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oD lh‘iil‘il‘miii:il UNIT 3 — MILESTONE 3 21/27 < that's 78% ‘ RETAKE ‘ @ 21 questions were answered correctly. 6 questions were answered incorrectly. T Q@ John makes random guesses on his multiple-choice test, which has five options for each question. Let the random variable X be the number of guesses taken before guessing correctly. Assuming the guesses are independent, find the probability that he doesn't guess correctly until his 6th guess. © O 0.0655 O 0.3277 o O 0.3521 O 0.0789 RATIONALE Since we are looking for the probability until the first success, we will use the following Geometric distribution formula: 21/27
o lh‘iil‘il‘miii:il UNIT 3 — MILESTONE 3 The variable p is the probability of success, which in this case, a success is considered getting a question right on a multiple- choice test with five options, which would be 1/5 or 0.2. Px=6)=(1-0.2°"10.2)=(0.8)°(0.2) = (0.32768)(0.2) = 0.0655 CONCEPT — Geometric Distribution Report an issue with this question . S P16 | 68 66 00 (60 00 oo XA g R . N YRR Y SRR YIRE SN S AN AR el e ew eel eel eel ewl e o . : : : S N " 8 ' Y N 'Y d» ’o: ' ' 3 3 ‘."o** 4 3 : S FS S FPS FYFS »> o b 04 % @ é 8 ~e . ¥ : v Y: Y v ww: ww: ww: wew <ew; . g ; $ ie ‘e ivw| foe [fow ve | oo [low ‘-°v'v " 8 B v ve v e » v v ve ve veY b R 9 &b Gy s 4 ' Aa: (aa: a0 a4 | aa: 'Y 4% : 8 s A B 2 R 3 , ) 4 . CON s . (7, , ) (8, . ) f(o. ., ) 10 ) ;) [(a K + + & + ¢+ +4 4 R + 4 4 b 4 +4 ¢ + 4+ N N + + 4 o s *e? +*e A L4 ! + + ¢ P E \V4 Vv ‘ 4! + ! + 4+ + 4! Y ¢4 e + 4 +te; ’ 3 : Annika was having fun playing poker. She needed the next two cards dealt to be diamonds so she could make a flush (five cards of the same suit). There are 15 cards left in the deck, and five are diamonds. What is the probability that the two cards dealt to Annika (without replacement) will both be diamonds? Answer choices are in percentage format, rounded to the nearest whole number. 21/27
oD lh‘iil‘il‘miii:il UNIT 3 — MILESTONE 3 © 0 o 29% o 33 RATIONALE If there are 15 cards left in the deck with 5 diamonds, the probability of being dealt 2 diamonds if they are dealt without replacement means that we have dependent events because the outcome of the first card will affect the probability of the second card. We can use the following formula: P(1st card diamond and 2nd card diamond) = P(1st card diamand) ®* P(2nd card diamond | 1st card diamond) The probability that the first card is a diamond would be 5 out 2 of 15, or 15 The probability that the second card is a diamond, given that the first card was also a diamond, would b because we now have only 14 cards remaining and only 13 4 of those cards are diamond (since the first card was a diamond). So we can use these probabilities to find the probability that the two cards will both be diamonds: P(1st card diamond and 2nd card diamond) = % - % = % = (0.095 or roughly 10% CONCEPT — "And" Probability for Dependent Events Report an issue with this question 21/27
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lh‘iil‘il‘miii:il UNIT 3 — MILESTONE 3 o distribution? A probability distribution of the workers who arrive late to work O each day. A probability distribution of the average time it takes employees O o .g pPloYy to drive to work. A probability distribution showing O the number of minutes employees spend at lunch. A probability distribution showing O the number of pages employees read during the workday. RATIONALE For a distribution to be continuous, there must be an infinite number of possibilities. Since we are measuring the time to drive to work, there are an infinite number of values we might observe, for example: 2 hours, 30 minutes, 40 seconds, etc. CONCEPT — Probability Distribution Report an issue with this question 21/27
o lh‘iil‘il‘miii:il UNIT 3 — MILESTONE 3 sided die? RATIONALE Recall that the probability of a complement, or the probability of something NOT happening, can be calculated by finding the probability of that event happening, and then subtracting from 1. Note that the probability of rolling a four would be 1/6. So the probability of NOT rolling a four is equivalent to: FINOT four)="1 L %— Ll B B 6 CONCEPT — Complement of an Event Report an issue with this question 21/27
