Chapter 12.1 Sample Spaces and Probability

12.1 Sample Spaces and Probability Exercise 1 A number that describe the likelihood of an event is the Probability of the event. Exercise 2 The theoretical probability of an event is defined using the following formula, 𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁 𝑜𝑜𝑜𝑜 𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜 Theoretical Probability = 𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛 𝑜𝑜𝑜𝑜 𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜 An experimental probability is based on repeated trails of a probability experiment. The number of trails is the number of times the probability experiment is performed. Each trail in which a favourable outcome occurs is called a success. The experimental probability can be found using the following, Number of success Experimental Probability = Number of trails Exercise 3 The number of outcomes in rolling a die is 6. And the sample space is {1, 2, 3, 4, 5, 6}. The number of outcomes in tossing three coins is 8. And the sample space is {𝐻𝐻𝐻𝐻𝐻𝐻, 𝐻𝐻𝐻𝐻𝐻𝐻, 𝑇𝑇𝑇𝑇𝑇𝑇, 𝐻𝐻𝐻𝐻𝐻𝐻, 𝑇𝑇𝑇𝑇𝑇𝑇, 𝑇𝑇𝑇𝑇𝑇𝑇, 𝑇𝑇𝑇𝑇𝑇𝑇, 𝐻𝐻𝐻𝐻𝐻𝐻}. So, when you roll a die and toss three coins the number outcome will be 6*8=48. And the required sample is, { 1𝐻𝐻𝐻𝐻𝐻𝐻, 1𝐻𝐻𝐻𝐻𝐻𝐻, 1𝑇𝑇𝑇𝑇𝑇𝑇, 1𝐻𝐻𝐻𝐻𝐻𝐻, 1𝑇𝑇𝑇𝑇𝑇𝑇, 1𝑇𝑇𝑇𝑇𝑇𝑇, 1𝑇𝑇𝑇𝑇𝑇𝑇, 1𝐻𝐻𝐻𝐻𝐻𝐻, 2𝐻𝐻𝐻𝐻𝐻𝐻, 2𝐻𝐻𝐻𝐻𝐻𝐻, 2𝑇𝑇𝑇𝑇𝑇𝑇, 2𝐻𝐻𝐻𝐻𝐻𝐻, 2𝑇𝑇𝑇𝑇𝑇𝑇, 2𝑇𝑇𝑇𝑇𝑇𝑇, 2𝑇𝑇𝑇𝑇𝑇𝑇, 2𝐻𝐻𝐻𝐻𝐻𝐻, 3𝐻𝐻𝐻𝐻𝐻𝐻, 3𝐻𝐻𝐻𝐻𝐻𝐻, 3𝑇𝑇𝑇𝑇𝑇𝑇, 3𝐻𝐻𝐻𝐻𝐻𝐻, 3𝑇𝑇𝑇𝑇𝑇𝑇, 3𝑇𝑇𝑇𝑇𝑇𝑇, 3𝑇𝑇𝑇𝑇𝑇𝑇, 3𝐻𝐻𝐻𝐻𝐻𝐻 4𝐻𝐻𝐻𝐻𝐻𝐻, 4𝐻𝐻𝐻𝐻𝐻𝐻, 4𝑇𝑇𝑇𝑇𝑇𝑇, 4𝐻𝐻𝐻𝐻𝐻𝐻, 4𝑇𝑇𝑇𝑇𝑇𝑇, 4𝑇𝑇𝑇𝑇𝑇𝑇, 4𝑇𝑇𝑇𝑇𝑇𝑇, 4𝐻𝐻𝐻𝐻𝐻𝐻, 5𝐻𝐻𝐻𝐻𝐻𝐻, 5𝐻𝐻𝐻𝐻𝐻𝐻, 5𝑇𝑇𝑇𝑇𝑇𝑇, 5𝐻𝐻𝐻𝐻𝐻𝐻, 5𝑇𝑇𝑇𝑇𝑇𝑇, 5𝑇𝑇𝑇𝑇𝑇𝑇, 5𝑇𝑇𝑇𝑇𝑇𝑇, 5𝐻𝐻𝐻𝐻𝐻𝐻, 6𝐻𝐻𝐻𝐻𝐻𝐻, 6𝐻𝐻𝐻𝐻𝐻𝐻, 6𝑇𝑇𝑇𝑇𝑇𝑇, 6𝐻𝐻𝐻𝐻𝐻𝐻, 6𝑇𝑇𝑇𝑇𝑇𝑇, 6𝑇𝑇𝑇𝑇𝑇𝑇, 6𝑇𝑇𝑇𝑇𝑇𝑇, 6𝐻𝐻𝐻𝐻𝐻𝐻} Exercise 4 The number of possible outcomes in flipping a coin is 2. And the sample space is { 𝐻𝐻, 𝑇𝑇, }. The number of possible outcomes in drawing a ball from bag with two purple marbles and one white marbles is 2, namely, Purple and White. So, the sample space is 𝑃𝑃, 𝑊𝑊. Where P stands for Purple and W stands for White marble. So, when you flip a coin and draw a marble from a bag with two purple marbles and one white marbles, the number of outcome will be 2*2=4. And the required sample space is , {𝐻𝐻𝐻𝐻, 𝐻𝐻𝐻𝐻, 𝑇𝑇𝑇𝑇, 𝑇𝑇𝑇𝑇}. Exercise 5 Let { 𝑅𝑅1, 𝑅𝑅2, 𝑅𝑅3, 𝑅𝑅4} represents the red cards numbered 1 through 4. Let { 𝑊𝑊1, 𝑊𝑊2, 𝑊𝑊3, 𝑊𝑊4} represents the white cards numbered 1 through 4. Let {𝐵𝐵1, 𝐵𝐵2, 𝐵𝐵3, 𝐵𝐵4} represents the black cards numbered 1 through 4. So, total number of outcomes is 12 and the required sample space is {𝑅𝑅1, 𝑅𝑅2, 𝑅𝑅3, 𝑅𝑅4, 𝑊𝑊1, 𝑊𝑊2, 𝑊𝑊3, 𝑊𝑊4, 𝐵𝐵1, 𝐵𝐵2, 𝐵𝐵3, 𝐵𝐵4}. Exercise 6 When you draw two marbles without replacement from a bag containing three