S1 Chapter 5 :: Probability

S1 Chapter 5 :: Probability
Dr J Frost Last modified: 17th November 2015

For Teacher Use Recommended lesson structure:
Lesson 1: Finding probabilities by enumerating outcomes (Ex 1) Lesson 2: Venn Diagrams involving Frequencies (Ex 2) (may be able to start Lesson 3 in same period so you can spend more on Lesson 4) Lesson 3: Events, ∪/∩/complements, identifying regions on a Venn Diagram, determining probabilities using Venn Diagrams. (Ex 3) Lesson 4: Independent and mutually exclusive events. Conditional probabilities. Laws of probability. (Ex 4) Lesson 5: Tree Diagrams. (Ex 5) Lesson 6: Consolidation/Assessment Go > Go > Go > Go > Go >

Overview In the exam you’re expected to be able to solve the following types of questions: (and I’ve checked since 2002!) #1: Enumerate matching outcomes to calculate probabilities. #2: Construct a Venn diagram (of frequencies) given information and calculate probabilities from it.

Overview In the exam you’re expected to be able to solve the following types of questions: (and I’ve checked since 2002!) #3: Construct and use Venn Diagrams involving Probabilities #4: Use laws of probability (independent, mutually exclusive and conditional events)

Overview #5: Deal with tree diagrams.

#1 :: Enumerate matching outcomes
? A sample space is the set of all possible outcomes. At GCSE you should be familiar with listing outcomes and then finding the probability of each. I spin a spinner twice, where 𝑆 1 is the outcome of the first spinner and 𝑆 2 the outcome of the second. The probability of the outcomes on each spin is given below: 𝑺 𝟏 𝟐 𝟑 𝟒 𝑃(𝑆) 0.1 0.2 0.3 0.4 Bro Tip: It’s sometimes helpful to ‘group’ outcomes (e.g. “≥4”). What is the probability of: Spinning two 3s? 𝟎.𝟑×𝟎.𝟑=𝟎.𝟎𝟗 Spinning a 2 and 4 in either order: 𝟎.𝟐×𝟎.𝟒 + 𝟎.𝟒×𝟎.𝟐 =𝟎.𝟏𝟔 Let 𝑋 be the product of my two spins, i.e. 𝑋= 𝑆 1 × 𝑆 2 . Find the probability that: c) 𝑃(𝑋=4) d) 𝑃(𝑋>2) ? ? ? 𝑆 𝑆 𝑃 𝑆 1 × 𝑆 𝑆 𝑆 𝑃 𝑆 1 × 𝑆 ≥ ×0.7= ≥ ×0.9= ≥ 𝑎𝑛𝑦 ×1=0.7 ? ∴𝑷 𝑿=𝟒 =𝟎.𝟏𝟐 ∴𝑃 𝑋>2 =𝟎.𝟗𝟓

Test Your Understanding
May 2013 (R) The score S when a spinner is spun has the following probability distribution. (Parts (a) to (e) we are skipping) The spinner is spun twice. The score from the first spin is S1 and the score from the second spin is S2. The random variables S1 and S2 are independent and the random variable X = S1 × S2. (f) Show that P({S1 = 1} ∩ X < 5) = (2) (g) Find P(X < 5) (3) s 1 2 4 5 P(S = s) 0.2 0.1 0.3 This effectively means “and” ? f 𝑆 𝑆 𝑃 𝑆 1 × 𝑆 < ×0.8=0.16 g 𝑆 𝑆 𝑃 𝑆 1 × 𝑆 2 𝑎𝑛𝑦 ×1=0.2 ≤ ×0.8= ≤ ×0.5= ≤ ×0.4= ×0.2=0.04 ∴𝑷 𝑿<𝟓 =𝟎.𝟓𝟕 ?