lh‘iil‘il‘miii:il UNIT 3 — MILESTONE 3 o All observations made are o O independent of each other. All observations are made O randomly. All observations are mutually O exclusive. All observations made are O dependent on each other. RATIONALE In the binomial distribution we always assume independence of trials. This is why we simply multiply the probability of successes and failures directly to find the overall probability. CONCEPT — Binomial Distribution Report an issue with this question 21/27
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o lh‘iil‘il‘miii:il UNIT 3 — MILESTONE 3 - b - - - ~h o = - ¥ : v Y: YW ww: e ww: ww: ww <ew ' 3 : . v e ) ‘v w vy ‘v v 3v'v L X ‘-"v'v | {3 v . : v v ve @ v - ¥ ] ¥ ve ve veY W R 9 <& R\ s S . 'Y 'Y 'Y A b | &a: 'Y 4% ¢ 8 : i I B# | [§#% oo | oo ool foo] oo Lol Poe)l [, 16, ) + + & # Pt > *, +*e $ & : ® + 4+ o’o A \V4 V; ¢ + T R B RARS: ¢4 + 4 +Tes ¢ 3 : Select the following statement that describes non-overlapping events. To win, Jon needs a red card. He O receives a Queen of Diamonds. Jon wants a face card so he can have a winning hand, and he receives the eight of clubs. Jon needs to roll an even number to win. When it’s his turn, he rolls a two. Receiving the King of Hearts O fulfills Jon's need of getting both a face card and a heart. RATIONALE Events are non-overlapping if the two events cannot both occur in a single trial of a chance experiment. Since he wants a face card {Jack, Queen or King} and he got an eight {8}, there is no overlap.
o lh‘iil‘il‘miii:il UNIT 3 — MILESTONE 3 7 Q@ Kendra was trying to decide which type of frozen yogurt to restock based on popularity: flavors with chocolate or flavors without chocolate. After studying the data, she noticed that chocolate flavors sold best on the weekdays and on the weekends, but not best overall. Which paradox has Kendra encountered? Simpson's Paradox False Negative Benford's Law O False Positive O RATIONALE This is an example of Simpson's paradox, which is when the trend overall is not the same that is examined in smaller groups. Since the sale of chocolate flavors is larger on the weekends, but this trend changes when looking at sales overall, this is a reversal of the trend. CONCEPT — Paradoxes 21/27
o lh‘iil‘il‘miii:il UNIT 3 — MILESTONE 3 8 Colleen has 6 eggs, one of which is hard-boiled while the rest are raw. Colleen can't remember which of the eggs are raw. Which of the following statements is true? If Colleen selected one egg, cracked it open and found out it was raw, the probability of selecting the hard-boiled egg on her second pick is 1/6. If Colleen selected one egg, cracked it open and found out it was raw, the probability of selecting the hard-boiled egg on her second pick is 1/5. The probability of Colleen selecting the hard-boiled egg on her first try is 1/5. The probability of Colleen selecting a raw egg on her first try is 1/6. RATIONALE The probability of choosing the hard-boiled egg is 1/6. If she cracks an egg and it is not the hard-boiled egg, then it 21/27
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lh‘iil‘il‘miii:il UNIT 3 — MILESTONE 3 o e’ e’ 1| U %’ i |} — Independent vs. Dependent Events Report an issue with this question ° @ Using the Venn Diagram below, what is the conditional probability of event A occurring, assuming that event B has already occurred [P(AIB)]? A B 21/27
o lh‘iil‘il‘miii:il UNIT 3 — MILESTONE 3 RAIIONALE To get the probability of A given B has occurred, we can use the following conditional formula: P(Aand B) 0.1 PA|B)=—p@y— = gas =022 The probability of A and B is the intersection, or overlap, of the Venn diagram, which is O.1. The probability of B is all of Circle B, or 0.1+ 0.35 =0.45. CONCEPT — Conditional Probability Report an issue with this question 0 @ Two sets A and B are shown in the Venn diagram below. 21/27
o lh‘iil‘il‘miii:il UNIT 3 — MILESTONE 3 21/27 Which statement is TRUE? Set A has 8 elements. Sets A and B have 3 common Q 0 elements. There are a total of 2 elements O shown in the Venn diagram. Set B has 7 elements. O RATIONALE The intersection, or middle section, would show the common elements, which is 3. The number of elements of Set A is everything in Circle A, or 8+3 = 11 elements, not 8 elements.
