GCNU311A Lecturer 2019 Probability PDF

**Unit Title Solve problems involving uncertainty 不確定性 using basic principles of probability 概率** **Unit Code GCNU311A** **Warmup with Gamification "Tong Pak Fu and Chou Heung -- The Probabilistic Fantasy (唐伯虎點秋香 -- 概率幻想)"** [***http://appsrv.cse.cuhk.edu.hk/\~mhp/***](http://apps...

**Unit Title Solve problems involving uncertainty 不確定性 using basic principles of probability 概率** **Unit Code GCNU311A** **Warmup with Gamification "Tong Pak Fu and Chou Heung -- The Probabilistic Fantasy (唐伯虎點秋香 -- 概率幻想)"** [***http://appsrv.cse.cuhk.edu.hk/\~mhp/***](http://appsrv.cse.cuhk.edu.hk/~mhp/) **Chapter 1: The First Sight (第一幕：邂逅)** **1. Definition of Probability 概率/機率/機會率/或然率** The probability (*P*) that an event that may happen is given by: \ [\$\$\\mathrm{\\text{Probability\\ of\\ an\\ event}} = P(\\ \\text{event}\\ ) = \\frac{\\mathrm{\\text{Number\\ of}}\\mathrm{\\ }\\mathrm{\\text{Favourable}}\\mathrm{\\ }\\mathrm{\\text{Outcomes\\ }}\\mathrm{合適的結果}}{\\mathrm{\\text{Total\\ Number\\ of\\ }}\\mathrm{\\text{Possible\\ Outcomes\\ }}\\mathrm{可能的結果}}\$\$]{.math.display}\ *In order to use this formula, we need to be able to **count 數 the number of outcomes correctly**.* From the formula, we can realise some basic properties of probability: [**0** ≤ *P*( event ) ≤ **1**]{.math.inline} Probability **0** means an event that will never happen Probability **1** means an event that is certain to happen **1.1. Empirical Probability 實驗概率** If the number of favourable outcomes is collected from observations during an experiment, the probability calculated is called empirical (or experimental) probability. **1.2. Theoretical Probability 理論概率** With theoretical probability, we would not actually conduct an experiment. Based upon knowledge of the situation, we arrive at the "expected 預期的" answer of the number of favourable outcomes to the number of possible outcomes. **2. Method of Counting**  We use the curly brackets **{ }** to enclose the outcomes. The possible outcomes of tossing/throwing/rolling 擲 a die 一顆骰子 are **{ 1, 2, 3, 4, 5, 6 }**.  The possible outcomes of flipping/tossing 擲 a coin 一枚硬幣 are **{ H, T }**, that is head 公 or tail 字. **\ ** **Example** **1.** Consider flipping of a fair 公平的 **coin twice** and we want to have **one head only**. \(a) List all the possible outcomes and find the number of outcomes. \(b) List all the favourable outcomes and find the number of outcomes. \(c) Find the probability of getting one head only. +-----------------------+-----------------------+-----------------------+ | **Solution:** | (a) | Possible outcomes are | | | | **{** HH, HT, TH, TT | | | | **}** | | | | | | | | ∴The total number of | | | | possible outcomes is | | | | 4. | +=======================+=======================+=======================+ | | (b) | As we are only | | | | interested in having | | | | one head, favourable | | | | outcomes are **{** | | | | HT, TH **}** | | | | | | | | ∴The number of | | | | favourable outcomes | | | | is 2. | +-----------------------+-----------------------+-----------------------+ | | (c) | \ | | | | [\$\$P(\\mathrm{\\tex | | | | t{only\\ | | | | one\\ head\\ by\\ | | | | tossing\\ a\\ coin\\ | | | | twice}}) = | | | | \\frac{2}{4} = | | | | \\frac{1}{2} = | | | | 0.5\$\$]{.math | | | |.display}\ | | | | | | | | ∴The probability of | | | | getting only one head | | | | is 0.5. | +-----------------------+-----------------------+-----------------------+ **Example** **2.** Consider tossing of a fair die.  (a) What is the probability of obtaining an even 雙 number? \(b) Find the probability of getting an odd 單 number. \(c) Calculate the probability of getting the number "6". \(d) Determine the probability of getting a number smaller than 3. +-----------------------+-----------------------+-----------------------+ | **Solution:** | (a) | Possible outcomes are | | | | **{** , | | | |  | | | | , | | | | , | | | | , | | | | , | | | |  | | | | **}** | | | | | | | | ∴The total number of | | | | possible outcomes is | | | | 6. | +=======================+=======================+=======================+ | | | As we want to find | | | | out the probability | | | | of getting an even | | | | number, so the | | | | favourable outcomes | | | | are **{** , | | | | , | | | | **}** | | | | | | | | ∴The number of | | | | favourable outcomes | | | | is 3. | +-----------------------+-----------------------+-----------------------+ | | | \ | | | | [\$\$P(\\mathrm{\\tex | | | | t{even}}) | | | | = \\frac{3}{6} = | | | | \\frac{1}{2} = | | | | 0.5\$\$]{.math | | | |.display}\ | | | | | | | | ∴The probability of | | | | obtaining an even | | | | number is 0.5. | +-----------------------+-----------------------+-----------------------+ | | (b) | Favourable outcomes | | | | are **{** 1, 3, 5 | | | | **}** and the number | | | | of outcomes = 3 | | | | | | | | \ | | | | [\$\$\\therefore | | | | P(\\mathrm{\\text{odd | | | | }}) | | | | = \\frac{3}{6} = | | | | \\frac{1}{2} = | | | | 0.5\$\$]{.math | | | |.display}\ | | | | | | | | The probability of | | | | getting an odd number | | | | is 0.5. | +-----------------------+-----------------------+-----------------------+ | | (c) | Favourable outcome is | | | | **{** 6 **}** and the | | | | number of outcome = 1 | | | | | | | | \ | | | | [\$\$\\therefore | | | | P(\"\\mathrm{6\"}) = | | | | \\frac{1}{6}\$\$]{.ma | | | | th | | | |.display}\ | | | | | | | | The probability of | | | | getting the number | | | | \"6\" is | | | | [\$\\frac{1}{6}\$]{.m | | | | ath | | | |.inline}. | +-----------------------+-----------------------+-----------------------+ | | (d) | Favourable outcomes | | | | are **{** 1, 2 **}** | | | | and the number of | | | | outcomes = 2 | | | | | | | | \ | | | | [\$\$\\therefore | | | | P\\left( | | | | \\mathrm{less\\ | | | | than\\ 3} \\right) = | | | | \\frac{2}{6} = | | | | \\frac{1}{3}\$\$]{.ma | | | | th | | | |.display}\ | | | | | | | | The probability of | | | | getting a number less | | | | than 3 is | | | | [\$\\frac{1}{3}\$]{.m | | | | ath | | | |.inline}. | +-----------------------+-----------------------+-----------------------+ **Example** **3.