Answer Key

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, Theoretical Probability = ππππππππππππ ππππ ππππππππππππππππππ ππππππππππππππππ ππππππππππ ππππππππππππ ππππ ππππππππππππππππ 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, Experimental Probability = Number of success Number of trails Exercise 3 The number of outcomes in rolling a die is 6. And the sample space is {1, β2 , β3 , β4 , β5 , β6 } . The number of outcomes in tossing three coins is 8. And the sample space is { π»π»π»π»π»π» , π»π»π»π»π»π» , π»π»π»π»π»π» , π»π»π»π»π»π» , π»π»π»π»π»π» , π»π»π»π»π»π» , π»π»π»π»π»π» , π»π»π»π»π»π» } . So, when you roll a die and toss three coins the number outcome will be 6*8=48. And the required sample is, { β 1 π»π»π»π»π»π» , β1π»π»π»π»π»π» , 1 π»π»π»π»π»π» , β1π»π»π»π»π»π» , β1π»π»π»π»π»π» , β1π»π»π»π»π»π» , β1π»π»π»π»π»π» , β1π»π»π»π»π»π» , β2π»π»π»π»π»π» , β2π»π»π»π»π»π» , 2π»π»π»π»π»π» , β2π»π»π»π»π»π» , β2π»π»π»π»π»π» , β2π»π»π»π»π»π» , β2π»π»π»π»π»π» , β2π»π»π»π»π»π» , β3π»π»π»π»π»π» , β3π»π»π»π»π»π» , 3π»π»π»π»π»π» , β3π»π»π»π»π»π» , β3π»π»π»π»π»π» , β3π»π»π»π»π»π» , β3π»π»π»π»π»π» , β3π»π»π»π»π»π»β β4π»π»π»π»π»π» , β4π»π»π»π»π»π» , 4π»π»π»π»π»π» , β4π»π»π»π»π»π» , β4π»π»π»π»π»π» , β4π»π»π»π»π»π» , β4π»π»π»π»π»π» , β4π»π»π»π»π»π» , β5π»π»π»π»π»π» , β5π»π»π»π»π»π» , 5π»π»π»π»π»π» , β5π»π»π»π»π»π» , β5π»π»π»π»π»π» , β5π»π»π»π»π»π» , β5π»π»π»π»π»π» , β5π»π»π»π»π»π» , β6π»π»π»π»π»π» , β6π»π»π»π»π»π» , 6π»π»π»π»π»π» , β6π»π»π»π»π»π» , β6π»π»π»π»π»π» , β6π»π»π»π»π»π» , β6π»π»π»π»π»π» , β6π»π»π»π»π»π» } Exercise 4 The number of possible outcomes in flipping a coin is 2. And the sample space is { π»π» , π»π» , } . 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 ππ , ππ . 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 , { π»π»ππ , π»π»ππ , π»π»ππ , π»π»ππ } . Exercise 5 Let { π π 1, π π 2 , π π 3 , π π 4 } represents the red cards numbered 1 through 4. Let { ππ 1, ππ2 , ππ3 , ππ4 } represents the white cards numbered 1 through 4. Let { π΅π΅ 1, π΅π΅2 , π΅π΅3 , π΅π΅4 } represents the black cards numbered 1 through 4. So, total number of outcomes is 12 and the required sample space is { π π 1, π π 2 , π π 3 , π π 4 , ππ 1, ππ2 , ππ3 , ππ4 , π΅π΅ 1, π΅π΅2 , π΅π΅3 , π΅π΅4 } . 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, πΊπΊπΊπΊ , π΅π΅π΅π΅ , πΊπΊπ΅π΅ , π΅π΅πΊπΊ , 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. So, the probability that exactly two of the five contestants win is 1032 Exercise 8 Each deck of 52 cards contains 13 spades each. So, the the probability of of getting a spade is 1352 . So, the probability that you will draw two spades is 1352 β 1352 , which is equal to 1 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) = 1 β 3 36 = 3336 = 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) = 1 β 10 36 = 2636

= 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 = 45 b) From the figure we see that, the probability that a person chosen at random is from 25 to 44 years old is, 13 100 β + 13 100 = 26 100 = 1350 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. So, the required probability is 14 Exercise 12 The probability that the number is less than 4 is 4 30 not 3 30 . So, the required probability is, 1 β 4 30