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lh‘iil‘il‘miii:il UNIT 3 — MILESTONE 3 o CICITICIILS, T10L £ CICelTICTILS. CONCEPT — Venn Diagrams Report an issue with this question 1 @ Ryan is playing a multiplication game with a pile of 26 cards, each with a number on them. Each turn, he flips over two of the cards, and has to multiply the numbers. How many possible outcomes are there on Ryan's first turn flipping two cards? O 26 O 52 © O 650 O 676 RATIONALE We can use the general counting principle and note that for each step, we simply multiply all the possibilities at each step to get the total number of outcomes. If we assume that the numbers are 1- 26, then the overall number of outcomes is: 21/27
o lh‘iil‘il‘miii:il UNIT 3 — MILESTONE 3 e ol &1 1NN LI I\&] WA AITNA Wil IS I\ UII] LAl A T 1TATITIN ] 1 il \J\.fllvll e uf o However, the second card chosen would only have 25 possible outcomes since the first card has already been drawn. CONCEPT — Fundamental Counting Principle Report an issue with this question . A Y Y 3 X ) H X X X H X X ) : : : & & 6 o0 4% @ v v v? L A M L A M e L A M e L K M L M M ! 3 M A e RS oan| loa| o]l laa (oaa| loasan| aas [ 9 ; SUEE I * * . - " e - o’ A A Q : v Y: e ww: wv v ww: ev: wew . g : . 9 K ‘v L X ‘v e Y 3v'v L X ‘-"v'v 3 v . : v v ve @ v - ¥ v » ve ve veY WM R 9 b G0 P P 4 'Y 'Y, W'Y a6 | aa: 'Y 4% . 8 : i R & | § & oo | ool Loo]l Boe]l ool Boeol ool 2, V0.1 [ + + + ‘4 P P e +*+ x A : + + ¢ * ¢ E AV4 Vv ¢ 4 T RSIERAS I RS ¢4 + 4 +tes ’ g : Eric is randomly drawing cards from a deck of 52. He first draws a red card, places it back in the deck, shuffles the deck, and then draws another card. What is the probability of drawing a red card, placing it back in the deck, and drawing another red card? Answer choices are in the form of a percentage, rounded to the nearest whole number. 04% 21/27
o lh‘iil‘il‘miii:il UNIT 3 — MILESTONE 3 O L L /0 25% @ © RATIONALE Since Eric puts the card back and re-shuffles, the two events (first draw and second draw) are independent of each other. To find the probability of red on the first draw and second draw, we can use the following formula: Flred first and red second) = Plred)*P(red) = 5*5 =20 259%. 26 1 Note that the probability of drawing a red card is o) or 5 for each event. CONCEPT — "And" Probability for Independent Events Report an issue with this question 3 @ The average number of babies born at a private hospital's maternity wing is 6 per hour. What is the probability that three babies are born during a particular 1-hour period in this maternity wing? 21/27
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lh‘iil‘il‘miii:il UNIT 3 — MILESTONE 3 o RATIONALE Since we are finding the probability of a given number of events happening in a fixed interval when the events occur independently and the average rate of occurrence is known, we can use the following Poisson distribution formula: L kl P(X = k) = The variable k is the given number of occurrences, which in this case, is 3 babies. The variable A is the average rate of event occurrences, which in this case, is 6 babies. 3_—h , sflx=3%=6§| =E€?§=Dfl9 CONCEPT — Poisson Distribution Report an issue with this question 14 @ A magician asks an audience member to pick any number from 6 to 15. 21/27
o lh‘iil‘il‘miii:il UNIT 3 — MILESTONE 3 O 9 s O 15 6 O 15 s ® O 1 RATIONALE If we suppose that the card chosen by the magician is fixed, then there are 10 possible values, {6, 7, 8, 9, 10, 11, 12, 13, 14, or 15}, that are all equally likely. So, the probability that a specific value is chosenis: specific card _ 1 total number of cards 10 CONCEPT — Theoretical Probability/A Priori Method Report an issue with this question 5 @ 21/27