** A card is drawn 抽 randomly 隨機地 from a standard deck of cards 一副紙牌. Find the probability of getting [a King] **OR** [a Heart greater than 10]?  +-----------------------------------+-----------------------------------+ | **Solution:** | A standard deck of cards has 52 | | | cards. | | | | | | ∴The total number of possible | | | outcomes is 52. | +===================================+===================================+ | | Favourable outcomes are **{** ♠K, | | | ♦K, ♥K, ♣K, ♥A, ♥Q, ♥J **}\ | | | **and the number of outcomes = 7. | +-----------------------------------+-----------------------------------+ | | \ | | | [\$\$P(\\mathrm{a\\ King\\ OR\\ | | | a\\ Heart\\ greater\\ than\\ 10}) | | | = \\frac{7}{52}\$\$]{.math | | |.display}\ | | | | | | ∴The probability of getting a | | | King OR a Heart greater than 10 | | | is [\$\\frac{7}{52}\$]{.math | | |.inline}. | +-----------------------------------+-----------------------------------+ **3. (Probability) Tree Diagram 樹形圖** A probability tree diagram helps us to determine all the possible outcomes of an experiment/event. The diagram is started by a dot. From the dot, branches are drawn to represent all possible outcomes of the experiment/event. The number of leaves/branches at the end is the number of possible outcomes. *Note: The probability of each outcome can also be written on a branch.* **Example** **4.** Construct a tree diagram to determine the number of outcomes of tossing two coins. +-----------------------------------+-----------------------------------+ | **Solution:** | **1st Coin** | | | | | | | | | **2nd Coin** | | | | | | | | | | | | **Outcomes** | | | -- ---------------------------- | | | --------------------------------- | | | --------------------------------- | | | ----------------- -- ------------ | | | --------------------------------- | | | --------------------------------- | | | --------------------------------- | | | -- ----------------------------- | | | --------------------------------- | | | --------------------------------- | | | --------------------------------- | | | --------------------------------- | | | --------------------------------- | | | ---- | | | | | | | | | | | | C:\\Users\\a | | | dmin\\AppData\\Local\\Microsoft\\ | | | Windows\\INetCache\\Content.Word\ | | | \Head.png | | | C:\\Users\\admin\\A | | | ppData\\Local\\Microsoft\\Windows | | | \\INetCache\\Content.Word\\Head.p | | | ng | | |  | | | | | | | | | | | | | | | | | | | | | | | | C:\\Users\\a | | | dmin\\AppData\\Local\\Microsoft\\ | | | Windows\\INetCache\\Content.Word\ | | | \Tail.png | | | C:\\Users\\admin\\A | | | ppData\\Local\\Microsoft\\Windows | | | \\INetCache\\Content.Word\\Tail.p | | | ng | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | |  | | | C:\\Users\\admin\\AppData\\Lo | | | cal\\Microsoft\\Windows\\INetCach | | | e\\Content.Word\\Tail.png | | | C:\\Users\\admin\\AppData\\L | | | ocal\\Microsoft\\Windows\\INetCac | | | he\\Content.Word\\Tail.png | | | | | | | | | | | | | | | | | | | | | | | | | | |  | | | C:\\Users\\admin\\AppData\\Lo | | | cal\\Microsoft\\Windows\\INetCach | | | e\\Content.Word\\Tail.png | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | ∴The possible outcomes are **{** | | | HH, HT, TH, TT **}** and the | | | total number is 4. | +-----------------------------------+-----------------------------------+ **Example** **5.** Mary plans to meet her friends at Admiralty and go to The Peak together. She can take the bus or MTR to Admiralty. They can then go to The Peak on foot, by peak tram or by bus. Draw a tree diagram to show how Mary goes to The Peak. +-----------------------------------+-----------------------------------+ | **Solution:** | **Admiralty** **The Pea | | | k** **Outcomes** | | | -- --------------- -- --------- | | | ----- -- ------------------ | | | On Foot | | | (Bus, On Foot) | | | | | | | | | Bus Peak Tram | | | (Bus, Peak Tram) | | | | | | | | | Bus | | | (Bus, Bus) | | | | | | | | | On Foot | | | (MTR, On Foot) | | | | | | | | | MTR Peak Tram | | | (MTR, Peak Tram) | | | | | | | | | Bus | | | (MTR, Bus) | | | | | | | | | | | | ∴The possible outcomes are **{** | | | (Bus, On Foot), (Bus, Peak Tram), | | | (Bus, Bus), (MTR, On Foot), (MTR, | | | Peak Tram), (MTR, Bus) **}** and | | | the total number is 6. | +-----------------------------------+-----------------------------------+ **Example** **6.