100 8 100 ? ? 3 100 𝑫 7 100 𝑭 ? 26 100 ? 6 100 ? 4 100 Dr Frost’s cat “Pippin”

Exercises Page 84 Exercise 5B Q6

Recap ? ? ? If 𝐴 and 𝐵 are mutually exclusive, this means: they can not happen at the same time. ? On a Venn Diagram… the circles appear separately. ? 𝐴 𝐵 If 𝐴 and 𝐵 are independent, this means: one does not affect the other. Note that these 2 things are ENTIRELY DIFFERENT, they are not ‘opposites’. If 𝐴 and 𝐵 are not mutually exclusive, that doesn’t necessarily mean they are independent. The Venn Diagram is NOT AFFECTED BY INDEPENDENCE. ?

𝑃 𝐴∪𝐵 =𝑃 𝐴 +𝑃(𝐵) 𝑃 𝐴∩𝐵 =0 𝑃 𝐴∩𝐵 =𝑃 𝐴 ×𝑃 𝐵 Recap If events A and B are mutually exclusive, then: 𝑃 𝐴∪𝐵 =𝑃 𝐴 +𝑃(𝐵) ? 𝑃 𝐴∩𝐵 =0 ? If events A and B are independent, then: 𝑃 𝐴∩𝐵 =𝑃 𝐴 ×𝑃 𝐵 But we’re interested in how we can calculate probabilities when events are not mutually exclusive, or not independent.

Addition Law ? ? 𝑨 𝑩 𝑃 𝐴∪𝐵 =𝑃 𝐴 +𝑃(𝐵) 𝑩 𝑨 𝑃 𝐴∪𝐵 =𝑃 𝐴 +𝑃 𝐵 −𝑃 𝐴∩𝐵 Mutually Exclusive Think about the areas… 𝑨 𝑩 𝑃 𝐴∪𝐵 =𝑃 𝐴 +𝑃(𝐵) ? Not Mutually Exclusive 𝑩 𝑨 𝑃 𝐴∪𝐵 =𝑃 𝐴 +𝑃 𝐵 −𝑃 𝐴∩𝐵 ?

Example 𝑃 𝐴∩𝐵 =0.4 ? 𝑃 𝐴′ =0.4 ? 𝑃 𝐴 ′ ∪𝐵 =0.8 ? 𝑃 𝐴 ′ ∩𝐵 =0.3 ? 𝐴 and 𝐵 are two events such that 𝑃 𝐴 =0.6, 𝑃 𝐵 =0.7 and 𝑃 𝐴∪𝐵 =0.9. Find: 𝑃 𝐴∩𝐵 =0.4 ? 𝑃 𝐴′ =0.4 ? Bro Tip: You could use a Venn Diagram here. 𝑃 𝐴 ′ ∪𝐵 =0.8 ? 𝑃 𝐴 ′ ∩𝐵 =0.3 ?

Click to reveal Venn Diagram Check your understanding The events 𝐸 and 𝐹 are such that 𝑃 𝐸 =0.28 𝑃 𝐸∪𝐹 =0.76 𝑃 𝐸∩ 𝐹 ′ =0.11 Find a) 𝑃 𝐸∩𝐹 =0.17 b) 𝑃 𝐹 =0.65 c) 𝑃 𝐸 ′ 𝐹 ′ = 𝑃 𝐸 ′ ∩ 𝐹 ′ 𝑃 𝐹 ′ = 0.24 0.35 = 24 35 ? ? ? Click to reveal Venn Diagram 𝑺 𝑬 𝑭 0.11 0.17 0.48 Bro Tip: Venn Diagrams can typically be used when you have intersections involving ‘not’s. 0.24

Exercises Page 86 Exercise 5C Q1, 3, 5

Conditional Probability Think about how we formed a probability tree at GCSE: 𝑃 𝐴∩𝐵 =𝑃 𝐴 ×𝑃 𝐵 𝐴 P 𝐵|𝐴 ? 𝐵 ? 𝑃 𝐴 𝐴 𝐵′ 𝐵 𝐴′ 𝐵′ Alternatively: Bro Tip: You’re dividing by the event you’re conditioning on. 𝑃 𝐵 𝐴 = 𝑃 𝐴∩𝐵 𝑃 𝐴 ?