green marbles and four black marbles, there are 4 possible outcomes, namely, Both of them will be green, or Both of them will be black, or First one is green and second one is black, or First one is black and second one is green. So, the required sample space is, 𝐺𝐺𝐺𝐺, 𝐵𝐵𝐵𝐵, 𝐺𝐺𝐺𝐺, 𝐵𝐵𝐵𝐵, Where G represents green marble and B represents white marble. Exercise 7 Here it is given that, In a game show that airs on television five days per week, a prize is placed behind one of the two doors. The contestant wins the prize by selecting the correct door. We need to find the probability that exactly two of the five contestants win a prize during a week. First let us write down all the possible outcomes of the sample space. First Case: When no one wins. {LLLLL} Second Case: When exactly one contestant win. {WLLLL, LWLLL, LLWLL, LLLWL, LLLLW} Third Case: When exactly two contestants win. {WWLLL, LWWLL, LLWWL, LLLWW, WLLLW, WLWLL, LWLLW, LLWLW, WLLWL, LWLWL} Fourth Case: When exactly three contestants win. {WWWLL, LWWWL, LLWWW, WLLWW, LWLWW, WWLLW, WWLWL, WLWLW, LWWLW, WLWWL} Fifth Case: When exactly four contestant win. {WWWWL, WWWLW, WWLWW, WLWWW, LWWWW} Sixth Case: When all the contestants win, {WWWWW} Here, L represent a loss and W represent a win. So, total number of outcome is 32 and total number of favourable outcome is 10. 10 So, the probability that exactly two of the five contestants win is 32 Exercise 8 Each deck of 52 cards contains 13 spades each. So, the the probability of of getting a 13 spade is 52. 13 13 1 So, the probability that you will draw two spades is 52 ∗ 52, which is equal to 16. Exercise 9 a) P(Sum is not 4)= 1- P(Sum is 4) Favourable outcomes for the event that the sum is 4 is {(1,3), (2,2), (3,1)}. So, the required probability is, P(Sum is not 4)= 1- P(Sum is 4) 3 =1− 36 33 = 36 11 = 12 b) P(Sum is greater than 5)= 1- P(Sum is less than or equal to 5) Favourable outcomes for the event that the sum is less than or equal to 5 is {(1,1), (1,2), (1,3), (1,4), (2,1), (2,2), (2,3), (3,1), (3,2), (4,1)}. So, the required probability is, P(Sum is greater than 5)= 1- P(Sum is less than or equal to 5) 10 =1− 36 26 = 36 = 13 18 Exercise 10 a) From the figure we see that, the probability that a person chosen at random is at least 15 years old is, 80 100 4 = 5 b) From the figure we see that, the probability that a person chosen at random is from 25 to 44 years old is, 13 13 + 100 100 26 = 100 13 = 50 Exercise 11 There are four possibilities. 1) Both the guesses are correct. 2) Both the guesses are incorrect. 3) First guess is correct and the second guess is incorrect. 