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A certain company plans to sell Product X for p dollars per
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Updated on: 27 May 2013, 05:45
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29% (02:19) correct 71% (02:38) wrong based on 258 sessions
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A certain company plans to sell Product X for p dollars per unit, where p is randomly chosen from all possible positive values not greater than 100. The monthly manufacturing cost for Product X (in thousands of dollars) is 12  p, and the projected monthly revenue from Product X (in thousands of dollars) is p(6  p). If the projected revenue is realized, what is the probability that the company will NOT see a profit on sales of Product X in the first month of sales? A. 0 B. 1/100 C. 1/25 D. 99/100 E. 1 My ANS : E1, Explanation – Profit = Revenue cost = p(6p)(12p) = (p^27p+12) = (p3)(p4) There will be no profit for 1 =< p =< 100 So probability = 1. Hence E
But the ans in given in the solution list is 99/100
Please explain which one is correct .
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Originally posted by verycoolguy33 on 23 Feb 2012, 21:24.
Last edited by Bunuel on 27 May 2013, 05:45, edited 2 times in total.
Edited the question and added the OA




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Re: A certain company plans to sell Product X for p dollars per
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24 Feb 2012, 02:09
A certain company plans to sell Product X for p dollars per unit, where p is randomly chosen from all possible positive values not greater than 100. The monthly manufacturing cost for Product X (in thousands of dollars) is 12  p, and the projected monthly revenue from Product X (in thousands of dollars) is p(6  p). If the projected revenue is realized, what is the probability that the company will NOT see a profit on sales of Product X in the first month of sales?A. 0 B. 1/100 C. 1/25 D. 99/100 E. 1 Profit=RevenueCost=p(6  p)(12p). Let's see for which range of p company will see a profit on sales, so for which range of p the following holds true: \(p(6  p)(12p)>0\) > \(6pp^212+p>0\) > \(p^27p+12<0\) > \((p4)(p3)<0\) > \(3<p<4\). So the probability that company will see a profit on sales is \(\frac{1}{100}\) (1 unit range of p out of 100). The probability that the company will NOT see a profit on sales is \(1\frac{1}{100}=\frac{99}{100}\). Answer: D. Hope it's clear.
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Re: A certain company plans to sell Product X for p dollars per
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25 Feb 2012, 07:11
Thanks Bubuel , I understand your explanation . But not sure whether we can suddenly shift our question from finding probablity of values for P to probablity of range for P . I understand there has a range within which there has a profit , but if we think about possible values for P within that range there has infinite values and so it is TURE for the total possible values from 1 to 100  in line to this . So isnt it out of the question to find out the probablity of the range in which the values of P satisfy the condition instead of the probablity for the eaxct values of P .
Thanks.



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Re: A certain company plans to sell Product X for p dollars per
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25 Feb 2012, 08:50
verycoolguy33 wrote: Thanks Bubuel , I understand your explanation . But not sure whether we can suddenly shift our question from finding probablity of values for P to probablity of range for P . I understand there has a range within which there has a profit , but if we think about possible values for P within that range there has infinite values and so it is TURE for the total possible values from 1 to 100  in line to this . So isnt it out of the question to find out the probablity of the range in which the values of P satisfy the condition instead of the probablity for the eaxct values of P .
Thanks. I'm not sure I understand your reasoning above. Anyway there was no "sudden shift". We are asked to find the probability, not exact value in dollars, and that's exactly what we did. Consider this: what is the probability that computer will generate a random number from 3 to 4 if it's restricted to generate only numbers from 0 to 100. Favorable outcome is 1 unit out of 100, so P=1/100. There is a little assumption though concerning the value of p, which should be some number in dollars but it has nothing to do with your doubt.
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Re: A certain company plans to sell Product X for p dollars per
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26 Feb 2012, 10:36
Quote: (p4)(p3)<0 > 3<p<4. So the probability that company will see a profit on sales is \frac{1}{100} (1 unit range of p out of 100). Hi buneuel, Can you throw some light on solving efficietly equations like these (p4) (p3) < 0 Though im able to pick values and justify these but in a times scenario that will not work out also we have 3<p<4 a number less than 4 and greater than 3 can only be decimals say 3.45,3.55,3.65? how do we coem to conclusion that it is (1/100) please advice.



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Re: A certain company plans to sell Product X for p dollars per
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26 Feb 2012, 11:00



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Re: A certain company plans to sell Product X for p dollars per
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23 Jun 2014, 02:31
A certain company plans to sell Product X for p dollars per unit, where p is randomly chosen from all possible positive values not greater than 100. The monthly manufacturing cost for Product X (in thousands of dollars) is 12  p, and the projected monthly revenue from Product X (in thousands of dollars) is p(6  p). If the projected revenue is realized, what is the probability that the company will NOT see a profit on sales of Product X in the first month of sales? Quote: Profit=RevenueCost=p(6  p)(12p).
Let's see for which range of p company will see a profit on sales, so for which range of p the following holds true: p(6  p)(12p)>0 > 6pp^212+p>0 > p^27p+12<0 > (p4)(p3)<0 > 3<p<4. So the probability that company will see a profit on sales is \frac{1}{100} (1 unit range of p out of 100).
The probability that the company will NOT see a profit on sales is 1\frac{1}{100}=\frac{99}{100}.
Answer: D.
Hope it's clear. Dear Bunuel, For the revenue to be realized the value of p can not be greater than 6, as negative revenue is not possible. So shouldn't the answer to the question be 1/6. Thanks !



