Bunuel wrote:
If the probability that there is exactly 1 defective pen in a box is 1 percent, what is the average (arithmetic mean) number of defective pens in the box of pens?
(1) The probability that all the pens in the box are not defective is 96%
(2) The probability that there is more than 1 defective pen in the box is 3%
Are You Up For the Challenge: 700 Level Questions Analyzing the question:Note the question uses the term "exactly 1", this means we might have 0, 1, 2, or more defective pens for each box. We need to find the probability for 0, 1, 2, 3 ... defective pens in order to answer the question.
Statement 1:96% there are 0 defective, 1% there is 1 defective, then the other 3% must have 2 or more defective pens. We don't know exactly how the 3% is distributed yet so this is insufficient.
Statement 2:This was the conclusion from statement 1. We can infer there are 96% boxes with 0 defective pens with this information but we still don't know the details of the 3%. Insufficient.
Combined:We have duplicate information so we can automatically choose E.
Let us assume either one of the statements told us the probability for 2 defective pens was 3%, hence the entire probability set for "more than 1" is covered by the probability for 2 defective pens. Then the average amount of defective pens per box would be calculated by: 96%* 0 + 1%*1 + 3%*2 = 7% = 0.07. So for each box, there is an average of 0.07 pens that are defective, or for every 100 boxes, there are 7 pens defective.
Ans: E
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