Quickfire Examples Given that P(A) = 0.5 and 𝑃 𝐴∩𝐵 =0.3, what is P(B | A)? 𝑷 𝑩 𝑨 = 𝑷 𝑨∩𝑩 𝑷 𝑨 = 𝟎.𝟑 𝟎.𝟓 =𝟎.𝟔 Given that P(Y) = 0.6 and 𝑃 𝑋∩𝑌 =0.4, what is 𝑃 𝑋 ′ 𝑌 ? P(X’ | Z) = 1 – P(X | Z) = 1 – (0.4/0.6) = 0.33 Given that P(A) = 0.5, P(B) = 0.5 and 𝑃 𝐴∩𝐵 =0.4, what is 𝑃 𝐵 𝐴 ′ ? (Hint: you’ll likely need a Venn Diagram for this!) 𝑷 𝑩 𝑨 ′ =𝑷( 𝑨 ′ ∩𝑩)/𝑷 𝑨 ′ = 0.1 / 0.5 = 0.2 ? ? ? Bro Tip: Note that P(A | B’) + P(A’ | B’) = 1 It is NOT in general true that: P(A’ | B’) = 1 – P(A | B)

Summary so far ? ? ? ? ? ? P 𝐴∩𝐵 =𝑃 𝐴 ×𝑃 𝐵 P 𝐴 𝐵 =𝑃(𝐴) P 𝐴∩𝐵 =0 If events 𝑨 and 𝑩 are independent. P 𝐴∩𝐵 =𝑃 𝐴 ×𝑃 𝐵 P 𝐴 𝐵 =𝑃(𝐴) ? ? If events 𝑨 and 𝑩 are mutually exclusive: P 𝐴∩𝐵 =0 P 𝐴∪𝐵 =𝑃 𝐴 +𝑃 𝐵 ? ? In general: ? P 𝐴 𝐵 = 𝑃 𝐴∩𝐵 𝑃 𝐵 P 𝐴∪𝐵 =𝑝 𝐴 +𝑝 𝐵 −𝑝 𝐴∩𝐵 ?

More difficult Venn Diagrams based on mutual exclusivity (Page 102) Events 𝐴, 𝐵 and 𝐶 are defined in the sample space 𝑆 such that 𝑝 𝐴 =0.4, 𝑝 𝐵 =0.2, 𝑃 𝐴∩𝐶 =0.04 and 𝑃 𝐵∪𝐶 =0.44. The events 𝐴 and 𝐵 are mutually exclusive and 𝐵 and 𝐶 are independent. a) Draw a Venn Diagram to illustrate the relationship between the three events and the sample space. [We’ll work out the probabilities later] ? 𝑆 Key Points: Recall that only mutual exclusivity affects the Venn Diagram. You will lose a mark if you forget the outer rectangle. 𝐶 𝐵 𝐴

More difficult Venn Diagrams based on mutual exclusivity (Page 102) Events 𝐴, 𝐵 and 𝐶 are defined in the sample space 𝑆 such that 𝑝 𝐴 =0.4, 𝑝 𝐵 =0.2, 𝑃 𝐴∩𝐶 =0.04 and 𝑃 𝐵∪𝐶 =0.44. The events 𝐴 and 𝐵 are mutually exclusive and 𝐵 and 𝐶 are independent. b) Find 𝑷(𝑩|𝑪), 𝑷(𝑪), 𝑷 𝑩∩𝑪 , 𝑷 𝑨 ′ ∩ 𝑩 ′ ∩ 𝑪 ′ , 𝑷 𝑪∩ 𝑩 ′ Bro Tip: Use the last sentence about mutual exclusivity/independence to immediately write out some extra information, e.g. 𝑃 𝐵∩𝐶 =0.2𝑃(𝐶) 𝑆 𝐶 𝐵 𝐴 0.36 ? 0.04 ? 0.2 0.06 ? 0.14 ? ? ? ? 𝑃 𝐵 𝐶 =0.2 𝑃 𝐶 =0.3 ? 𝑃 𝐴 ′ ∩ 𝐵 ′ ∩ 𝐶 ′ =0.2 ? 𝑃 𝐶∩ 𝐵 ′ =0.24

June 2013 Answer to (d): = 11 20 ? ? = 99 100 ? = 3 8 ? = 1 4 ?

Provided sheet of past paper questions! Exercises Provided sheet of past paper questions!

Exercises (on worksheet) a ? P(A u B) = P(A) + P(B) – P(A n B) = 0.67 P(A’ | B’) = P(A’ n B’) / P(B’) = 0.33 / 0.55 = 0.6 (We can see that P(A’ n B’) = 1 – P(A u B) by a quick sketch of a Venn Diagram) P(B n C) = P(B)P(C) = 0.09 (we can directly multiply because they’re independent) b ? c ? ? 0.22 d e ? Using a Venn Diagram, we can see that: P([B u C]’) = P(A n B’) + P(A’ n B’ n C’) = 0.22 + 0.22 = 0.44 0.22 0.13 0.09 0.11 0.23 A C B S

Exercises (on worksheet) B and W, or T and W. Because the circles don’t overlap/the events can’t happen at the same time. P(B n T) = 5/25 = 1/5 P(B)P(T) = 9/25 x 8/25 = 72/625 These are not the same so not independent. P(W) = 7/25 P(B n T) = 5/25 P(T | B) = 5 / 9 (either using the Venn Diagram directly, or by using P(T n B) / P(B) a ? b ? c ? d ? e ?

