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26 May 2017, 04:41
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The probability that event \(A\) occurs is \(0.8\) and the probability that event \(B\) occurs is \(0.55\). What is the probability that event \(A\) occurs but not event \(B\)?
(1) Event \(A\) and event \(B\) are independent.
(2) The probability that neither event \(A\) nor event \(B\) occurs is \(0.09\).
(1) Event \(A\) and event \(B\) are independent.
(2) The probability that neither event \(A\) nor event \(B\) occurs is \(0.09\).
26 May 2017, 04:41
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Official Solution:
If you use the 1st step of the variable approach and modify the original condition and the question, you get the probability of Event A occurring \(= P(A)\), the probability of event B occurring \(= P(B)\), and the probability of event A or event B occurring \(= P(A \cap B)\). Then, you get \(P(A \cup B) = P(A) + P(B) - P(A \cap B)\). Since you already have \(P(A) = 0.8\) and \(P(B) = 0.55\), you just need to know \(P(A \cap B)\), and there is 1 variable. In order to match the number of variables to the number of equations, there must be 1 equation. Therefore, D is most likely to be the answer.
In the case of con 1), if event A and event B are independent, \(P(A \cap B) = P(A)P(B)\), so \(P(A \cap B) = P(A)P(B) = (0.8)(0.55) = 0.44\), and the probability that event A occurs but not event \(B=P(A \cap B^{c}) = P(A) - P(A \cap B) = 0.8 - 0.44 = 0.36\), hence it is unique and sufficient.
In the case of con 2), \(P((A \cup B)^{c}) = 1 - P(A \cup B) = 0.09\), then \(P(A \cup B) = 0.91\), and by substituting into \(P(A \cup B) = P(A) + P(B) - P(A \cap B)\), you get \(0.91 = 0.8 + 0.55 - P(A \cap B)\). Since you get \(P(A \cap B) = 1.35 - 0.91 = 0.44\), it becomes the same with con 1), hence it is sufficient. The answer is D.
Answer: D
If you use the 1st step of the variable approach and modify the original condition and the question, you get the probability of Event A occurring \(= P(A)\), the probability of event B occurring \(= P(B)\), and the probability of event A or event B occurring \(= P(A \cap B)\). Then, you get \(P(A \cup B) = P(A) + P(B) - P(A \cap B)\). Since you already have \(P(A) = 0.8\) and \(P(B) = 0.55\), you just need to know \(P(A \cap B)\), and there is 1 variable. In order to match the number of variables to the number of equations, there must be 1 equation. Therefore, D is most likely to be the answer.
In the case of con 1), if event A and event B are independent, \(P(A \cap B) = P(A)P(B)\), so \(P(A \cap B) = P(A)P(B) = (0.8)(0.55) = 0.44\), and the probability that event A occurs but not event \(B=P(A \cap B^{c}) = P(A) - P(A \cap B) = 0.8 - 0.44 = 0.36\), hence it is unique and sufficient.
In the case of con 2), \(P((A \cup B)^{c}) = 1 - P(A \cup B) = 0.09\), then \(P(A \cup B) = 0.91\), and by substituting into \(P(A \cup B) = P(A) + P(B) - P(A \cap B)\), you get \(0.91 = 0.8 + 0.55 - P(A \cap B)\). Since you get \(P(A \cap B) = 1.35 - 0.91 = 0.44\), it becomes the same with con 1), hence it is sufficient. The answer is D.
Answer: D
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