4) First guess is incorrect and the second guess is correct. So, the number of total outcomes are 4 and the number of favourable outcome is 1. 1 So, the required probability is 4 Exercise 12 4 3 The probability that the number is less than 4 is 30 not 30. So, the required probability is, 4 1− 30 26 30 13 = 15 = Exercise 13 From the figure we see that, Area of the entire board, 𝐴𝐴 = 18 ∗ 18 = 324 1 Area of the yellow region, 𝑌𝑌 = π ∗ 92 − 62 + 2 ∗ 2 ∗ 6 ∗ 6 So, the required probability is, 𝑌𝑌 P(Hitting the yellow region) = 𝐴𝐴 ≈ 0.56. Exercise 14 Total length of the shoreline = 367+397+44+53+770=1631mi a) The length of shoreline along the Gulf of Mexico for Texas is 367 mi. 367 So, the required probability is 1631 ≈ 0.23 b) The length of shoreline along the Gulf of Mexico for Alabama is 53 mi. 53 So, the required probability is 1631 ≈ 0.03 c) The length of shoreline along the Gulf of Mexico for Florida is 770 mi. 770 So, the required probability is 1631 ≈ 0.47 d) The length of shoreline along the Gulf of Mexico for Louisiana is 397 mi. 397 So, the required probability is 1631 ≈ 0.24 Exercise 15 1 The theoretical probability of rolling each number is 6 From the table let us calculate the experimental probability for each number. 11 𝑃𝑃(1) = 60 14 𝑃𝑃(2) = 60 7 𝑃𝑃(3) = 60 10 𝑃𝑃(4) = 60 16 𝑃𝑃(5) = 60 12 𝑃𝑃(6) = 60 1 So, We get experimental probability for rolling number 4 is 6 which is equal to the theoretical probability of rolling number 4. Exercise 16 1 The theoretical probability of getting a marble of any color is 5. Let us calculate the experimental probability for each coloured marble. 5 𝑃𝑃(𝑊𝑊ℎ𝑖𝑖𝑖𝑖𝑖𝑖) = 30 6 𝑃𝑃(𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵) = 30 8 𝑃𝑃(𝑅𝑅𝑅𝑅𝑅𝑅) = 30 2 𝑃𝑃(𝐺𝐺𝐺𝐺𝐺𝐺𝐺𝐺𝐺𝐺) = 30 9 𝑃𝑃(𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵) = 30 1 So, We get experimental probability for getting a black coloured marble is 5 which is equal to the theoretical probability. Exercise 17 a) Total number of outcomes is 10. Favourable outcomes that the number is multiple of 3 are { 6, 9, 12, 15, 18, 21, 24, 27, 30}. 9 So, the required probability is 10. b) Here, the number of success that is the spinner stops on a multiple of 3 is 20. The number of trails is 30. 20 So, the required experimental probability is 30. c) In case of (b), the value of the required probability is based on number of trails, not possible outcomes. Exercise 18 The probability of getting a 0 with a 6-sided die is 0. The probability of of getting a head or tails after tossing a coin is 1. Exercise 19 Here, the number of trails is 2237. The number of success is 179. 179 So, the required probability is 2237 Exercise 20 Here, the number of trails is 2392. The number of success is 526. 