o lh‘iil‘il‘miii:il UNIT 3 — MILESTONE 3 21/27 - b - - - ~h o = - ¥ w - @ > o - o - w v e E ) ‘v w veY ve ve -v'v L X ‘v"v'v‘ i' : v v ve @ v i ¥ ] ¥ ve ve v‘v W R 9 <& NG Y Y 'Y 'Y Y W'Y a6 Y 'Y 4% . \ : A (2 , (3 . s ™™ (6 2 . ) T (8 2 9. . 1 fo N () (@ Kk N PO 4+ 4+ 4+ 4+ 4 4 ¢+ 4 + b 4 + b ¢ 'S R 2 ‘ Py ’*Q . 4.4 +* . ‘r-‘vr 1‘f sr-’f.r‘ + + + + 4 43 MM R % O ¥ + T + 4+ + 4 + 4 + 4 ¢4 + 4 +te; ¢ + Asmita went to a blackjack table at the casino. At the table, the dealer has just shuffled a standard deck of 52 cards. Asmita has had good luck at blackjack in the past, and she actually got three blackjacks with Aces in a row the last time she played. Because of this lucky run, Asmita thinks that Ace is the luckiest card. The dealer deals the first card to her. In a split second, she can see that it is a non-face card, but she is unsure if it is an Ace. What is the probability of the card being an Ace, given that it is a non-face card? Answer choices are in a percentage format, rounded to the nearest whole number. O 10% 77% 69% 8% RATIONALE
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lh‘iil‘il‘miii:il UNIT 3 — MILESTONE 3 o Note, that in a standard deck of 52 cards, there are 12 face cards, so 40 non-face cards. Of those non-face cards, there are only 4 Aces. CONCEPT — Conditional Probability Report an issue with this question 6 @ Dan is playing a game where he selects a card from a deck of four cards, labeled 1, 2, 3, or 4. The possible cards and probabilities are shown in the probability distribution below. P(x) 0.4 O W 1 Probability o N | 01 0 1 2 3 4 Card Value What is the expected value for the card that Dan selects? 2.5 21/27
o lh‘iil‘il‘miii:il UNIT 3 — MILESTONE 3 RATIONALE The expected value, also called the mean of a probability distribution, is found by adding the products of each individual outcome and its probability. We can use the following formula to calculate the expected value, E(X): > xip;=1%0.2+2*0.25+3*0.4+4*0.15=0.2+0.5+1.2+0.6 = 2.5 ] CONCEPT — Expected Value Report an issue with this question 17 @ Mark looked at the statistics for his favorite baseball player, Jose Bautista. Mark looked at seasons when Bautista played 100 or more games and found that Bautista's probability of hitting a home run in a game is 0.173. If Mark uses the normal approximation of the binomial distribution, what will be the variance of the number of home runs Bautista is projected to hit in 100 games? Answer choices are rounded to the tenths place. 0.8 21/27
lh‘iil‘il‘miii:il UNIT 3 — MILESTONE 3 oD O 1/7.9 14.3 @ © RATIONALE In this situation, we know: n = sample size =100 P = success probability = 0.173 We can also say that g, or the complement of p, equals: q=1-p=1-0.173=0.827 The variance is equivalent to n*p*q: Variance =ne*peg="100*0.173*0.827/=14.3 CONCEPT — Normal Distribution Approximation of the Binomial Distribution Report an issue with this question 13 @ La'Vonn rolled a die 100 times. His results are below. Number Times Rolled 1 18 2 20 3 15 21/27
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o lh‘iil‘il‘miii:il UNIT 3 — MILESTONE 3 21/27 What is the relative frequency for La'vonn rolling a 37 Answer choices are rounded to the hundredths place. © O 0.15 O 0.01 O 0.07 O 0.38 RATIONALE The relative frequency of a 3 is: number of 3's _ 15 _ 15 015 total number of rolls 18+20+15+17+14+16 100 ' CONCEPT — Relative Frequency Probability/Empirical Method Report an issue with this question 19 @
