2015 Question 3 Intent of Question   The primary goals of this question were to assess a student’s ability to perform a probability calculation from a discrete random variable; calculate the expected value of a discrete random variable; perform a conditional probability calculation from a discrete random variable; and use probabilistic thinking to make a prediction about how an expected value will change given a condition about the random variable. For the question, scoring rubric, and sample papers: Copyright © 2015  College Board. The College Board was not involved in the production of and does not endorse this material. Used with permission.

Number of ATMs working when mall opens A shopping mall has three automated teller machines (ATMs). Because the machines receive heavy use, they sometimes stop working and need to be repaired. Let the random variable X represent the number of ATMs that are working when the mall opens on a randomly selected day. The table shows the probability distribution of X. Number of ATMs working when mall opens 1 2 3 Probability 0.15 0.21 0.40 0.24

Number of ATMs working when mall opens 1 2 3 Probability 0.15 0.21 0.40 0.24 What is the probability that at least one ATM is working when the mall opens? Solution – Part (a)   The probability that at least one ATM is working when the mall opens is: P(X ≥ 1) = 0.21 + 0.40 + 0.24 = 0.85

Parts (a), (b), (c), and (d) are scored as E - Essentially correct P - Partially correct or I - Incorrect

Scoring – Part (a)   E – if the probability is computed correctly with work shown P – if the correct answer is given, but no work is shown OR if appropriate work is shown but the answer is incorrect or missing If one of the incorrect cumulative probabilities P(X < 1), P(X ≤ 1), or P(X > 1) is stated AND computed correctly. I – Otherwise Note: The probability can be calculated as 1 – P(X = 0) = 1 – 0.15 = 0.85

Number of ATMs working when mall opens 1 2 3 Probability 0.15 0.21 0.40 0.24 (b) What is the expected value of the number of ATMs that are working when the mall opens? Solution – Part (b)   The expected value of the number of ATMs that are working when the mall opens is: E(X) = 0(0.15) + 1(0.21) + 2(0.40) + 3(0.24) = 1.73 machines

Scoring Part (b)   E – if the expected value is computed correctly with work shown P – if the correct answer is given, but no work is shown OR if appropriate work is shown but the answer is incorrect or missing I – Otherwise

Number of ATMs working when mall opens 1 2 3 Probability 0.15 0.21 0.40 0.24 (c) What is the probability that all three ATMs are working when the mall opens, given that at least one ATM is working? Solution Part(c) The probability that all three ATMs are working when the mall opens, given that at least one ATM is working is:   𝑃(𝑋=3│𝑋≥1)= 𝑃(𝑋=3 𝑎𝑛𝑑 𝑋 ≥1) 𝑃(𝑋≥1) = 𝑃(𝑋=3) 𝑃 𝑋≥1 = 0.24 0.85 ≈ 0.282

Scoring Part(c)   E – if the probability is computed correctly, with work shown that includes correct numerical values for both the numerator and denominator. P – if the response includes a numerator and denominator in calculating the conditional probability, with one (numerator or denominator) correct in numerical value and the other incorrect; OR If the correct answer is given, but no work is shown I – Otherwise

Number of ATMs working when mall opens 1 2 3 Probability 0.15 0.21 0.40 0.24 (d)Given that at least one ATM is working when the mall opens, would the expected value of the number of ATMs that are working be less than, equal to, or greater than the expected value from part (b)? Explain. Solution Part (d) Given that at least one ATM is working when the mall opens, the expected value of the number of working ATMs would be greater than the expected value calculated in part (b). By eliminating the possibility of 0 working ATMs, the probabilities for 1, 2, and 3 working ATMs all increase proportionally, so the expected value must increase.

Number of ATMs working when mall opens 1 2 3 Probability 0.15 0.21 0.40 0.24 E – if the response provides the correct answer (greater than) with a reasonable explanation based on the fact that with X = 0 eliminated, the probabilities for X = 1, X = 2, and X = 3 ALL increase; OR If the response provides the correct answer (greater than) with a the balance point of the distribution increases; If the response provides the correct answer (greater than) and the conditional expected value is computed correctly with work shown;

Number of ATMs working when mall opens 1 2 3 Probability 0.15 0.21 0.40 0.24 P – if the response provides the correct answer (greater than) with a weak explanation such as “Yes, because 0 is eliminated”; OR If the response provides the correct answer (greater than) with a reasonable but incorrect attempt to calculate the conditional expected value using a revised probability distribution with only the values X = 1, X = 2, and X = 3, and corresponding probabilities greater than or equal to those in the original probability distribution.