** For a family of four, there are two children. Use tree diagram to show the possible genders 性别 of the children. +-----------------------------------+-----------------------------------+ | **Solution:** | **1st Child** **2nd Chi | | | ld** **Outcomes** | | | -- --------------- -- --------- | | | ------ -- -------------- | | | M | | | (M, M) | | | M | | | | | | F | | | (M, F) | | | | | | | | | M | | | (F, M) | | | F | | | | | | F | | | (F, F) | | | | | | | | | | | | ∴The possible outcomes are **{** | | | (M, M), (M, F), (F, M), (F, F) | | | **}** and the total number is 4. | +-----------------------------------+-----------------------------------+ **\ ** **Example** **7.** Consider tossing of a fair die and a coin. \(a) Draw a tree diagram to show all the possible outcomes. \(b) Find the probability of getting [an even number and a head]. +-----------------------+-----------------------+-----------------------+ | **Solution:** | (a) | **Die** **C | | | | oin** **Outcomes | | | | ** | | | | -- --------- -- --- | | | | ------- -- ---------- | | | | ---- | | | | 1 H | | | | **1H** | | | | | | | | | | | | T | | | | **1T** | | | | | | | | | | | | 2 H | | | | **2H** | | | | | | | | | | | | T | | | | **2T** | | | | | | | | | | | | 3 H | | | | **3H** | | | | | | | | | | | | T | | | | **3T** | | | | | | | | | | | | 4 H | | | | **4H** | | | | | | | | | | | | T | | | | **4T** | | | | | | | | | | | | 5 H | | | | **5H** | | | | | | | | | | | | T | | | | **5T** | | | | | | | | | | | | 6 H | | | | **6H** | | | | | | | | | | | | T | | | | **6T** | | | | | | | | | +=======================+=======================+=======================+ | | (b) | Possible outcomes are | | | | **{** 1H, 1T, 2H, 2T, | | | | 3H, 3T, 4H, 4T, 5H, | | | | 5T, 6H, 6T **}** | | | | | | | | ∴The total number of | | | | possible outcomes is | | | | 12. | | | | | | | | Favourable outcomes | | | | are **{** 2H, 4H, 6H | | | | **}** and the number | | | | of outcomes = 3 | | | | | | | | \ | | | | [\$\$\\therefore | | | | P(\\mathrm{\\text{eve | | | | n\\ | | | | number\\ and\\ a\\ | | | | head}}) = | | | | \\frac{3}{12} = | | | | \\frac{1}{4} = | | | | 0.25\$\$]{.math | | | |.display}\ | | | | | | | | The probability of | | | | getting an even | | | | number and a head is | | | | 0.25. | +-----------------------+-----------------------+-----------------------+ **Example** **8.** When two dice 骰子 are rolled, list the possible outcomes. Hence find the probability of getting [a sum of 4] from the two dice. +-----------------------------------+-----------------------------------+ | **Solution:** | Possible outcomes are | | | [\$\\begin{Bmatrix} | | | \\begin{matrix} (1,\\ 1) & (1,\\ | | | 2) \\\\ (2,\\ 1) & (2,\\ 2) \\\\ | | | \\end{matrix} & \\begin{matrix} | | | \\cdots & (1,\\ 6) \\\\ \\cdots & | | | (2,\\ 6) \\\\ \\end{matrix} \\\\ | | | \\begin{matrix} \\vdots & \\vdots | | | \\\\ (6,\\ 1) & (6,\\ 2) \\\\ | | | \\end{matrix} & \\begin{matrix} | | | \\ddots & \\vdots \\\\ \\cdots & | | | (6,\\ 6) \\\\ \\end{matrix} \\\\ | | | \\end{Bmatrix}\$]{.math.inline} | | | | | | ∴The total number of possible | | | outcomes is 36. | +===================================+===================================+ | | Favourable outcomes are **{** (1, | | | 3), (2, 2), (3, 1) **}** and the | | | number of outcomes = 3 | +-----------------------------------+-----------------------------------+ | | \ | | | [\$\$\\therefore P(\\mathrm{sum\\ | | | is\\ 4}) = \\frac{3}{36} = | | | \\frac{1}{12}\$\$]{.math | | |.display}\ | | | | | | The probability of getting a sum | | | of 4 is [\$\\frac{1}{12}\$]{.math | | |.inline}. | +-----------------------------------+-----------------------------------+ **4. Fundamental Counting Principle 基本計數原理** The methods of listing and tree diagram always work. But it is troublesome 麻煩的 to do the full listing. For simple independent events 獨立事件, we may calculate the total number of outcomes easily by multiplying each event's number of possibilities: \ [\$\$\\mathrm{\\text{Total\\ Number\\ of\\ Outcomes}} = \\begin{pmatrix} \\mathrm{\\text{Total\\ No.\\ of\\ Outcomes}} \\\\ \\mathrm{\\text{of\\ One\\ Event}} \\\\ \\end{pmatrix}\\begin{pmatrix} \\mathrm{\\text{Total\\ No.\\ of\\ Outcomes}} \\\\ \\mathrm{\\text{of\\ Another\\ Event}} \\\\ \\end{pmatrix}\$\$]{.math.display}\ Consider the Example 7, tossing of a fair die and a coin. \ [\$\${\\mathrm{\\text{The\\ total\\ number\\ of\\ possible\\ outcomes}} = \\begin{pmatrix} \\mathrm{\\text{The\\ total\\ no.\\ of\\ outcomes}} \\\\ \\mathrm{\\text{of\\ tossing\\ a\\ die}} \\\\ \\end{pmatrix}\\begin{pmatrix} \\mathrm{\\text{The\\ total\\ no.\\ of\\ outcomes}} \\\\ \\mathrm{\\text{of\\ tossing\\ a\\ coin}} \\\\ \\end{pmatrix} }{= (6)(2) }{= 12}\$\$]{.math.display}\ **Example** **9.** Calculate the number of possible outcomes of the following events: \(a) Flipping two coins \(b) Rolling two dice \(c) Throwing a die twice \(d) Tossing a coin three times **Solution:** (a) Total Number of Possible outcomes = (2)(2) = 4 --------------- ----- --------------------------------------------------- (b) Total Number of Possible outcomes = (6)(6) = 36 (c) Total Number of Possible outcomes = (6)(6) = 36 (d) Total Number of Possible outcomes = (2)(2)(2) = 8 **Example** **10.** A local ice-cream shop offers four flavours 味道 of ice-cream (*vanilla, mango, chocolate, strawberry*), two types of cones 筒 (*cake, waffle*) and three toppings 甜品醬 (*chocolate syrup 糖漿, caramel 焦糖 syrup, rainbow sprinkles 彩虹朱克力碎*). How many different types of ice-cream cones can we order? --------------- -------------------------------------------------------- **Solution:** The total number of possible outcomes = (4)(2)(3) = 24 --------------- -------------------------------------------------------- Once we are familiar with the counting, there is **no need to list out the outcomes for simple events**. **\ ** **Example** **11.