Exercise 1 ? ? ? ? (on provided sheet) 0.3×0.2=0.06
Possibilities: (1, 1), (1, 2), (2, 1) 𝑃 𝐷 1 × 𝐷 2 ≤3 = ×0.4×0.3=0.4 Possibilities: 4, 6 , 5, ≥5 , 6, ≥4 𝑃 𝐷 1 × 𝐷 2 ≥10 = 0.05× × ×0.2 =0.0325 Possibilities: 1,6 , 2,3 , 3,2 , 6,1 𝑃 𝐷 1 × 𝐷 2 =6 =2 0.4× ×0.1 =0.14 ? ? ? ?

Exercise 1 (on provided sheet) ?

Exercise 1 (on provided sheet) ? ? ?

Exercise 1 ? (on provided sheet) 𝑅 1 𝑅 2 𝑃( 𝑅 1 + 𝑅 2 )
𝑅 𝑅 𝑃( 𝑅 1 + 𝑅 2 ) × 1 2 = × 1 2 = 1 4 𝑃 𝑅 1 + 𝑅 2 ≥5 = = 7 12

#2 :: Venn Diagrams involving Frequencies
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) 𝝃 𝑪 Bro Note: The rectangular box represents all clients, i.e. the sample space. 𝜉 (Greek letter ‘xi’) is sometimes used to represent this set. ? ? 35 11 ? ? 8 3 𝑫 7 𝑭 ? ? ? 26 6 4 Dr Frost’s cat “Pippin”

#2 :: Venn Diagrams involving Frequencies
Conditional Probabilities Given that a randomly chosen person owns a cat, what’s the probability they own a dog? ? Now our choice is being restricted just to those who owns cats (i.e. 53). Thus the probability is 𝑪 𝑺 35 11 8 3 𝑫 7 𝑭 26 6 4 Dr Frost’s cat “Pippin”

Jan 2012 Q6 Bro Tip: You’ll lose a mark if you don’t have a box! The following shows the results of a survey on the types of exercise taken by a group of 100 people. 65 run 48 swim 60 cycle 40 run and swim 30 swim and cycle 35 run and cycle 25 do all three (a) Draw a Venn Diagram to represent these data (4) Find the probability that a randomly selected person from the survey (b) takes none of these types of exercise, (2) (c) swims but does not run, (2) takes at least two of these types of exercise (2) Jason is one of the above group. Given that Jason runs, (e) find the probability that he swims but does not cycle (3) ? ? ? ? ? (e) Uses an alternative method which we’ll learn later when we encounter conditional probabilities.

Exercise 2 (on provided sheet) ? ? ? ?

Exercise 2 (on provided sheet) ? ? ? ? ? ?

Exercise 2 (on provided sheet) a ? b ? c ? d ? e ?

In a Venn diagram an event is represented by a circle.
Events 𝑃 𝑒𝑣𝑒𝑛𝑡 =0.3 An event is a set of (one or more) outcomes. ? In a Venn diagram an event is represented by a circle. 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 (1 to 6) 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 (1 to 6) 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 ′ ?

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

B A C Probabilities in Venn Diagrams 0.1 0.2 0.3 0.1 0.3 ?
Instead of frequencies, we can also put probabilities into a Venn Diagram. The sum of all the probabilities must obviously be 1. Why is the event 𝐶 separated from the other two? C is mutually exclusive from both A and B, i.e. they cannot happen at the same time. Therefore 𝑷 𝑨∩𝑪 =𝟎 and 𝑷 𝑩∩𝑪 =𝟎. Note that if you are not told two events are mutually exclusive, you must assume they could happen at the same time, so ensure they overlap. ?

Construct Venn Diagrams that incorporate the following relationships between the events: Based on May 2013 Q3 There are three events 𝐴, 𝐵 and 𝐶. 𝐴 is mutually exclusive from 𝐶. The events F, H and C are that an employee is a full-time worker, part-time worker or contractor respectively. Let B be the event that an employee uses the bus. Some full-time workers use the bus, some part-time workers use the bus and some contractors use the bus. Draw a Venn diagram to represent the events F, H, C and B. (Put a 0 in any region that won’t be used) ? 𝑆 𝐴 ? 𝑆 𝐵 𝐹 𝐻 𝐵 𝐶 𝐶

