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ConcordiaUniversity
Department of Economics
ECON 221 – Sections A, B, BB, C
Instructors: G. Fisher, N. Islam, N. Debiparshad
Fall 2008 - ASSIGNMENT 1 – THEORY - ANSWERS
- Second-year economics students wishing to take Econ 222 must have first completed either a QM course at the College, or Econ 221 at the University. In the beginning of a specific academic year, 35% of second-year economics students had completed the QM course, while 55% of the second-year economics students had completed the Econ 221. In that specific academic year, Econ 222 was taken by 80% of all second-year economics students. (10 points)
a)If taking a QM course and Econ 221 are mutually exclusive, what is the probability that a randomly chosen student entering his/her second-year economics studies has completed at least one of the two courses? (3 points)
P (QM) = 0.35P (Econ 221) = 0.55P (Econ 222) = 0.8
Events QM and Econ 221 are mutually exclusive. Therefore, P (QM Econ 221) = 0
AndP (QM Econ221) = P (QM) + P (Econ 221) – P (QM Econ 221)= 0.35 + 0.55 – 0.0 = 0.9, or 90%
b)If having taken a QM course and taking Econ 222 during the specific academic year are statistically independent, what is the probability that a randomly chosen student will have completed at least one of these two courses during his/her second-year economics studies? (3 points)
Events QM and Econ 222 are independent. Therefore, P (QM Econ 222) = P(QM) * P(Econ 222) = 0.35 * 0.8 = 0.28
P (QM Econ222) = P (QM) + P (Econ 222) – P (QM Econ 222) = 0.35 + 0.8 – 0.28 = 0.87 or 87%
Econ 222 / QM / No QM /Total
Yes Econ 222No Econ 222 / 0.28
0.07 / 0.52
0.13 / 0.80
0.20
Total / 0.35 / 0.65 / 1.00
c)Of the students who had taken Econ 221, 70% also took Econ 222 in the specific academic year. What is the probability that a randomly chosen student will have completed at least one of these two courses during his/her second-year economics studies? (4 points)
P (Econ 222 | 221) = 0.70thenP (Econ 221 Econ 222) = P (Econ 221) * P (Econ 222 | 221) = 0.55 * 0.70 = 0.385
AndP (Econ221 Econ222) = P (Econ221) + P (Econ222) – P (Econ221 Econ222) = 0.55 + 0.8 – 0.385 = 0.965 or 96.5%
Econ 222 / Econ 221 / No Econ 221 /Total
Yes Econ 222No Econ 222 / 0.385
0.165 / 0.415
0.035 / 0.80
0.20
Total / 0.550 / 0.450 / 1.00
- A recent survey examined the working arrangement of married households. It was found that in 30% of all households the man does not work, while the corresponding proportion of households where the woman does not work was twice as much. It was also found that in 60% of the households where the man does not work the woman does not work either. (20 points)
a)Construct a table of joint probabilities. (4 points)
Married Households / Woman does not work(W NW) / Woman works
(W W) /
Total
Man Does not work (M NW)Man Works (M W) / 0.18
0.42 / 0.12
0.28 / 0.30
0.70
Total / 0.60 / 0.40 / 1.00
b)What is the probability that in a randomly chosen household both man and woman work? (3 points)
P (MW WW) = P (MW) * P (WW │MW) = 0.70 * (0.28/ 0.70) = 0.28 = 28%
c)What is the probability that in two randomly chosen households both man and woman work? (4 points)
The two households are independent of each other.
For each one the probability of both man and woman worki is 0.28
Therefore, the probability that two independent households both work is
P (1st Household 2nd household) = P (1sthousehold) * P (2nd household) = 0.28 * 0.28 = 0.0784 = 7.84%
d)What is the probability that a randomly chosen household has either person working (or both)? (3 points)
P (MW WW) = P (MW) + P (WW) – P (MW WW) = 0.70 + 0.4 – 0.28 = 0.82 = 82%
e)What is the probability that, in a household where the man works, the woman does not? (3 points)
P (WNW│MW) = P (WNW MW) / P (MW) = 0.42 / 0.7 = 0.6 = 60%
f)Are working arrangements of the two genders in married households statistically independent? (3 points)
For independence we need thatP (MW WW) = P (MW) * P (WW)
or0.28 = 0.70 * 0.40
Therefore, the two variables are statisticallyindependent.
- A market research group specializes in providing assessments of the prospects of sites for new clothing stores in shopping centres. The group assesses prospects as “good”, “fair”, or “poor”. From the records of this group it was found that for all stores that turned out to be successful, the assessment was “good” for 70%, “fair” for 20%, and “poor” for 10%. For all stores that turned out to be unsuccessful, the assessment was “good” for 20%, “fair” for 30%, and “poor” for 50%. It is known that 60% of new clothing stores are successful and 40% are unsuccessful. (20 points)
a)Construct a table diagram of joint probabilities. (4 points)
Stores /Assessment
Good(G) / Fair
(F) / Poor
(P) / Total
Successful (S)
Unsuccessful (UN) / 0.42
0.08 / 0.12
0.12 / 0.06
0.20 / 0.60
0.40
Total / 0.50 / 0.24 / 0.26 / 1.00
b)What is the probability that a randomly chosen store will be assessed as having “good prospects”? (3 points)
P (G) = P (G S) + P (G UN) = 0.42 + 0.08 = 0.50 = 50%
c)If the prospects for a store are assessed as “good”, what is the probability that this store will be successful? (3 points)
P (S│G) = P (S G) / P (G) = 0.42 / 0.50 = 0.84 = 84%
d)Are the events “Prospects assessed as poor” and “Store is not successful” mutually exclusive? (3 points)
For mutually exclusive events we need thatP (P UN) = 0
In our caseP (P UN) = 0.20 0. Therefore, the two events are not mutually exclusive
e)Are the events “Prospects assessed as good” and “Store is successful” collectively exhaustive? (3 points)
For two events to be collectively exhaustive, their union should produce a probability equal to 1.
In our case P (G S) = P (G) + P (S) – P (G S) = 0.5 + 0.6 – 0.42 = 0.68 < 1.0
Therefore, the two events are not collectively exhaustive.
f)Suppose that five stores are chosen at random, what is the probability that at least one will be successful? (4 points)
P (UN) = 0.4and P (S) = 0.6
To have “at least one successful store” within 5 stores means one or more stores (up to all 5 of them) being successful. Thus, the probability of having at least one successful store is equal to the sum of probabilities of having one out of 5 stores being successful (any one), or two out of 5 stores being successful (any two), and so on up to all 5 stores being successful.
Therefore, the only scenario that is left out is that of having all five stores being unsuccessful.
Then, the probability of all unsuccessful stores is
P (UN UN UN UN UN) = [P (UN)]5 = (0.4)5 = 0.01024
And the probability of one or more successes is 1 – 0.01024 = 0.98976 or 98.976%