1. 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.

2. 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

3. Number of ATMs working when mall opens1 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) = = 0.85

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

5. 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

6. Number of ATMs working when mall opens1 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

7. 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

8. Number of ATMs working when mall opens1 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 = ≈ 0.282

9. 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

10. Number of ATMs working when mall opens1 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.

11. Number of ATMs working when mall opens1 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;

12. Number of ATMs working when mall opens1 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.

13. Number of ATMs working when mall opens1 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(0.247) + 2(0.471) + 3(0.282) = 2.04 machines

14. 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.

15. Paper A
Probability correct
Work shown  E 

16. Paper A
Expected value correct
Work shown  E 

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

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

19. Paper A
E E E E 4

20. Paper B
Probability correct
Work shown  E 

21. 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 

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

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

24. Paper B
E P E E 3

25. Paper C
Probability correct
Work shown  E 

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

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

28. 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

29. Paper C
E E E I 3

30. Paper D
Probability correct
Work shown  P X

31. Paper D
Expected value correct
Work shown  E 

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

33. 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

34. Paper D
P E P I 2

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

36. 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.

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

38. 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

39. 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.

40. 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 /.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

41. Paper F
E E E E 4

42. 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).

43. 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”.

44. 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 instead of
Looking holistically at the paper, the response demonstrates a thorough understanding of how to perform the required calculations for all four parts.

45. 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.

46. Questions?