QUESTION IMAGE
Question
a student entering a doctoral program in educational psychology is required to select three courses from the list of courses provided as part of his or her program.
(a) list all possible three - course selections.
(b) comment on the likelihood that epr 669, epr 610, and epr 645 will be selected.
click the icon to view the course list.
(a) select all the possible three - course selections below.
a. 669, 645, 655
b. 669, 660, 610
c. 655, 653, 655
d. 610, 645, 655
e. 669, 610, 655
f. 610, 606, 645
g. 606, 645, 655
h. 669, 610, 606
i. 606, 655, 606
j. 669, 606, 655
k. 669, 606, 645
l. 669, 669
m. 610, 606, 655
n. 669, 610, 645
course list
epr 669, interpretive methods in educational research
epr 610, theory of measurement
epr 606, methods of multivariate analysis
epr 645, educational research planning and interpretation
epr 655, research in child development
Part (a)
First, we identify the course list: EPR 669, EPR 610, EPR 606, EPR 645, EPR 655. We need to find all combinations of 3 courses from these 5.
The formula for combinations of \( n \) items taken \( r \) at a time is \( C(n, r)=\frac{n!}{r!(n - r)!} \), here \( n = 5 \), \( r=3 \), so \( C(5,3)=\frac{5!}{3!2!}=\frac{5\times4}{2\times 1}=10 \) combinations.
Let's list all possible 3 - course combinations:
- 669, 610, 606 (Option H)
- 669, 610, 645 (Option N)
- 669, 610, 655 (Option E)
- 669, 606, 645 (Option K)
- 669, 606, 655 (Option J)
- 669, 645, 655 (Option A)
- 610, 606, 645 (Option F)
- 610, 606, 655 (Option I)
- 610, 645, 655 (Option D)
- 606, 645, 655 (Option G)
Now we check the options:
- Option A: 669, 645, 655 - Valid (one of the combinations)
- Option B: 669, 660, 610 - 660 is not in the course list - Invalid
- Option C: 655, 653, 655 - 653 is not in the course list - Invalid
- Option D: 610, 645, 655 - Valid (one of the combinations)
- Option E: 669, 610, 655 - Valid (one of the combinations)
- Option F: 610, 606, 645 - Valid (one of the combinations)
- Option G: 606, 645, 655 - Valid (one of the combinations)
- Option H: 669, 610, 606 - Valid (one of the combinations)
- Option I: 606, 655, 606 - Repeated course (606 appears twice) - Invalid (Wait, the option I is 606, 655, 606? No, looking back, option I is 606, 655, 606? Wait, no, the option I is 606, 655, 606? Wait, the user's option I is 606, 655, 606? No, maybe a typo, but according to the course list, the correct combination 610, 606, 655 is not listed as I. Wait, no, the course list has 5 courses: 669, 610, 606, 645, 655. So 606, 655, 606 is invalid (repeated). Wait, maybe the option I is 606, 655, 610? No, the user's option I is 606, 655, 606. So Option I is invalid.
- Option J: 669, 606, 655 - Valid (one of the combinations)
- Option K: 669, 606, 645 - Valid (one of the combinations)
- Option L: 669, 669 - Only two courses - Invalid
- Option M: 610, 606, 655 - Wait, 610, 606, 655 is a valid combination, but is it in the options? Option M is 610, 606, 655 - Yes, it is a valid combination. Wait, earlier when we calculated \( C(5,3)=10 \), let's re - list:
The 10 combinations are:
- {669, 610, 606} (H)
- {669, 610, 645} (N)
- {669, 610, 655} (E)
- {669, 606, 645} (K)
- {669, 606, 655} (J)
- {669, 645, 655} (A)
- {610, 606, 645} (F)
- {610, 606, 655} (M)
- {610, 645, 655} (D)
- {606, 645, 655} (G)
So the valid options are:
A. 669, 645, 655
D. 610, 645, 655
E. 669, 610, 655
F. 610, 606, 645
G. 606, 645, 655
H. 669, 610, 606
J. 669, 606, 655
K. 669, 606, 645
M. 610, 606, 655
N. 669, 610, 645
Part (b)
The total number of possible 3 - course selections is \( C(5,3) = 10 \) (as calculated above). The number of ways to select EPR 669, EPR 610, and EPR 645 is 1 (since there is only one combination that includes these three courses).
The probability of selecting these three courses is \( \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}=\frac{1}{10} = 0.1 \). Since the probability is relatively low (0.1), the likelihood that EPR 669, EPR 610, and EPR 645 will be selected is low.
