Answer:
- Question 9: They are not independent.
Explanation:
<u />
<u>1. Build a two-way frequency table filling only the initial information:</u>
Take AMS1303 No take AMS1303 Total
Take FIN1001 25 40
No take FIN1001
Total 50 100
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<u>2. Complete the table calculating the missing values by differences:</u>
- Take AMS1303 only: 50 - 25 = 25
- No take AMS1303 and take FIN1001: 40 - 25 = 15
- No take AMS1303: 100 - 50 = 50
- No take FIN1001: 100 - 40 = 60
- No take AMS and no take FIN: 60 - 25 = 35
Take AMS1303 No take AMS1303 Total
Take FIN1001 25 15 40 (F)
No take FIN1001 25 35 60
Total 50 50 100 (S)
(M)
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<u>3. Question 6: Probability that the chosen student takes only AMS1303</u>
- P(M) - P(M∩F) = 50/100 - 25/100 = 25/100 = 1/4
<u>4. Question 7: Probability that the chosen student takes only FIN1001</u>
- P(F) - P(M∩F) = 40/100 - 25/100 = 15/100 = 3/20
<u>5. Question 8: Probability that the chosen student takes AMS1303 or FIN1001</u>
- P(MUF) = P(M) + P(F) - P(M∩F) = 50/100 + 40/100 - 25/100 = 65/100 = 13/20
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<u>6. Question 9: Is the event of selecting a student taking AMS1303 independent of the event of selecting a student taking FIN1001?</u>
We will use contidional probability. Two events are independent only if the probability of one event is equal to the probability is the same as the probability of the event given that the other event occurs.
P(M|F) = P(M∩F) / P(F) = (25/100) / (40/100) = 25/40 = 5/8
But P(M) = 50/100 = 1/2
Then, P(M|F) ≠ P(M) , therefore they are not independent
Also, you might determine:
P(F|M) = P(M∩F) / P(M) = (25/100) / (50/100) = 25/50 = 1/2
P(F) = 40/100 = 2/5
Again, of course, P(F|M) ≠ P(F), meaning that they are not independent.