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A certain company plans to sell Product X for p dollars per
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25 Jul 2015, 22:59
parassagi wrote: A certain company plans to sell Product X for p dollars per unit, where p is randomly chosen from all possible positive values not greater than 100. The monthly manufacturing cost for Product X (in thousands of dollars) is 12  p, and the projected monthly revenue from Product X (in thousands of dollars) is p(6  p). If the projected revenue is realized, what is the probability that the company will NOT see a profit on sales of Product X in the first month of sales? Quote: Profit=RevenueCost=p(6  p)(12p).
Let's see for which range of p company will see a profit on sales, so for which range of p the following holds true: p(6  p)(12p)>0 > 6pp^212+p>0 > p^27p+12<0 > (p4)(p3)<0 > 3<p<4. So the probability that company will see a profit on sales is \frac{1}{100} (1 unit range of p out of 100).
The probability that the company will NOT see a profit on sales is 1\frac{1}{100}=\frac{99}{100}.
Answer: D.
Hope it's clear. Hi Bunuel, I have a query on the above post. I think the sample space or possible values of p need to be from 0 to 6 only as beyond 6, 6p becomes negative and therefore revenue becomes negative, which is not possible. So the correct answer needs to be 5/6 rather than 99/100  what do you think? Please clarify and do correct me if I am wrong, thanks.



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A certain company plans to sell Product X for p dollars per
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05 Apr 2017, 01:24
Bunuel wrote: A certain company plans to sell Product X for p dollars per unit, where p is randomly chosen from all possible positive values not greater than 100. The monthly manufacturing cost for Product X (in thousands of dollars) is 12  p, and the projected monthly revenue from Product X (in thousands of dollars) is p(6  p). If the projected revenue is realized, what is the probability that the company will NOT see a profit on sales of Product X in the first month of sales? A. 0 B. 1/100 C. 1/25 D. 99/100 E. 1
Profit=RevenueCost=p(6  p)(12p).
Let's see for which range of p company will see a profit on sales, so for which range of p the following holds true: \(p(6  p)(12p)>0\) > \(6pp^212+p>0\) > \(p^27p+12<0\) > \((p4)(p3)<0\) > \(3<p<4\). So the probability that company will see a profit on sales is \(\frac{1}{100}\) (1 unit range of p out of 100).
The probability that the company will NOT see a profit on sales is \(1\frac{1}{100}=\frac{99}{100}\).
Answer: D.
Hope it's clear. Dear Bunuel, Why the probability that company will see a profit on sales is \(\frac{1}{100}\) since \(3 < p < 4\) and p is neither 3 nor 4?
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Re: A certain company plans to sell Product X for p dollars per
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05 Apr 2017, 02:00
ziyuen wrote: Bunuel wrote: A certain company plans to sell Product X for p dollars per unit, where p is randomly chosen from all possible positive values not greater than 100. The monthly manufacturing cost for Product X (in thousands of dollars) is 12  p, and the projected monthly revenue from Product X (in thousands of dollars) is p(6  p). If the projected revenue is realized, what is the probability that the company will NOT see a profit on sales of Product X in the first month of sales? A. 0 B. 1/100 C. 1/25 D. 99/100 E. 1
Profit=RevenueCost=p(6  p)(12p).
Let's see for which range of p company will see a profit on sales, so for which range of p the following holds true: \(p(6  p)(12p)>0\) > \(6pp^212+p>0\) > \(p^27p+12<0\) > \((p4)(p3)<0\) > \(3<p<4\). So the probability that company will see a profit on sales is \(\frac{1}{100}\) (1 unit range of p out of 100).
The probability that the company will NOT see a profit on sales is \(1\frac{1}{100}=\frac{99}{100}\).
Answer: D.
Hope it's clear. Dear Bunuel, Why the probability that company will see a profit on sales is \(\frac{1}{100}\) since \(3 < p < 4\) and p is neither 3 nor 4? Are you asking why is the probability 1/100 even though it's \(3 < p < 4\) and not \(3 \leq p \leq 4\)? If so, then 3 and 4 are points with no dimension, thus the probaiblity that p is in \(3 < p < 4\) is the same as the probability that p is in \(3 \leq p \leq 4\).
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Re: A certain company plans to sell Product X for p dollars per
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20 Jul 2018, 02:33
how do we find out the level of difficulty (600 level or 700 level) of this question?
thanks



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Re: A certain company plans to sell Product X for p dollars per
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