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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
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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