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26 May 2017, 04:41
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The probability that event \(A\) occurs is \(P(A)\) and the probability that event \(B\) occurs is \(P(B)\). If event \(A\) and event \(B\) are independent, is the probability \(P(A \cap B)\) less than \(0.6\)?
(1) \(P(A)\) is \(0.59\)
(2) The probability that event \(B\) does not occur is \(0.41\)
(1) \(P(A)\) is \(0.59\)
(2) The probability that event \(B\) does not occur is \(0.41\)
26 May 2017, 04:41
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Official Solution:
If you take the 1st step of the variable approach and modify the original condition and the question, you get the probability of \(P(A) \le 1\) and \(P(B) \le 1\).
In other words, the probability is always less than or equal to \(1\), and since event \(A\) and event \(B\) are independent, \(P(A \cap B) = P(A)P(B)\). In the original condition, there are 2 variables \(([M]P(A)\) and $P(B)[/M]
)\(, and in order to match the number of variables to the number of equations, there must be 2 equations. Therefore, C is most likely to be the answer.\\
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By solving con 1) and con 2), in case of con 1), you can use\)P(B) \le 1\(and get\)P(A \cap B) = P(A)P(B) = 0.59*P(B) \le 0.59*1 < 0.59 < 0.6\(, hence yes, it is sufficient. \\
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In the case of con 2), you get\)P(B) = 1 - P(B^{c}) = 1 - 0.41 = 0.59\(, and if you use\)P(A) \le 1\(, you get\)P(A \cap B) = P(A)P(B) = P(A)*0.59 \le 1*0.59 = 0.59 < 0.6$, hence yes, it is sufficient. Therefore con 1) = con 2). The answer is D.
Answer: D
If you take the 1st step of the variable approach and modify the original condition and the question, you get the probability of \(P(A) \le 1\) and \(P(B) \le 1\).
In other words, the probability is always less than or equal to \(1\), and since event \(A\) and event \(B\) are independent, \(P(A \cap B) = P(A)P(B)\). In the original condition, there are 2 variables \(([M]P(A)\) and $P(B)[/M]
)\(, and in order to match the number of variables to the number of equations, there must be 2 equations. Therefore, C is most likely to be the answer.\\
\\
By solving con 1) and con 2), in case of con 1), you can use\)P(B) \le 1\(and get\)P(A \cap B) = P(A)P(B) = 0.59*P(B) \le 0.59*1 < 0.59 < 0.6\(, hence yes, it is sufficient. \\
\\
In the case of con 2), you get\)P(B) = 1 - P(B^{c}) = 1 - 0.41 = 0.59\(, and if you use\)P(A) \le 1\(, you get\)P(A \cap B) = P(A)P(B) = P(A)*0.59 \le 1*0.59 = 0.59 < 0.6$, hence yes, it is sufficient. Therefore con 1) = con 2). The answer is D.
Answer: D
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