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o lh‘iil‘il‘miii:il UNIT 3 — MILESTONE 3 - b - - - ~h o = - ¥ : v Y YW, W W v ww: ww: ww; ' 3 $ . v e ) ‘v w vy ‘e v e L X ‘v"v'v‘ i' . : v v v ve @ v - ¥ ] ¥ ve ve ve W R 9 <& NG s S O |68 | aa 'Y, A b | &a: 'Y 4% ¢ 8 : 5 R & | [¢ @ oo | oo oo (oo Boe)] Boe)] oo P, ), ) [ . . so| [t |0 b | | % ! & : ¢ + 4+ e A \V4 Vv ¢ + T R B RARS: ¢4 + 4 +Tes ¢ 3 : Zhi and her friends moved on to the card tables at the casino. Zhi wanted to figure out the probability of drawing a King of clubs or an Ace of clubs. Choose the correct probability of drawing a King of clubs or an Ace of clubs. Answer choices are in the form of a percentage, rounded to the nearest whole number. O 8% © 0 O 6% © 0% RATIONALE Since the two events, drawing a King of Clubs and drawing an Ace of Clubs, are non-overlapping, we can use the following formula: 1 2 P(King of clubs or Ace of clubs) = P(King of clubs) + P(Ace of clubs) = % e 0.038 or 4%
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o lh‘iil‘il‘miii:il UNIT 3 — MILESTONE 3 21/27 20 @ The gender and age of Acme Painting Company's employees are shown below. Age Gender 23 Female 23 Male 24 Female 26 Female 27 Male 28 Male 30 Male 31 Female 33 Male 33 Female 33 Female 34 Male 36 Male 37 Male 38 Female 40 Female 42 Male 44 Female
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oD lh‘iil‘il‘miii:il UNIT 3 — MILESTONE 3 21/27 © O 1118 O 1118 © O 1/2 O 1/3 RATIONALE Since it is possible for an employee to be a male and a person in their 40s, these two events are overlapping. We can use the following formula: P(Male or Forties) = P(Male) + P(Forties) — P(Male and Forties) = % T8 18 T8 Of the 18 employees, there are 9 females and 9 males, so g F(Male) = TR There are a total of 3 people in their 40s, so 3 F(Forties) = g Of the people in their 40s, only one is male 1 so P(Male and Forties) = X CONCEPT — "Either/Or" Probability for Overlapping Events Report an issue with this question
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lh‘iil‘il‘miii:il UNIT 3 — MILESTONE 3 o Test results indicate that a patient does not have cancer when, in fact, he does. Test results indicate that a patient has cancer when, in fact, he does not. Test results confirm that a patient O has cancer. Test results confirm that a patient O does not have cancer. RATIONALE Since the test results indicate positively that the patient has cancer, when in fact cancer is not present, this is a false positive. CONCEPT — False Positives/False Negatives Report an issue with this question 22 @ A credit card company surveys 125 of its customers to ask about satisfaction with customer service. The results of the survey, 21/27
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oD lh‘iil‘il‘miii:il UNIT 3 — MILESTONE 3 o Satisfied 21 13 Neutral 13 16 Dissatisfied 9 14 Extremely Dissatisfied 2 5 If a survey is selected at random, what is the probability that the person is a female with neutral feelings about customer service? Answer choices are rounded to the hundredths place. 0.13 0.19 0.5 0.81 0.29 O RATIONALE If we want the probability of selecting a survey that is from a female who marked "neutral feelings," we just need to look at the box that is associated with both categories, or 16. To calculate the probability, we can use the following formula: 21/27
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lh‘iil‘il‘miii:il UNIT 3 — MILESTONE 3 o [ 4 L4 [ 4 Report an issue with this question 23 @ There is a 30% chance of rain tomorrow. What are the odds in favor of it raining? O 7:3 © O 3:7 O 10:3 O 3:10 RATIONALE A Recall that we if we have a probability "E"’ we can formally express odds as "A: B — A" So if the probability is 30%, or 30/100 in fraction form, then we can express odds as: odds =30 100-30=30: 7jO=3: 7 21/27