526 So, the required probability is 2392 Exercise 21 From the figure we see that, Area of the entire board, 𝐴𝐴 = 18 ∗ 18 = 324 1 a) Area of the green region, 𝐺𝐺 = 2 ∗ 2 ∗ 6 ∗ 6. So, the probability of hitting the green region, 𝐺𝐺 P(Hitting the green region)= 𝐴𝐴 ≈ 0.11. b) Area of the not blue region, 𝐵𝐵 = π ∗ 92 . So, the probability of not hitting the blue region, 𝐵𝐵 P(not Hitting the blue region)= 𝐴𝐴 ≈ 0.79. c) Area of the red region, 𝑅𝑅 = π ∗ 32 . So, the probability of hitting the red region, 𝑅𝑅 P(Hitting the red region)= 𝐴𝐴 ≈ 0.09. 1 d) Area of the not yellow region, 𝑌𝑌 = 182 − π ∗ 92 + 62 + 2 ∗ 2 ∗ 6 ∗ 6 So, the required probability is, 𝑌𝑌 P(Not Hitting the yellow region) = 𝐴𝐴 ≈ 0.44. So, the required order is c,a,d,b. Exercise 22 From the given chart, we see that, a) The probability that rains on Sunday is 80%. b) The probability that it does not rain on Saturday is 70%. c)The probability that rains on Monday is 90%. d) The probability that it does not rain on Friday is 95%. So, the required order is b,a,c,d. Exercise 23 a) The possible sums are { 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}. b) Let us calculate the theoretical probability for each sum. 1 . 𝑃𝑃(2) = 36 2 𝑃𝑃(3) = . 36 3 𝑃𝑃(4) = . 36 4 𝑃𝑃(5) = . 36 5 𝑃𝑃(6) = . 36 6 𝑃𝑃(7) = . 36 5 . 𝑃𝑃(8) = 36 6 𝑃𝑃(9) = . 36 3 𝑃𝑃(10) = . 36 2 𝑃𝑃(11) = . 36 1 𝑃𝑃(12) = 36. c) The theoretical and experimental probabilities are similar. Exercise 24 The given probability of the coin landing heads up is experimental probability for the 1 particular experiment. The theoretical probability of landing heads up is 2. Exercise 25 The probability that a point chosen at random inside the cube is also inside the sphere is given by the formula, Volume of the sphere P(Point is inside the sphere) = Volume of the cube Exercise 26 The graph of the function y=f(x)+c will intersect the x-axis if the value of c is 1,2,3 or 4. So, the number of favourable case is 4 and the number of total case 6. 4 So, the required probability is 6. Exercise 27 The probability of defective computer is, 9 P(defective computer)= 1200 . So, the probability in percentage is 0.75%. So, In a shipment of 15000 computers, the number od defective computer will be 15000 ∗ 0.75% ≈ 113. Exercise 28 Problem: Box A contains three balls numbered 1 through 3 and Box B contains two balls numbered 1 through 2. You draw one ball from each box. Find the probability that you draw the ball numbered 2 from box A and ball numbered 1 from box B. Answer: Here, the sample space is, (1,1), (1,2), (2,1), (2,2), (3,1), (3,2). 1 So, the required probability is 6.