= 2630 = 1315 Exercise 13 From the figure we see that, Area of the entire board, π΄π΄ = 18 β 18 = 324 Area of the yellow region, ππ = Ο β 9 2 β 6 2 + 2 β 12 β 6 β 6 So, the required probability is, P(Hitting the yellow region) = πππ΄π΄ β 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. So, the required probability is 367 1631 β 0. 23 b) The length of shoreline along the Gulf of Mexico for Alabama is 53 mi. So, the required probability is 53 1631 β 0. 03 c) The length of shoreline along the Gulf of Mexico for Florida is 770 mi. So, the required probability is 770 1631 β 0. 47 d) The length of shoreline along the Gulf of Mexico for Louisiana is 397 mi. So, the required probability is 397 1631 β 0. 24 Exercise 15 The theoretical probability of rolling each number is 16 From the table let us calculate the experimental probability for each number. ππ (1) = 11 60 ππ ( 2 ) = 1460 ππ ( 3 ) = 7 60 ππ ( 4 ) = 10 60 ππ ( 5 ) = 1660 ππ ( 6 ) = 1260

So, We get experimental probability for rolling number 4 is 16 which is equal to the theoretical probability of rolling number 4. Exercise 16 The theoretical probability of getting a marble of any color is 15 . Let us calculate the experimental probability for each coloured marble. ππ ( ππβππππππ ) = 5 30 ππ ( π΅π΅π΅π΅π΅π΅π΅π΅π΅π΅ ) = 6 30 ππ ( π π πππ π ) = 8 30 ππ ( πΊπΊπΊπΊπππππΊπΊ ) = 2 30 ππ ( π΅π΅π΅π΅π΅π΅ππ ) = 9 30 So, We get experimental probability for getting a black coloured marble is 15 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 { β6 , β9 , β12 , β15 , β18 , β21 , β24 , β27 , β30 } .

So, the required probability is 9 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. So, the required experimental probability is 2030 . 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. So, the required probability is 179 2237 Exercise 20 Here, the number of trails is 2392. The number of success is 526. So, the required probability is 526 2392 Exercise 21 From the figure we see that, Area of the entire board, π΄π΄ = 18 β 18 = 324 a) Area of the green region, πΊπΊ = 2 β 12 β 6 β 6 . So, the probability of hitting the green region, P(Hitting the green region)= πΊπΊπ΄π΄ β 0.11 . b) Area of the not blue region, π΅π΅ = Ο β 9 2 . So, the probability of not hitting the blue region, P(not Hitting the blue region) = π΅π΅π΄π΄ β 0. 79 . c) Area of the red region, π π = Ο β 3 2 . So, the probability of hitting the red region, P(Hitting the red region) = π π π΄π΄ β 0.09 .

d) Area of the not yellow region, ππ = 18 2 β Ο β 9 2 + 6 2 + 2 β 12 β 6 β 6 So, the required probability is, P(Not Hitting the yellow region) = πππ΄π΄ β 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 { β2 , β3 , β4 , β5 , β6 , β7 , β8 , β9 , β10 , β11 , β12 } . b) Let us calculate the theoretical probability for each sum. ππ ( 2 ) = 1 36 . ππ ( 3 ) = 2 36 . ππ ( 4 ) = 3 36 . ππ ( 5 ) = 4 36 . ππ ( 6 ) = 5 36 . ππ ( 7 ) = 6 36 . ππ (8) = 5 36 . ππ (9) = 6 36 . ππ (10) = 3 36 . ππ (11) = 2 36 . ππ ( 12 ) = 1 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 particular experiment. The theoretical probability of landing heads up is 12 . Exercise 25 The probability that a point chosen at random inside the cube is also inside the sphere is given by the formula, P(Point is inside the sphere) = Volume of 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. So, the required probability is 46 . Exercise 27 The probability of defective computer is, P(defective computer) = 9 1200 . So, the probability in percentage is 0.75%. So, In a shipment of 15000 computers, the number od defective computer will be 15000 β 0. 75% β 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 ) . So, the required probability is 16 .

- Chapter 9.1 The Pythagorean theorem
- 2.3 - Continuity
- 2.2 - Infinite Limits
- 2.1 - Introduction to Limits
- Tabular Integrations
- Rules of Differentiation
- Linear Function
- General Form Function
- Define Integrals
- Algebraic Expressions
- General Prime Numbers
- Primality Test
- Imaginary Numbers
- The Matrix (Mathematics)

Chapter 12.1 Sample Spaces and Probability

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