** When three coins are tossed, find the probability of getting more heads than tails. +-----------------------------------+-----------------------------------+ | **Solution:** | The total number of possible | | | outcomes = (2)(2)(2) = 8 | +===================================+===================================+ | | Favourable outcomes are **{** | | | HHH, HHT, HTH, THH **}** and the | | | number of outcomes = 4 | +-----------------------------------+-----------------------------------+ | | \ | | | [\$\$\\therefore P\\left( | | | \\mathrm{\\text{more\\ heads\\ | | | than\\ tails}} \\right) = | | | \\frac{4}{8} = \\frac{1}{2} = | | | 0.5\$\$]{.math.display}\ | | | | | | The probability of getting more | | | heads than tails is 0.5. | +-----------------------------------+-----------------------------------+ **Example** **12.** A family has three children. Find the probability that \(a) the first child is a boy [and] the second child is a girl \(b) the first child is a boy [or] the second child is a girl +-----------------------+-----------------------+-----------------------+ | **Solution:** | (a) | The total number of | | | | possible outcomes = | | | | (2)(2)(2) = 8 | +=======================+=======================+=======================+ | | | Favourable outcomes | | | | are **{** BGB, BGG | | | | **}** and the number | | | | of outcomes is 2 | +-----------------------+-----------------------+-----------------------+ | | | \ | | | | [\$\$P\\left( | | | | \\mathrm{1st\\ | | | | child\\ is\\ boy\\ | | | | and\\ 2nd\\ child\\ | | | | is\\ girl} \\right) = | | | | \\frac{2}{8} = | | | | \\frac{1}{4} = | | | | 0.25\$\$]{.math | | | |.display}\ | | | | | | | | ∴The probability that | | | | the first child is a | | | | boy and the second | | | | child is a girl is | | | | 0.25. | +-----------------------+-----------------------+-----------------------+ | | (b) | Outcomes of 1st child | | | | is a boy are **{** | | | | BBB, BBG, BGB, BGG | | | | **}** | | | | | | | | Outcomes of 2nd child | | | | is a girl are **{** | | | | BGB, BGG, GGB, GGG | | | | **}** | | | | | | | | ∴Favourable outcomes | | | | are **{** BBB, BBG, | | | | BGB, BGG, GGB, GGG | | | | **}\ | | | | ** and the number of | | | | outcomes = 6 | +-----------------------+-----------------------+-----------------------+ | | | \ | | | | [\$\$P\\left( | | | | \\mathrm{1st\\ | | | | child\\ is\\ boy\\ | | | | or\\ 2nd\\ child\\ | | | | is\\ girl} \\right) = | | | | \\frac{6}{8} = | | | | \\frac{3}{4} = | | | | 0.75\$\$]{.math | | | |.display}\ | | | | | | | | ∴The probability that | | | | the first child is a | | | | boy or the second | | | | child is a girl is | | | | 0.75. | +-----------------------+-----------------------+-----------------------+ **5. Complementary Event 互補事件** In a random experiment/event, the probabilities of all possible outcomes must total to 1. The complement of an event is all outcomes that are **NOT the event**. So the complement of an event is all the other outcomes (*not the ones we want*). The probability (*P*) of an event may also be found by its complement: \ [Probability of an event = *P*( event ) = 1 − *P*( not event )]{.math.display}\ *This formula is useful when the calculation of **P*****( not *event* )** *is easier than that of **P*****( *event* )***.* Consider the Example 12(b), the probability that the first child is a boy or the second is a girl. \ [\$\${P\\left( \\mathrm{1st\\ child\\ is\\ boy\\ or\\ 2nd\\ child\\ is\\ girl} \\right) = 1 - P\\left( \\mathrm{\\text{not}}\\mathrm{\\ \"1st\\ child\\ is\\ boy\\ or\\ 2nd\\ child\\ is\\ girl\"} \\right) }{= 1 - P\\left( \\mathrm{1st\\ child\\ is\\ girl\\ and\\ 2nd\\ child\\ is\\ boy} \\right) }{= 1 - P\\left( \\mathrm{\\{}\\mathrm{\\ }\\mathrm{\\text{GB}}\\mathrm{B,\\ }\\mathrm{\\text{GB}}\\mathrm{\\text{G\\ }}\\mathrm{\\}} \\right) }{= 1 - \\frac{2}{(2)\\left( 2 \\right)\\left( 2 \\right)} }{= 1 - \\frac{2}{8} }{= \\frac{3}{4}}\$\$]{.math.display}\ **Example** **13.** Three identical 相同的 unbiased 公正的 coins are tossed. Find the probability of getting \(a) 3 tails \(b) **exactly** 2 tails \(c) **at least** one head \(d) **at least** one head and one tail \(e) **at least** one head or one tail +-----------------------+-----------------------+-----------------------+ | **Solution:** | (a) | The total number of | | | | possible outcomes = | | | | (2)(2)(2) = 8 | +=======================+=======================+=======================+ | | | \ | | | | [\$\$P\\left( | | | | \\mathrm{3\\ tails} | | | | \\right) = | | | | P(\\mathrm{\\{}\\math | | | | rm{\\text{TTT}}\\math | | | | rm{\\}}) | | | | = | | | | \\frac{1}{8}\$\$]{.ma | | | | th | | | |.display}\ | | | | | | | | ∴The probability of | | | | getting 3 tails is | | | | [\$\\frac{1}{8}\$]{.m | | | | ath | | | |.inline}. | +-----------------------+-----------------------+-----------------------+ | | (b) | \ | | | | [\$\$P\\left( | | | | \\mathrm{exactly\\ | | | | 2\\ tails} \\right) = | | | | P(\\mathrm{\\{}\\math | | | | rm{TTH,\\ | | | | THT,\\ | | | | HTT}\\mathrm{\\}}) = | | | | \\frac{3}{8}\$\$]{.ma | | | | th | | | |.display}\ | | | | | | | | ∴The probability of | | | | getting exactly 2 | | | | tails is | | | | [\$\\frac{3}{8}\$]{.m | | | | ath | | | |.inline}. | +-----------------------+-----------------------+-----------------------+ | | (c) | \ | | | | [\$\${P\\left( | | | | \\mathrm{at\\ least\\ | | | | 1\\ head} \\right) = | | | | 1 - P\\left( | | | | \\mathrm{\\text{not}} | | | | \\mathrm{\\ | | | | \"at\\ least\\ 1\\ | | | | head\"} \\right) }{= | | | | 1 - P\\left( | | | | \\mathrm{\\text{no\\ | | | | head}} \\right) }{= 1 | | | | - P\\left( | | | | \\mathrm{\\{}\\mathrm | | | | {\\text{TTT}}\\mathrm | | | | {\\}} | | | | \\right) }{= 1 - | | | | \\frac{1}{8} }{= | | | | \\frac{7}{8}}\$\$]{.m | | | | ath | | | |.display}\ | | | | | | | | ∴The probability of | | | | getting at least 1 | | | | head is | | | | [\$\\frac{7}{8}\$]{.m | | | | ath | | | |.inline}. | +-----------------------+-----------------------+-----------------------+ +-----------------------+-----------------------+-----------------------+ | **Solution:** | (d) | \ | | | | [\$\${P\\left( | | | | \\mathrm{at\\ least\\ | | | | 1\\ head\\ and\\ 1\\ | | | | tail} \\right) = 1 - | | | | P\\left( | | | | \\mathrm{\\text{not}} | | | | \\mathrm{\\ | | | | \"at\\ least\\ 1\\ | | | | head\\ and\\ 1\\ | | | | tail\"} \\right) }{= | | | | 1 - P\\left( | | | | \\mathrm{\\text{no\\ | | | | head\\ | | | | }}\\mathrm{\\text{or} | | | | }\\mathrm{\\text{\\ | | | | no\\ tail}} \\right) | | | | }{= 1 - P\\left( | | | | \\mathrm{\\{}\\mathrm | | | | {TTT,\\ | | | | HHH}\\mathrm{\\}} | | | | \\right) }{= 1 - | | | | \\frac{2}{8} }{= | | | | \\frac{3}{4}}\$\$]{.m | | | | ath | | | |.display}\ | | | | | | | | ∴The probability of | | | | getting at least 1 | | | | head and 1 tail is | | | | [\$\\frac{3}{4}\$]{.m | | | | ath | | | |.inline}. | +=======================+=======================+=======================+ | | (e) | \ | | | | [\$\${P\\left( | | | | \\mathrm{at\\ least\\ | | | | 1\\ head\\ or\\ 1\\ | | | | tail} \\right) = 1 - | | | | P\\left( | | | | \\mathrm{\\text{not}} | | | | \\mathrm{\\ | | | | \"at\\ least\\ 1\\ | | | | head\\ or\\ 1\\ | | | | tail\"} \\right) }{= | | | | 1 - P\\left( | | | | \\mathrm{\\text{no\\ | | | | head\\ | | | | }}\\mathrm{\\text{and | | | | }}\\mathrm{\\text{\\ | | | | no\\ tail}} \\right) | | | | }{= 1 - P\\left( | | | | \\mathrm{\\{}\\mathrm | | | | {\\varnothing}\\mathr | | | | m{\\}} | | | | \\right) }{= 1 - 0 | | | | }{= 1}\$\$]{.math | | | |.display}\ | | | | | | | | ∴The probability of | | | | getting at least 1 | | | | head or 1 tail is **1 | | | | (*100% sure to | | | | happen*)**. | +-----------------------+-----------------------+-----------------------+ *In most of the "at least" cases, there may be many favourable outcomes. It's all right if you can find the correct number of favourable outcomes and then get the probability directly.* - Outcomes of at least 1 head = **{** HTT, THT, TTH, HHT, HTH, THH, HHH **}** - Outcomes of at least 1 head and 1 tail = **{** HTT, THT, TTH, HHT, HTH, THH **}** **Example** **14.** A study was carried out in an IVE campus to examine the preferences of students on choosing the type of transportation for their travels to the campus. **Mode of Transport** On Foot Bus Minibus MTR Others ----------------------- --------- ----- --------- ----- -------- **No. of Students** 130 225 165 445 85 If a student is randomly selected from the campus, \(a) what is the probability that he/she will go to institute by MTR? \(b) what is the probability that he/she will go to institute by either bus or minibus? \(c) what is the probability that he/she will take [certain type of vehicles 交通工具] to institute? +-----------------------+-----------------------+-----------------------+ | **Solution:** | | The total number of | | | | possible outcomes = | | | | 130 + 225 + 165 +445 | | | | + 85 = 1050 | +=======================+=======================+=======================+ | | (a) | \ | | | | [\$\$P\\left( | | | | \\mathrm{\\text{By\\ | | | | MTR}} \\right) = | | | | \\frac{445}{1050} = | | | | \\frac{89}{210}\$\$]{ | | | |.math | | | |.display}\ | | | | | | | | ∴The probability of | | | | going to institute by | | | | MTR is | | | | [\$\\frac{89}{210}\$] | | | | {.math | | | |.inline}. | +-----------------------+-----------------------+-----------------------+ | | (b) | \ | | | | [\$\$P\\left( | | | | \\mathrm{\\text{By\\ | | | | Bus\\ or\\ Minibus}} | | | | \\right) = \\frac{225 | | | | + 165}{1050} = | | | | \\frac{13}{35}\$\$]{. | | | | math | | | |.display}\ | | | | | | | | ∴The probability of | | | | going to institute by | | | | either bus or minibus | | | | is | | | | [\$\\frac{13}{35}\$]{ | | | |.math | | | |.inline}. | +-----------------------+-----------------------+-----------------------+ | | (c) | \ | | | | [\$\$P\\left( | | | | \\mathrm{\\text{By\\ | | | | Vehicles}} \\right) = | | | | 1 - P\\left( | | | | \\mathrm{\\text{On\\ | | | | Foot}} \\right) = 1 - | | | | \\frac{130}{1050} = | | | | \\frac{92}{105}\$\$]{ | | | |.math | | | |.display}\ | | | | | | | | ∴The probability of | | | | going to institute by | | | | either bus or minibus | | | | is | | | | [\$\\frac{92}{105}\$] | | | | {.math | | | |.inline}. | +-----------------------+-----------------------+-----------------------+ **6. Expected Values 期望值 (⇒ 預期結果)** In probability theory, the expected value is the long-run average value 長遠平均數 of repetitions 重複 of the experiment/event. It is calculated by multiplying each of the possible outcomes by the probability each outcome will occur, and summing all of those values. \ [\$\$\\mathrm{\\text{Expected\\ value\\ of\\ an\\ event}} = E\\left( \\ \\text{event}\\ \\right) = \\sum\_{}\^{}{(\\text{event}\\ value) \\bullet P\\left( \\ \\text{event}\\ \\right)}\$\$]{.math.display}\ **Example** **15.