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o lh‘iil‘il‘miii:il UNIT 3 — MILESTONE 3 21/27 24 © Using this Venn diagram, what is the probability that event A or event B occurs? 077 © O 0.68 O 0.41 © O
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oD lh‘iil‘il‘miii:il UNIT 3 — MILESTONE 3 P(A or B)=FP(A)+P(B)-P(A and B)=0.65+0.53-0.41=0.77. The probability of event A is ALL of circle A, or 0.24 + 0.41 = 0.65. The probability of event B is ALL of circle B, or 012 + 0.41= 0.53. The probability of event A and B is the intersection of the Venn diagram, or 0.41. We can also simply add up all the parts =0.24 + 0.41+ 012 = 0.77. CONCEPT — "Either/Or" Probability for Overlapping Events Report an issue with this question 25 @ A survey asked 1,000 people which magazine they preferred, given three choices. The table below breaks the votes down by magazine and age group. Age Below 40 Age 40 and Above The National Journal 104 200 Newsday 120 230 The Month 240 106 If you randomly select a person under the age of 40 from the group, what is the probability that they voted for "The Month?" Answer choices are rounded to the hundredths place. 21/27
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o lh‘iil‘il‘miii:il UNIT 3 — MILESTONE 3 21/27 © O o 048 o 020 RATIONALE The probability of a person voting for "The Month" given he or she is younger than 40 is a conditional probability. We can use the following formula: " " 240 P(" The Month" and Younger than 40) _ /1000 _ 240 — 0517 or 52% P(" The Month" | than 40) = - ("The Month" | Younger than 40) P(Younger than 40) 464/100p 464 Remember, to find the total number of Younger than 40, we need to add all values in this column: 104 + 120 + 240 = 464. CONCEPT — Conditional Probability and Contingency Tables Report an issue with this question 26 @ Tracie spins the four-colored spinner shown below. She records the total number of times the spinner lands on the color red and constructs a graph to visualize her results.
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o lh‘iil‘il‘miii:il UNIT 3 — MILESTONE 3 21/27 Which of the following statements is TRUE? If Tracie spins the spinner 4 O times, it will land on red at least once. If Tracie spins the spinner 1,000 times, the relative frequency of it O landing on red will remain constant. If Tracie spins the spinner 1,000 times, it would land on red close to 250 times.
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o lh‘iil‘il‘miii:il UNIT 3 — MILESTONE 3 RATIONALE If we make the assumption that the area of the colors represents the true proportion, then each color is equally weighted. Since there are four colors we would expect them to come up roughly 1/4 of the time. So on 1000 rolls the expected value = n*p = 1000*0.25 = 250. CONCEPT — Law of Large Numbers/Law of Averages Report an issue with this question 27 @ Sadie is selecting two pieces of paper at random from the stack of colored paper in her closet. The stack contains several sheets of each of the standard colors: red, orange, yellow, green, blue, and violet. All of the following are possible outcomes for Sadie's selection, EXCEPT: Red, red O Green, violet O 21/27
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lh‘iil‘il‘miii:il UNIT 3 — MILESTONE 3 oD 21/27 O Vidlilyc, yc<livvy RATIONALE Since black is not part of the original set, it cannot be chosen into the sample. CONCEPT — Outcomes and Events Report an issue with this question 69 About Contact Us Privacy Policy Cookie Policy Terms of Use Your Privacy Choices © 2023 SOPHIA Learning, LLC. SOPHIA is a registered trademark of SOPHIA Learning, LLC.
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