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

Question

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 :
E-1,
Explanation – 
Profit = Revenue -cost
= p(6-p)-(12-p)
= -(p^2-7p+12)
= -(p-3)(p-4)
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 .

Bunuel · Accepted Answer

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=Revenue-Cost=p(6 - p)-(12-p).

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:

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.

Bunuel · Answer

Solving inequalities: x2-4x-94661.html#p731476 ( check this one first ) inequalities-trick-91482.html data-suff-inequalities-109078.html range-for-variable-x-in-a-given-inequality-109468.html?hilit=extreme#p873535 everything-is-less-than-zero-108884.html?hilit=extreme#p868863 As for your another question: p can be decimal. Why not? The distance of a range from 3 to 4 is 1 unit, there are total of 100 units, the probability that p is in this range is favorable/total=1/100. Hope it helps.

verycoolguy33 · Answer

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.

Bunuel · Answer

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.

shankar245 · Answer

Hi buneuel,

Can you throw some light on solving efficietly equations like these

(p-4) (p-3) < 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.

parassagi · Answer

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=Revenue-Cost=p(6 - p)-(12-p).

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:

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.

Thanks !

joyandbliss · Answer

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, 6-p 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.

BillyZ · Answer

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?

Bunuel · Answer

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 .

Mansoor50 · Answer

how do we find out the level of difficulty (600 level or 700 level) of this question?

thanks

Bunuel · Answer

The difficulty level of a question is calculated automatically based on the timer stats from the users which attempted the question. So our stats say that it's 700 level.

Somieboy · Answer

What if the options have 98/100. having picked the range 3 and 4, we know that the company will make a profit at and inside these values, so 1- 2/100=98/100. what are your views on this?

Vinayak1996 · Answer

Why isn't option C the answer.

I substituted the value for p in both manufacturing cost and revenue. Took the range for value p as 12<=p<0(as p should be a positive number and 0 is not a positive number).Got 4 values for which the manufacturing cost > revenue(hence loss). ie when p = 1,2,4,5.So shouldn't the answer be 4/100 or 1/25?

Please advice.

Kinshook · Answer

Given: 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).

Asked: 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?

Selling price = p dollars/unit
p<100
Monthly Revenues = 1000p(6-p) dollars
Monthly Manufacturing cost = 1000(12-p) dollars
Units sold = 1000(6-p)
Profits = Revenues - Costs = 1000p(6-p) - 1000(12-p) = 1000(6p - p^2-12+p) = 1000(7p - p^2 - 12) = - 1000(p-3)(p-4)
Profits < 0 if p>4 or p<3
Profits > 0 if 3<p<4
Since 0<p<100

Probability that the company will NOT see a profit on sales of Product X in the first month of sales = 99/100

IMO D

MissionAdmit · Answer

Between 3.1 and 3.99 there are 99 units not 100. P does not include value of extremities 3 and 4. Can you please help me understand this

Bunuel · Answer

Where do 3.1 and 3.99 come from? Why 3.1? Why not 3.0000001 or 3.000000000000000000000000001? Anyway, it's as simple as this, the distance between 3 and 4 is 1 unit and the distance between 0 and 100 is 100 units, so P = 1/100.

SudsMeister · Answer

p = (0,100] i.e. it is a positive value that is more than 0 but less than 100. So, 100 units = distance between 0 (not inclusive) and 100 (inclusive).

If we were to be pedantic and divide p in 100 units, then shouldn't the 1st unit = (0,1] i.e. the distance between 0 (not inclusive) and 1 (inclusive) is 1 unit, while the 2nd unit = (1,2] and so on, the 100th unit = (99,100].

So, if p = (3,4) then isn't the distance between 3 (not inclusive) and 4 (not inclusive) < 1 unit? For it to be 1 unit, either of 3 or 4 needs to be inclusive.

Let's say distance between (3,4) is 1 unit. Then, the distance between (0,1) = 1 unit = (1,2) = ..... = (99,100). In all these cases the positive integers aren't considered, so 100 integers are missing in the total 100 units. Won't that mean that either our definition of 1 unit is wrong, or we need to increase the denominator by some value as 100 units doesn't include all possible outcomes of p (positive integers are missing).

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

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

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