Click to Broculate probabilities >
Using a Venn Diagram to Find Probabilities 𝑆 Click to Broculate probabilities > 𝐵 𝐴 The principle of adding/subtracting probabilities is exactly the same as when you were doing it with frequencies. 0.15 0.6 0.25 Given that 𝑃 𝐴∩𝐵 =0.6 and 𝑃 𝐴 =0.75 and 𝑃 𝐴∪𝐵 =1, determine: (using a suitable Venn Diagram) 𝑃 𝐴∩ 𝐵 ′ =𝟎.𝟏𝟓 𝑃 𝐴 ′ ∩ 𝐵 ′ =𝟎 𝑃 𝐵 =𝟎.𝟖𝟓 ? a b ? c ?

? 𝑆 𝐵 𝐴 0.33 0.32 0.3 0.05 Given that 𝑃 𝐴 ′ ∩𝐵 =0.3 and 𝑃 [𝐴∪𝐵]′ =0.05 and 𝑃 𝐵 =0.62, determine: (using a suitable Venn Diagram) 𝑃 𝐴 =𝟎.𝟔𝟓 𝑃 𝐴∩𝐵 =𝟎.𝟑𝟐 𝑃 𝐴∩𝐵′ =𝟎.𝟑𝟑 𝑃 𝐴∪𝐵 =𝟎.𝟗𝟓 ? a b ? c ? d ?

Exercise 3 ? ? ? ? ? ? ? ? ? ? ? ? ? (on provided sheet) 2 1 𝑆 𝐴 2 1 a
Construct Venn Diagrams which match the following descriptions. Put a 0 in any regions which are not used. 𝐴, 𝐵, 𝐶 are events. 𝐴 and 𝐵 are mutually exclusive. 𝑀, 𝐴, 𝐵, 𝐶 are events. All animals in our study can either be an Aardvark (A), Beetle (B) or Caterpillar (C). 𝑀 is the event that they are male. 𝐴, 𝐵, 𝐶, 𝐷 are events. 𝐴 and 𝐵 are mutually exclusive, and 𝐶 and 𝐷 are mutually exclusive. Given the Venn Diagram, find the total value in each region (or sketch the region). 2 1 𝑆 𝐴 2 1 a 𝑆 ? 𝐴 𝐶 𝐵 16 64 128 4 32 8 𝐵 𝐶 b 𝐴∩𝐶 → 192 𝐴∩𝐵∩ 𝐶 ′ → 16 𝐴∩ 𝐵∪𝐶 →208 𝐴 ′ ∪ 𝐵 ′ → 𝐴∪𝐶 ′ → 5 𝐴 ′ ∩ 𝐶 ′ → 5 𝐴∩𝐵 ′ ∩ 𝐵∩𝐶 ′ →79 𝐴∩ 𝐵 ′ → 66 𝐴 ′ ∪ 𝐵 ′ ∪𝐶′ → 127 𝐶∩ 𝐴∪𝐵 ∩ 𝐴∩𝐵∩𝐶 ′ →96 ? a ? b 𝑆 𝐴 𝐵 𝐶 ? c ? d ? e ? 𝑀 f ? c g ? h ? 𝑆 𝐶 ? i ? j 𝐴 𝐵 ? 𝐷