Final Answers
(a)
The valid three - course selections are:
A. 669, 645, 655
D. 610, 645, 655
E. 669, 610, 655
F. 610, 606, 645
G. 606, 645, 655
H. 669, 610, 606
J. 669, 606, 655
K. 669, 606, 645
M. 610, 606, 655
N. 669, 610, 645
(b)
The probability that EPR 669, EPR 610, and EPR 645 are selected is \( \frac{1}{10} \) or 0.1, so the likelihood is low (since there are 10 possible selections and only 1 favo…
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Part (a)
First, we identify the course list: EPR 669, EPR 610, EPR 606, EPR 645, EPR 655. We need to find all combinations of 3 courses from these 5.
The formula for combinations of \( n \) items taken \( r \) at a time is \( C(n, r)=\frac{n!}{r!(n - r)!} \), here \( n = 5 \), \( r=3 \), so \( C(5,3)=\frac{5!}{3!2!}=\frac{5\times4}{2\times 1}=10 \) combinations.
Let's list all possible 3 - course combinations:
- 669, 610, 606 (Option H)
- 669, 610, 645 (Option N)
- 669, 610, 655 (Option E)
- 669, 606, 645 (Option K)
- 669, 606, 655 (Option J)
- 669, 645, 655 (Option A)
- 610, 606, 645 (Option F)
- 610, 606, 655 (Option I)
- 610, 645, 655 (Option D)
- 606, 645, 655 (Option G)
Now we check the options:
- Option A: 669, 645, 655 - Valid (one of the combinations)
- Option B: 669, 660, 610 - 660 is not in the course list - Invalid
- Option C: 655, 653, 655 - 653 is not in the course list - Invalid
- Option D: 610, 645, 655 - Valid (one of the combinations)
- Option E: 669, 610, 655 - Valid (one of the combinations)
- Option F: 610, 606, 645 - Valid (one of the combinations)
- Option G: 606, 645, 655 - Valid (one of the combinations)
- Option H: 669, 610, 606 - Valid (one of the combinations)
- Option I: 606, 655, 606 - Repeated course (606 appears twice) - Invalid (Wait, the option I is 606, 655, 606? No, looking back, option I is 606, 655, 606? Wait, no, the option I is 606, 655, 606? Wait, the user's option I is 606, 655, 606? No, maybe a typo, but according to the course list, the correct combination 610, 606, 655 is not listed as I. Wait, no, the course list has 5 courses: 669, 610, 606, 645, 655. So 606, 655, 606 is invalid (repeated). Wait, maybe the option I is 606, 655, 610? No, the user's option I is 606, 655, 606. So Option I is invalid.
- Option J: 669, 606, 655 - Valid (one of the combinations)
- Option K: 669, 606, 645 - Valid (one of the combinations)
- Option L: 669, 669 - Only two courses - Invalid
- Option M: 610, 606, 655 - Wait, 610, 606, 655 is a valid combination, but is it in the options? Option M is 610, 606, 655 - Yes, it is a valid combination. Wait, earlier when we calculated \( C(5,3)=10 \), let's re - list:
The 10 combinations are:
- {669, 610, 606} (H)
- {669, 610, 645} (N)
- {669, 610, 655} (E)
- {669, 606, 645} (K)
- {669, 606, 655} (J)
- {669, 645, 655} (A)
- {610, 606, 645} (F)
- {610, 606, 655} (M)
- {610, 645, 655} (D)
- {606, 645, 655} (G)
So the valid options are:
A. 669, 645, 655
D. 610, 645, 655
E. 669, 610, 655
F. 610, 606, 645
G. 606, 645, 655
H. 669, 610, 606
J. 669, 606, 655
K. 669, 606, 645
M. 610, 606, 655
N. 669, 610, 645
Part (b)
The total number of possible 3 - course selections is \( C(5,3) = 10 \) (as calculated above). The number of ways to select EPR 669, EPR 610, and EPR 645 is 1 (since there is only one combination that includes these three courses).
The probability of selecting these three courses is \( \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}=\frac{1}{10} = 0.1 \). Since the probability is relatively low (0.1), the likelihood that EPR 669, EPR 610, and EPR 645 will be selected is low.
Final Answers
(a)
The valid three - course selections are:
A. 669, 645, 655
D. 610, 645, 655
E. 669, 610, 655
F. 610, 606, 645
G. 606, 645, 655
H. 669, 610, 606
J. 669, 606, 655
K. 669, 606, 645
M. 610, 606, 655
N. 669, 610, 645
(b)
The probability that EPR 669, EPR 610, and EPR 645 are selected is \( \frac{1}{10} \) or 0.1, so the likelihood is low (since there are 10 possible selections and only 1 favorable selection).