** A fair die is thrown 120 times. How many times would you expect to get a "6"? +-----------------------------------+-----------------------------------+ | **Solution:** | \ | | | [\$\$P(\"6\") = | | | \\frac{1}{6}\$\$]{.math | | |.display}\ | | | | | | \ | | | [\$\$\\therefore E\\left( | | | \\mathrm{\"6\"\\ in\\ throwing\\ | | | 120\\ times} \\right) = \\left( | | | 120 \\right)\\left( \\frac{1}{6} | | | \\right) = 20\$\$]{.math | | |.display}\ | | | | | | The expected number of times to | | | get a \"6\" is 20 times. | +===================================+===================================+ | | *The probability of getting a | | | \"6\" happening in a single trial | | | 試驗 is* [\$\\frac{1}{6}\$]{.math | | |.inline}*.* | | | | | | *So we expect 預期 after 120 | | | trials, the event of getting a | | | \"6\" will occur 20 times.* | +-----------------------------------+-----------------------------------+ **Example** **16.** A bag contains three \$10 coins and two \$5 coins. A coin is picked out from the bag randomly, what is the expected value of the coin obtained? +-----------------------------------+-----------------------------------+ | **Solution:** | \ | | | [\$\$P\\left( \\mathrm{picking\\ | | | a\\ \\\$ 10\\ coin} \\right) = | | | \\frac{3}{5}\\mathrm{\\text{\\ \\ | | | \\ \\ \\ and\\ \\ \\ \\ \\ | | | }}P\\left( \\mathrm{picking\\ a\\ | | | \\\$ 5\\ coin} \\right) = | | | \\frac{2}{5}\$\$]{.math | | |.display}\ | | | | | | \ | | | [\$\$\\therefore E\\left( | | | \\mathrm{picking\\ 1\\ coin} | | | \\right) = \\left( 10 | | | \\right)\\left( \\frac{3}{5} | | | \\right) + \\left( 5 | | | \\right)\\left( \\frac{2}{5} | | | \\right) = 6 + 2 = 8\$\$]{.math | | |.display}\ | | | | | | The expected value of picking one | | | coin from the bag is \$8. | +===================================+===================================+ | | *Clearly it doesn't mean we would | | | get \$8 coin from the bag. If we | | | repeat the trial of picking a | | | coin from the bag again and | | | again, the average value we got | | | is \$8.* | +-----------------------------------+-----------------------------------+ The concept of expected value is very useful for the risk analysis. The most typical risk is gambling in which you risk a financial loss in the hope of a financial gain. **Example** **17.** Consider a bet of tossing two coins. I pay you \$2 if there are two heads but you pay me \$1 if there are one or two tails. Would you bet? +-----------------------------------+-----------------------------------+ | **Solution:** | \ | | | [\$\$P\\left( \\mathrm{2\\ heads} | | | \\right) = | | | \\frac{1}{4}\\mathrm{\\text{\\ \\ | | | \\ \\ \\ and\\ \\ \\ \\ \\ | | | }}P\\left( \\mathrm{1\\ or\\ 2\\ | | | tails} \\right) = | | | P(\\mathrm{\\{}\\mathrm{HT,\\ | | | TH,\\ TT}\\mathrm{\\}}) = | | | \\frac{3}{4}\$\$]{.math | | |.display}\ | | | | | | \ | | | [\$\$\\therefore E\\left( | | | \\mathrm{tossing\\ 2\\ coins} | | | \\right) = \\left( 2 | | | \\right)\\left( \\frac{1}{4} | | | \\right) + \\left( - 1 | | | \\right)\\left( \\frac{3}{4} | | | \\right) = - \\frac{1}{4} = - | | | 0.25\$\$]{.math.display}\ | | | | | | The expected value of the bet to | | | you is --\$0.25. So, you would | | | rather not to bet. | +===================================+===================================+ | | *You know that you will either | | | win \$2 or lose \$1. However, the | | | expected value tells you that you | | | may eventually lose \$0.25 if you | | | bet over and over again.* | +-----------------------------------+-----------------------------------+ **7. Addition Law 加法定律** Addition law is a method for finding the probability that either or both of two events happen in one task/situation. Of course, the counting method is always applied. \ [\$\$\\mathrm{Probability\\ of\\ 2\\ events\\ }\\mathrm{A}\\mathrm{\\ }\\mathrm{\\text{OR}}\\mathrm{\\ }\\mathrm{B}\\mathrm{\\ } = P\\left( \\ \\mathrm{A}\\mathrm{\\ }\\mathrm{\\text{OR}}\\mathrm{\\ }\\mathrm{B}\\ \\right) = \\frac{\\mathrm{\\text{Number\\ of}}\\mathrm{\\ }\\mathrm{\\text{Favourable}}\\mathrm{\\ }\\mathrm{\\text{Outcomes\\ of\\ }}\\mathrm{A}\\mathrm{\\ }\\mathrm{\\text{OR}}\\mathrm{\\ }\\mathrm{B}}{\\mathrm{\\text{Total\\ Number\\ of\\ Possible\\ Outcomes}}}\$\$]{.math.display}\ However, it may be more convenient to consider the probabilities of the two events separately in certain situations and the addition law may help: \ [*Probability* *of* 2 *events* *A* OR *B*  = *P*( *A* OR *B* ) = *P*( *A* ) + *P*( *B* ) − *P*( *A* AND *B* )]{.math.display}\ **7.1. Mutually Exclusive Events 互斥事件 (⇒ 兩個事件不能一起發生)** Suppose *A* and ***B*** are two events in a given situation. If *A* and ***B*** cannot occur simultaneously 同時, they are said to be **mutually exclusive events** and [*P*( *A* AND *B* ) **=** **0**]{.math.inline}. ∴Addition law for mutually exclusive events can be simplified to: \ [*P*( *A* OR *B* ) = *P*( *A* ) + *P*( *B* ) − 0 = *P*( *A* ) + *P*( *B* )]{.math.display}\ **7.2. NOT Mutually Exclusive Events** On the contrary, [*P*( *A* AND *B* ) **\>** **0**]{.math.inline} if *A* and ***B*** can possibly occur simultaneously. There would be overlapping area between *A* and ***B*** as illustrated in the diagram. ∴The general addition law applies: \ [*P*( *A* OR *B* ) = *P*( *A* ) + *P*( *B* ) − *P*( *A* AND *B* )]{.math.display}\ **Example** **18.