Exercise 3 ? ? ? ? ? ? ? ? ? ? ? (on provided sheet) 3 5 𝑆 𝐴 𝐵 6 4 7 𝑆
The events 𝐴 and 𝐵 are such that 𝑃 𝐴 = 1 2 , 𝑃 𝐵 = 1 3 and 𝑃 𝐴∩𝐵 = 1 4 . Represent these probabilities in a Venn Diagram. Hence determine 𝑃 𝐴∪𝐵 The events 𝐴 and 𝐵 are such that 𝑃 𝐴∩ 𝐵 ′ =0.12, 𝑃 𝐴∪𝐵 ′ =0.34 and 𝑃(𝐴)=0.51. Determine: 𝑃 𝐵 =𝟎.𝟓𝟒 𝑃 𝐴 ′ ∩𝐵 =𝟎.𝟏𝟓 𝑃 𝐴∪𝐵 =𝟎.𝟔𝟔 3 5 ? ? ? 𝑆 ? 𝐴 𝐵 1 4 1 4 1 12 𝑷 𝑨∪𝑩 = 𝟕 𝟏𝟐 The events 𝐴 and 𝐵 are such that 𝑃 𝐴∩𝐵 =0.2, 𝑃 𝐵 =0.33 and 𝑃(𝐴∪𝐵)=0.6. Determine: 𝑃 𝐴 ′ ∩𝐵′ =𝟎.𝟒 𝑃 𝐴 =𝟎.𝟒𝟕 𝑃 𝐴 ′ ∪𝐵′ =𝟎.𝟖 6 5 12 [Jan 2006 Q6a-b] For the events A and B, 𝑃 𝐴∩ 𝐵 ′ =0.32, 𝑃 𝐴 ′ ∩𝐵 =0.11, 𝑃 𝐴∪𝐵 =0.65 (a) Draw a Venn diagram to illustrate the complete sample space for the events A and B. (3) (b) Write down the value of P(A) and the value of P(B). (3) ? 4 ? ? Jeremy and Andy are playing archery. The probability Jeremy hits the target is 0.3. The probability neither hits the target is 0.2. The probability both hit the target is Determine the probability that: Andy hits the target and Jeremy doesn’t. 0.5 Exactly one of them hits the target. 0.69 Andy doesn’t hit the target 7 ? 𝑆 𝐴 𝐵 ? 𝑷 𝑨 =𝟎.𝟓𝟒 𝑷 𝑩 =𝟎.𝟑𝟑 0.32 0.22 0.11 ? 0.35 ?

𝑃 𝐴∪𝐵 =𝑃 𝐴 +𝑃(𝐵) 𝑃 𝐴∩𝐵 =0 𝑃 𝐴∩𝐵 =𝑃 𝐴 ×𝑃 𝐵 #4 :: Laws of Probability
If events A and B are mutually exclusive, then: they can’t happen at the same time. ? 𝑃 𝐴∪𝐵 =𝑃 𝐴 +𝑃(𝐵) ? 𝑃 𝐴∩𝐵 =0 ? If events A and B are independent, then: one doesn’t affect the other. ? 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 Known as the Addition Law. 𝑃 𝐴∪𝐵 =𝑃 𝐴 +𝑃 𝐵 −𝑃 𝐴∩𝐵 ?

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 Read the ‘|’ symbol as “given that”. i.e. “B occurred given that A occurred”. 𝐵′ 𝐵 𝐴′ 𝐵′ Alternatively (and more commonly): 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 = 𝟏 𝟑 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

Full Laws of Probability
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 ?

Click to reveal Venn Diagram
Check your understanding The events 𝐸 and 𝐹 are such that 𝑃 𝐸 = 𝑃 𝐸∪𝐹 = 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 ′ = = 24 35 ? Bro Tip: We know how do (a) and (b) from the previous exercise. Try to use Venn Diagrams in general – using probability laws when you require them. ? ? Click to reveal Venn Diagram 𝑺 𝑬 𝑭 0.11 0.17 0.48 0.24

Further Practice ? ? ? ? ? 𝑃 𝐴∩ 𝐵 ′ =0.4, 𝑃 𝐴∪𝐵 =0.75 1 Then:
𝑃 𝐴∩ 𝐵 ′ =0.4, 𝑃 𝐴∪𝐵 =0.75 Then: 𝑃 𝐵 =𝟎.𝟑𝟓 𝑃 𝐴 ′ ∩ 𝐵 ′ =𝟎.𝟐𝟓 𝑃 𝐴 =0.47 and 𝑃 𝐴∩𝐵 =0.12 and 𝑃 𝐴 ′ ∩ 𝐵 ′ =0.03 𝑃 𝐴 𝐵 ′ = 𝑷 𝑨∩ 𝑩 ′ 𝑷 𝑩 ′ = 𝟎.𝟑𝟓 𝟎.𝟑𝟖 = 𝟑𝟓 𝟑𝟖 𝑃 𝐴 ′ =0.7, 𝑃 𝐵 ′ =0.2, 𝑃 𝐴∩ 𝐵 ′ =0.1 𝑃 𝐴∪ 𝐵 ′ =𝑷 𝑨 +𝑷 𝑩 ′ −𝑷 𝑨∩ 𝑩 ′ =𝟎.𝟑+𝟎.𝟐−𝟎.𝟏 =𝟎.𝟒 𝑃 𝐵 𝐴 ′ = 𝑷 𝑨 ′ ∩𝑩 𝑷 𝑨 ′ = 𝟎.𝟔 𝟎.𝟕 = 𝟔 𝟕 1 ? ? 2 ? 3 ? ?