Probability Trees ? ? ? ? ? ? ? ? 1st pick 2nd pick 4 10 𝑅𝑒𝑑 1 2 𝑅𝑒𝑑 Trees are useful when you have later events conditioned on earlier ones, or in general when you have lots of conditional probabilities. Example: You have two bags, the first with 5 red balls and 5 blue balls, and the second with 3 red balls and 6 blue balls. You first pick a ball from the first bag, and place it in the second. You then pick a ball from the second bag. Complete the tree diagram. Hence find the probability that: You pick a red ball on your second pick. 𝑃 𝑅 2 =𝑃 𝑅 1 ∩ 𝑅 2 +𝑃 𝐵 1 ∩𝑅 2 = 1 5 + 3 20 = 7 20 Given that your second pick was red, the first pick was also red. 𝑃 𝑅 1 𝑅 2 = 𝑃 𝑅 1 ∩ 𝑅 2 𝑃 𝑅 2 = 1 5 7 20 = 4 7 1st pick 2nd pick 4 10 ? 𝑅𝑒𝑑 ? 1 2 ? 𝑅𝑒𝑑 6 10 ? 𝐵𝑙𝑢𝑒 3 10 ? 𝑅𝑒𝑑 1 2 ? 𝐵𝑙𝑢𝑒 ? 7 10 ? 𝐵𝑙𝑢𝑒

Probability Trees Key Point: When you need to find a probability using a tree, consider all possible paths in which that event is satisfied, and add the probabilities together. 𝐶 𝐵 𝑃 𝐶 =𝑃 𝐴∩𝐵∩𝐶 +𝑃 𝐴∩ 𝐵 ′ ∩𝐶 +𝑃 𝐴 ′ ∩𝐵∩𝐶 +𝑃 𝐴 ′ ∩ 𝐵 ′ ∩𝐶 𝑃 𝐵∩𝐶′ =𝑃 𝐴∩𝐵∩ 𝐶 ′ +𝑃 𝐴 ′ ∩𝐵∩ 𝐶 ′ 𝑃 𝐵 =𝑃 𝐴∩𝐵 +𝑃 𝐴 ′ ∩𝐵 (Notice that we can completely ignore C here) 𝑃 𝐴′∩𝐶 =𝑃 𝐴 ′ ∩𝐵∩𝐶 +𝑃 𝐴 ′ ∩ 𝐵 ′ ∩𝐶 ? 𝐶′ 𝐴 𝐶 𝐵′ ? 𝐶′ 𝐵 𝐶 ? 𝐴′ 𝐶′ 𝐶 ? 𝐵′ 𝐶′

Click to reveal Tree Diagram Check your understanding Of 120 competitors in a golf tournament, 68 reached the green with their tee shot on the first hole. Of these, 46 completed the hole in 3 shots or less. In total, 49 players took more than 3 shots on the first hole. Click to reveal Tree Diagram Find the probability that a player chosen at random: Reached the green with his tee shot and took more than 3 shots in total. P R∩ 𝐶 ′ = 68 120 × 22 68 = 22 120 Missed the green on his tee shot and took at most 3 shots. P 𝑅 ′ ∩𝐶 = 52 120 × 27 52 = 27 120 Took 3 shots or less in total, given that he missed the green with his tee shot. 𝑃 𝐶 𝑅 ′ = 𝑃 𝑅 ′ ∩𝐶 𝑃 𝑅 ′ = 27 120 52 120 = 27 52 46 68 𝐶 68 120 𝑅 22 68 𝐶′ 25 52 ? 𝐶 52 120 𝑅′ ? 27 52 𝐶′ ? I’ve used 𝑅 to represent the event “reached the green with tee shot on first hole” and 𝐶 to mean “completed shot in 3 shots or less”.

Exercises (on worksheet) P(H) = (5/12 x 2/3) + (7/12 x ½) = 41/72 P(R|H) = P(R n H) / P(H) = (5/18) / (41/72) = 20/41 P(RR or BB) = (5/12)2 + (7/12)2 = 37/72 b ? ? 2 3 c ? 5 12 ? 1 3 ? d ? ? 1 2 7 12 ? ? 1 2

Classic Conundrum I have two children. One of them is a boy. What is the probability the other is a boy? 𝐴𝑛𝑠𝑤𝑒𝑟= 1 3 ? ? The ‘restricted sample space’ method There’s four possibilities for the sex of the two children, but only 3 match the description. In 1 out of the 3 possibilities BB BG GB GG METHOD 1 ? Using conditional probability 𝑝 𝑜𝑡ℎ𝑒𝑟 𝑖𝑠 𝑏𝑜𝑦 𝑜𝑛𝑒 𝑖𝑠 𝑎 𝑏𝑜𝑦 = 𝑝 𝑜𝑛𝑒 𝑖𝑠 𝑎 𝑏𝑜𝑦 𝐴𝑁𝐷 𝑜𝑡ℎ𝑒𝑟 𝑖𝑠 𝑎 𝑏𝑜𝑦 𝑝 𝑜𝑛𝑒 𝑖𝑠 𝑎 𝑏𝑜𝑦 = 1/4 3/4 = 1 3 METHOD 2