** In drawing a card from a standard deck of playing cards, let's define 3 events: - *K* be the event that the card drawn is a "King" - *Q* be the event that the card drawn is a "Queen" - *H* be the event that the card drawn belongs to the "Heart" suit. \(a) Explain if the events are mutually exclusive with the help of a diagram. \(b) Find the probability of getting a "King". \(c) Find the probability of getting a "Queen". \(d) Find the probability of getting a "Heart". \(e) Find the probability of getting a "King" or a "Queen". \(f) Find the probability of getting a "King" or a "Heart". +-----------------+-----------------+-----------------+-----------------+ | **Solution:** | (a) | *K* and *Q* are | | | | | mutually | | | | | exclusive | | | | | (*there is no | | | | | common | | | | | element*). | | | | | | | | | | *K* and *H* are | | | | | not mutually | | | | | exclusive (♥K). | | | | | | | | | | *Q* and *H* are | | | | | not mutually | | | | | exclusive (♥Q). | | +=================+=================+=================+=================+ | | (b) | A standard | | | | | 52-card deck ⇒ | | | | | the total | | | | | number of | | | | | possible | | | | | outcomes = 52 | | | | | | | | | | \ | | | | | [\$\$P\\left( | | | | | \\mathrm{\\text | | | | | {King}} | | | | | \\right) = | | | | | \\frac{4}{52} = | | | | | \\frac{1}{13}\$ | | | | | \$]{.math | | | | |.display}\ | | | | | | | | | | ∴The | | | | | probability of | | | | | getting a | | | | | "King" is | | | | | [\$\\frac{1}{13 | | | | | }\$]{.math | | | | |.inline}. | | +-----------------+-----------------+-----------------+-----------------+ | | (c) | \ | | | | | [\$\$P\\left( | | | | | \\mathrm{\\text | | | | | {Queen}} | | | | | \\right) = | | | | | \\frac{4}{52} = | | | | | \\frac{1}{13}\$ | | | | | \$]{.math | | | | |.display}\ | | | | | | | | | | ∴The | | | | | probability of | | | | | getting a | | | | | "Queen" is | | | | | [\$\\frac{1}{13 | | | | | }\$]{.math | | | | |.inline}. | | +-----------------+-----------------+-----------------+-----------------+ | | (d) | \ | | | | | [\$\$P\\left( | | | | | \\mathrm{\\text | | | | | {Heart}} | | | | | \\right) = | | | | | \\frac{13}{52} | | | | | = | | | | | \\frac{1}{4}\$\ | | | | | $]{.math | | | | |.display}\ | | | | | | | | | | ∴The | | | | | probability of | | | | | getting a | | | | | "Heart" is | | | | | [\$\\frac{1}{4} | | | | | \$]{.math | | | | |.inline}. | | +-----------------+-----------------+-----------------+-----------------+ | | (e) | **Addition Law | **Counting | | | | (mutually | Method:** | | | | exclusive):** | | +-----------------+-----------------+-----------------+-----------------+ | | | \ | No. of | | | | [*P*(King OR | favourable | | | | Queen)]{.math | outcomes = 8 | | | |.display}\ | | | | | | \ | | | | \ | [\$\$\\therefor | | | | [ = *P*(King) + | e | | | |  *P*(Queen) − * | P\\left( | | | | P*(King | \\mathrm{\\text | | | | AND | {King\\ | | | | Queen)]{.math | }}\\mathrm{\\te | | | |.display}\ | xt{OR}}\\mathrm | | | | | {\\text{\\ | | | | \ | Queen}} | | | | [\$\$= | \\right) = | | | | \\frac{1}{13} + | \\frac{8}{52}\$ | | | | \\frac{1}{13}\\ | \$]{.math | | | | mathbf{- |.display}\ | | | | 0}\$\$]{.math | | | | |.display}\ | \ | | | | | [\$\$= | | | | \ | \\frac{2}{13}\$ | | | | [\$\$= | \$]{.math | | | | \\frac{2}{13}\$ |.display}\ | | | | \$]{.math | | | | |.display}\ | | +-----------------+-----------------+-----------------+-----------------+ | | | The probability | | | | | of getting a | | | | | "King" or | | | | | "Queen" is | | | | | [\$\\frac{2}{13 | | | | | }\$]{.math | | | | |.inline}. | | +-----------------+-----------------+-----------------+-----------------+ | | (f) | **Addition Law | **Counting | | | | (general):** | Method:** | +-----------------+-----------------+-----------------+-----------------+ | | | \ | No. of | | | | [*P*(King OR | favourable | | | | Heart)]{.math | outcomes = 16 | | | |.display}\ | | | | | | \ | | | | \ | [\$\$\\therefor | | | | [ = *P*(King) + | e | | | |  *P*(Heart) − * | P\\left( | | | | P*(King | \\mathrm{\\text | | | | AND | {King\\ | | | | Heart)]{.math | }}\\mathrm{\\te | | | |.display}\ | xt{OR}}\\mathrm | | | | | {\\text{\\ | | | | \ | Heart}} | | | | [\$\$= | \\right) = | | | | \\frac{1}{13} + | \\frac{16}{52}\ | | | | \\frac{1}{4}\\m | $\$]{.math | | | | athbf{-}\\frac{ |.display}\ | | | | \\mathbf{1}}{\\ | | | | | mathbf{52}}\$\$ | \ | | | | ]{.math | [\$\$= | | | |.display}\ | \\frac{4}{13}\$ | | | | | \$]{.math | | | | \ |.display}\ | | | | [\$\$= | | | | | \\frac{4}{13}\$ | | | | | \$]{.math | | | | |.display}\ | | +-----------------+-----------------+-----------------+-----------------+ | | | The probability | | | | | of getting a | | | | | "King" or | | | | | "Heart" is | | | | | [\$\\frac{4}{13 | | | | | }\$]{.math | | | | |.inline}. | | +-----------------+-----------------+-----------------+-----------------+ *When would you use the addition law or the basic counting method to calculate the probabilities? All depends on the information you got. The counting method is always worked but you have to have a clear picture of all the numbers.* **Example** **19.