Number of ATMs working when mall opens 1 2 3 Probability 0.15 0.21 0.40 0.24 I – If the response does not provide the correct answer (greater than); OR If the response provides the correct answer (greater than) with an incorrect explanation or no explanation    Note: The conditional expected value is: 1 0.21 0.85 +2 0.40 0.85 +3 0.24 0.85 ≈ 1(0.247) + 2(0.471) + 3(0.282) = 2.04 machines

Each essentially correct (E) part counts as 1 point. typo Each essentially correct (E) part counts as 1 point. Each partially correct (P) part counts as ½ point.   If a response is between two scores (for example, 2½ points), use a holistic approach to decide whether to score up or down, depending on the overall strength of the response and communication.

Paper A Probability correct Work shown  E 

Paper A Expected value correct Work shown  E 

E   Paper A Probability correct Work shown with correct numerator and denominator  E 

E   Paper A – part d Correct answer (greater than) Explanation: with 0 eliminated Prob. of x=1, x=2, x=3 increase  E 

Paper A E E E E 4

Paper B Probability correct Work shown  E 

Well, this is unfortunate…and a serious notation error Paper B Well, this is unfortunate…and a serious notation error Expected value correct Work shown  P 

E   Paper B Probability correct Work shown with correct numerator and denominator  E 

E   Paper B – part d Correct answer (greater than) Explanation: with 0 eliminated Prob. of x=1, x=2, x=3 increase  E 

Paper B E P E E 3

Paper C Probability correct Work shown  E 

Paper C – part (b) Expected value correct Work shown  E 

E   Paper C Probability correct Work shown with correct numerator and denominator  E 

I Paper C – part d Correct answer (greater than) Hmmm. 2 approaches here This is incorrect This one is good but…. I Correct answer (greater than) Explanation: with 0 eliminated Prob. of x=1, x=2, x=3 increase

Paper C E E E I 3

Paper D Probability correct Work shown  P X

Paper D Expected value correct Work shown  E 

P X  Paper D Probability correct We’re looking for 0.282 Probability correct Work shown with correct numerator and denominator X P 

I X Paper D – part d Correct answer (greater than) Explanation: with 0 eliminated Prob. of x=1, x=2, x=3 increase X I

Paper D P E P I 2

Paper E Probability correct Work shown  E  in symbolic form

This is incorrect. No summation sign. Paper E This is incorrect. No summation sign. ? Expected value correct Work shown P ? Had it but then rounded to 2. This is a misconception that the expected value needs to be a whole number.

I X X Paper E Probability correct Work shown with correct numerator and denominator X I X

I X Paper E – part d Correct answer (greater than) Explanation: with 0 eliminated Prob. of x=1, x=2, x=3 increase X game over

Paper E E P I I 1.5 1 The response clearly showed a misunderstanding of expected value in part (b). Looking holistically at the paper, the overall response was considered minimal but developing.

Get with a partner and grade Papers F – J. Your Turn Get with a partner and grade Papers F – J. If you finish before the others, discuss what is working well with your students and where you could use more help. Short Rubric probability correct (0.85) c) probability correct (0.282) work shown work shown .24/.85 b) expected value (1.73) d) correct answer (greater than) work shown explanation includes probability of x=1, x=2, x=3 all increasing

Paper F E E E E 4

E I I I 1 Paper G a) The initial response is 0.88 which is incorrect. But the student amends the answer to 0.85 in the following sentence. c) Incorrect numerator and incorrect denominator d) Response suggests the expected value would be less than the one found in part (b).

Paper H E E P P 3 c) Incorrect numerator but correct denominator for a P. d) Weak explanation of why the expected value from part (b) would increase: “because the probability of zero not working is taken out…” Nothing about the other probabilities increasing so the response earns a “P”.

Paper I E E E P 3.5 4 d) Response provides a valid method for calculating the conditional expected value but makes a minor computation error to obtain 2.282 instead of 2.035. Looking holistically at the paper, the response demonstrates a thorough understanding of how to perform the required calculations for all four parts.

E E P I 2.5 2 Paper J c) Incorrect numerator but correct denominator d) Incorrectly concludes the expected value would be the same as the one found in part (b). The response showed a major misconception in part (c) by multiplying the probabilities for X=1 and X=3 to get the numerator.

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