SUPER IMPORTANT TIPS If I were to identify one tip that will possible help you the most in probability questions, it’s this: If you see the words ‘given that’, Immediately write out the law for conditional probability. Example: “Given Bob walks to school, find the probability that he’s not late…” ? First thing you should write: 𝑃 𝐿 ′ 𝐵 = 𝑃 𝐿 ′ ∩𝐵 𝑃 𝐵 =… If you see the words ‘are independent’, Immediately write out the laws for independence. (Even before you’ve finished reading the question!) Example: “𝐴 is independent from 𝐵…” First thing you should write: 𝑃 𝐴∩𝐵 =𝑃 𝐴 𝑃 𝐵 𝑃 𝐴 𝐵 =𝑃(𝐴) ? If you’re stuck on a question where you have to find a probability given others, it’s probably because you’ve failed to take into account that two events are independent or mutually exclusive, or you need to use the conditional probability or additional law.

Before you start question! 𝑃 𝐴∩𝐵 =𝑃 𝐴 𝑃(𝐵) Using the above: 0.1= 𝑝+0.1 ×0.4 𝑝+0.1=0.25 𝑝=0.15 Your instinct should be: “I saw the ‘|’, so let’s use the conditional probability law!” 𝑃 𝐵 𝐶 = 𝑃 𝐵∩𝐶 𝑃 𝐶 = 𝑞 𝑞=0.24 𝑟=1−0.15−0.1−0.1−0.2−0.24 =0.21 𝑃 𝐴∪𝐶 𝐵 = 𝑃 𝐴∪𝐶 ∩𝐵 𝑃 𝐵 = =0.75 May 2013 (R) Q6 ? ? The Venn diagram in Figure 1 shows three events A, B and C and the probabilities associated with each region of B. The constants p, q and r each represent probabilities associated with the three separate regions outside B. The events A and B are independent. Find the value of p. (3) Given that 𝑃 𝐵 𝐶 = 5 11 , (b) find the value of 𝑞 and the value of 𝑟 (4) (c) Find 𝑃(𝐴∪𝐶|𝐵) (2) ?

There they are not independent.
One last very common question… 𝑃 𝐴 = 𝑃 𝐵 = 𝑃 𝐴∩𝐵 =0.1 Are 𝐴 and 𝐵 independent? (2) ? 0.4×0.3=0.12≠0.1 There they are not independent. (Note that we can sometimes also show they’re not independent if 𝑃 𝐴 𝐵 ≠𝑃(𝐴) )

Exercise 4 (on provided sheet) ? ? ? ? ?

Exercise 4 ? ? ? ? ? (on provided sheet) The set of all outcomes.
A set of one or more outcomes (that is a subset of the sample space). 𝑃 𝐴∩𝐵 = 1 3 × 1 4 = 1 12 𝑃 𝐴 𝐵 =𝑃 𝐴 = 1 3 𝑃 𝐴∪𝐵 = − 1 12 = 1 2 ? ? ? ? ?

Exercise 4 ? ? ? ? ? ? (on provided sheet)
(You could use a Venn Diagram for parts (ii) and (iii)) (i) 𝑃 𝐴∩ 𝐵 ′ =𝑃 𝐴 𝐵 ′ 𝑃 𝐵 ′ = 4 5 × 1 2 = 2 5 (ii) 𝑃 𝐴∩𝐵 =𝑃 𝐴 −𝑃 𝐴∩ 𝐵 ′ =0 (iii) 𝑃 𝐴∪𝐵 = −0= (iv) 𝑃 𝐴 𝐵 = 𝑃 𝐴∩𝐵 𝑃 𝐵 =0 i) Yes, because 𝑃 𝐴∩𝐵 =0 ii) No, because 𝑃 𝐴 𝐵 ≠𝑃(𝐴) or because 𝑃 𝐴∩𝐵 =0 whereas 𝑃 𝐴 𝑃 𝐵 = 1 5 ? ? ? ? ? ?