** In a football match 足球比賽, the probability that team *X* will win is ¼ and the probability that team *Y* will win is ⅓. Find the probability that \(a) List out all the possible outcomes of the football match. \(b) Find the probability that team *X* or *Y* will win the match. \(c) Find the probability that the match ended in a tie/draw 打和. +-----------------------+-----------------------+-----------------------+ | **Solution:** | (a) | Possible outcomes are | | | | **{** Team *X* win, | | | | Team *Y* win, Tie | | | | **}** | +=======================+=======================+=======================+ | | (b) | \ | | | | [*P*(Team *X* win OR | | | | Team *Y* win)]{.math | | | |.display}\ | | | | | | | | \ | | | | [ = *P*(Team *X* | | | | win) + *P*(Team *Y* | | | | win) − *P*(Team *X* | | | | win AND Team *Y* | | | | win)]{.math | | | |.display}\ | | | | | | | | \ | | | | [\$\$= \\frac{1}{4} + | | | | \\frac{1}{3}\\mathbf{ | | | | - | | | | 0}\$\$]{.math | | | |.display}\ | | | | | | | | \ | | | | [\$\$= | | | | \\frac{7}{12}\$\$]{.m | | | | ath | | | |.display}\ | | | | | | | | ∴The probability that | | | | team *X* or *Y* will | | | | win the match is | | | | [\$\\frac{7}{12}\$]{. | | | | math | | | |.inline}. | +-----------------------+-----------------------+-----------------------+ | | (c) | As there are only 3 | | | | possible outcomes, | | | | | | | | \ | | | | [*P*(Tie) = 1 − *P*(T | | | | eam | | | | *X* win OR Team *Y* | | | | win)]{.math | | | |.display}\ | | | | | | | | \ | | | | [\$\$= 1 - | | | | \\frac{7}{12}\$\$]{.m | | | | ath | | | |.display}\ | | | | | | | | \ | | | | [\$\$= | | | | \\frac{5}{12}\$\$]{.m | | | | ath | | | |.display}\ | | | | | | | | ∴The probability that | | | | the match ended in a | | | | tie is | | | | [\$\\frac{5}{12}\$]{. | | | | math | | | |.inline}. | +-----------------------+-----------------------+-----------------------+ **Example** **20.** In a survey, 200 IVE students are questioned about their newspaper reading habits. It is found that 120 students read Chinese newspapers, 85 students read English newspapers and 55 students read both Chinese and English newspapers. \(a) Calculate the number of students who \(i) read Chinese newspapers only. \(ii) read English newspapers only. \(iii) do not read any newspapers. \(b) If one of the 200 students is chosen at random, find the probabilities that \(i) the student reads exactly one kind of the newspapers. \(ii) the student reads either Chinese or English newspapers. \(iii) the student does not read any newspapers. **Solution:** (a) i) No. of students read Chinese newspapers only = 120 -- 55 = 65 --------------- ----- ------ ------------------------------------------------------------------------- ii) No. of students read English newspapers only = 85 -- 55 = 30 iii) No. of students do not read any newspapers = 200 -- 65 -- 30 -- 55 = 50 +-------------+-------------+-------------+-------------+-------------+ | **Solution: | | | | | | ** | | | | | +=============+=============+=============+=============+=============+ | | (b) | i) | \ | | | | | | [\$\${P\\le | | | | | | ft( | | | | | | \\mathrm{\\ | | | | | | text{exactl | | | | | | y\\ | | | | | | one\\ | | | | | | kind}} | | | | | | \\right) = | | | | | | P\\left( | | | | | | \\mathrm{\\ | | | | | | text{Chi\\ | | | | | | only\\ | | | | | | }}\\mathrm{ | | | | | | \\text{OR}} | | | | | | \\mathrm{\\ | | | | | | text{\\ | | | | | | Eng\\ | | | | | | only}} | | | | | | \\right) | | | | | | }{= | | | | | | \\frac{65 + | | | | | | 30}{200} | | | | | | }{= | | | | | | \\frac{19}{ | | | | | | 40}}\$\$]{. | | | | | | math | | | | | |.display}\ | | | | | | | | | | | | ∴The | | | | | | probability | | | | | | that the | | | | | | student | | | | | | reads | | | | | | exactly one | | | | | | kind of the | | | | | | newspapers | | | | | | is | | | | | | [\$\\frac{1 | | | | | | 9}{40}\$]{. | | | | | | math | | | | | |.inline}. | | +-------------+-------------+-------------+-------------+-------------+ | | | ii) | **Addition | **Counting | | | | | Law | Method:** | | | | | (general):* | | | | | | * | | +-------------+-------------+-------------+-------------+-------------+ | | | | \ | \ | | | | | [*P*(Chi OR | [*P*(Chi OR | | | | | Eng)]{.math | Eng)]{.math | | | | |.display}\ |.display}\ | | | | | | | | | | | \ | \ | | | | | [ = *P*(Chi | [\$\$= | | | | | ) + *P*(Eng | \\frac{65 + | | | | | ) − *P*(Chi | 55 + | | | | | AND | 30}{200}\$\ | | | | | Eng)]{.math | $]{.math | | | | |.display}\ |.display}\ | | | | | | | | | | | \ | \ | | | | | [\$\$= | [\$\$= | | | | | \\frac{120} | \\frac{3}{4 | | | | | {200} | }\$\$]{.mat | | | | | + | h | | | | | \\frac{85}{ |.display}\ | | | | | 200} | | | | | | - | | | | | | \\frac{55}{ | | | | | | 200}\$\$]{. | | | | | | math | | | | | |.display}\ | | | | | | | | | | | | \ | | | | | | [\$\$= | | | | | | \\frac{3}{4 | | | | | | }\$\$]{.mat | | | | | | h | | | | | |.display}\ | | +-------------+-------------+-------------+-------------+-------------+ | | | | ∴The | | | | | | probability | | | | | | that the | | | | | | student | | | | | | reads | | | | | | either | | | | | | Chinese or | | | | | | English | | | | | | newspapers | | | | | | is | | | | | | [\$\\frac{3 | | | | | | }{4}\$]{.ma | | | | | | th | | | | | |.inline}. | | +-------------+-------------+-------------+-------------+-------------+ | | | iii) | **Complemen | **Counting | | | | | tary | Method:** | | | | | Event \[& | | | | | | using | | | | | | (b)(ii)\]:* | | | | | | * | | +-------------+-------------+-------------+-------------+-------------+ | | | | \ | \ | |