Exercise 4 ? ? ? (on provided sheet)
(Parts (a) and (b) are the same as Q3 in Exercise 3) ? 𝜉 𝐴 𝐵 1 4 1 12 1 4 5 12 ? b) 𝑃 𝐴∪𝐵 = 7 12 c) 𝑃 𝐴 𝐵 ′ = 𝑃 𝐴∩ 𝐵 ′ 𝑃 𝐵 ′ = − 1 3 = 3 8 ?

Exercise 4 ? ? ? ? (on provided sheet) a) 𝜉 𝐶 𝐵 𝐴
I’d immediately write: 𝑃 𝐴∩𝐶 =𝑃 𝐴 𝑃 𝐶 =0.2𝑃(𝐶) b) 𝑃 𝐴 𝐶 =𝑃 𝐴 =0.2 c) 𝑃 𝐴∪𝐵 =𝑃 𝐴 +𝑃 𝐵 =0.6 d) 𝑃 𝐴∪𝐶 =𝑃 𝐴 +𝑃 𝐶 −𝑃 𝐴∩𝐶 0.7=0.2+𝑃 𝐶 −0.2𝑃 𝐶 0.5=0.8𝑃 𝐶 𝑃 𝐶 = 5 8 ? ? ?

#5 :: Probability Trees ? ? ? ? ? ? ? ? 1st pick 2nd pick 4 10 𝑅𝑒𝑑 1 2
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 = = 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 = = 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 ′ = × = Completed the hole in 3 shots or less. P 𝐶 = × × = Missed the green with his tee shot, given he took 3 shots or less in total. 𝑃 𝑅′ 𝐶 = 𝑃 𝑅 ′ ∩𝐶 𝑃 𝐶 = = 25 71 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 hole in 3 shots or less”.

Exercise 5 (on provided sheet) ?

Exercises (on worksheet)
2 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 ?

Exercise 5 ? ? ? ? ? (on provided sheet)
[Jan 2011 Q7] The bag P contains 6 balls of which 3 are red and 3 are yellow. The bag Q contains 7 balls of which 4 are red and 3 are yellow. A ball is drawn at random from bag P and placed in bag Q. A second ball is drawn at random from bag P and placed in bag Q. A third ball is then drawn at random from the 9 balls in bag Q. The event A occurs when the 2 balls drawn from bag P are of the same colour. The event B occurs when the ball drawn from bag Q is red. Copy and complete the tree diagram shown below. (4) (b) Find P(A). (3) (c) Show that 𝑃 𝐵 = (3) (d) Show that 𝑃 𝐴∩𝐵 = (2) (e) Hence find 𝑃 𝐴∪𝐵 . (2) (f) Given that all three balls drawn are the same colour, find the probability that they are all red. (3) ? ? ? ?

Exercise 5 ? ? ? (on provided sheet)
b) 0.2× × ×0.45 =0.355 c) 𝑃 𝑆𝑤 𝑆𝑎 = 𝑃 𝑆𝑤∩𝑆𝑎 𝑃 𝑆𝑎 = 0.2× = =0.197 d) 𝑃 𝑆𝑤 𝑆 𝑎 ′ = 𝑃 𝑆𝑤∩𝑆 𝑎 ′ 𝑃 𝑆 𝑎 ′ = 0.2× = =0.202 ? ?

Exercise 5 ? ? ? (on provided sheet) b) 𝑃 𝐾𝐾𝐾 = 2 3 × 1 2 × 1 4 = 1 12

To Finish Off :: A 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

S1 Chapter 5 :: Probability

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S1 Chapter 5 :: Probability

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