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Anyone want to take a stab at this stats problem?

M

ManBearPig

Diamond Member
An automobile manufacturer has three factories: A, B, and C. They produce 50%, 30%, and 20%, respectively, of a specific model of car. Thirty percent of the cars produced in factory A are white, 40 % of those produced in factory B are white, and 25 % produced in factory C are white. If an automobile produced by the manufacturer is selected at random, find the probability that it is white.

although i dont expect anyone to do it, i figured i'd try.

BONUS: Imagine your socks drawer contains 11 identical red socks and 8 identical black socks. Suppose also that you have overslept and are running late for school. In your morning rush you choose 2 socks at random.

What is the probability that you get a pair of red socks?
What is the probability that you get 2 unmatched socks?
 
Poorly worded question. Almost sounds like you could ignore the specific model info and just base probability off the second set of percentages.
 
Originally posted by: Mo0o
15+12+5=32

32/100 cars produced by the company are white. Thus 32%.
Click to expand...

That's what I initially thought, but it says *a* automobile selected at random, not of the particular model. Hence my thought that the problem is poorly worded.
 
Originally posted by: Spartan Niner
Originally posted by: Mo0o
15+12+5=32

32/100 cars produced by the company are white. Thus 32%.
Click to expand...

That's what I initially thought, but it says *a* automobile selected at random, not of the particular model. Hence my thought that the problem is poorly worded.
Click to expand...

Oh that's dumb if that difference matters. It woudl be impossible to solve without figuring out how many cars are produced in each factory
 
Originally posted by: Spartan Niner
Originally posted by: Mo0o
15+12+5=32

32/100 cars produced by the company are white. Thus 32%.
Click to expand...

That's what I initially thought, but it says *a* automobile selected at random, not of the particular model. Hence my thought that the problem is poorly worded.
Click to expand...

if that is the intention then this problem is impossible to solve.
 
thats kind of what i was thinking too...but i figured the first set of percentages would affect the pick somehow.

update: if anyone wants to have some bonus fun, theres another question up!
 
Originally posted by: Kazaam
thats kind of what i was thinking too...but i figured the first set of percentages would affect the pick somehow.

update: if anyone wants to have some bonus fun, theres another question up!
Click to expand...

But the first set of percentages do affect the pick. Otherwise its impossible to do the question in teh first place.
 
Are you sure they didn't just screw up the wording and mean to refer to the specific model? If they didn't, it can't be solved without some additional information.
 
Originally posted by: Cheesetogo
Are you sure they didn't just screw up the wording and mean to refer to the specific model? If they didn't, it can't be solved without some additional information.
Click to expand...

As I noted it's probably poorly worded, and as MoOo noted, you can pretty much assume that the way he solved it is the correct way.

Because otherwise you're missing information...
 
Yup. I'm going to assume that the "impossible" warning was to motivate people to do it, so they can show off how smart they are since the alternative means that you are not very good at math.

Sorry that sounds a bit harsh. It's one thing to not know how to do it, but to label those two problems as impossible is another.
 
Please tell me this is not a college level stats class?

My god what I would give to have had this for Stat class. I dropped mine the first time as I thought it be like this. The 2nd time I got the 2nd highest grade in the class.
 
Originally posted by: Mo0o
Matching red pair: (11/19) * (10/18)
Unmatched pair: (11/19) * (8/18) + (8/19) * (11/18)
Click to expand...

I disagree on the Unmatched Pair answer, in which you sum two terms. There are, I agree, two ways to get to an unmatched pair. If red is your first pick (11/19) then the second one MUST be a black (8/18), probability is the product as you say.
But if your first pick was black (8/19) then second MUST be red (11/18) and product of these two is the same as the first possibility. However, you do NOT need to add the two products. Basically, no matter which color your first pick is, the probability of an unmatched pair is the same (0.2573).

If you want it in harder math, the probability of the first situation (red is your first sock) is 11/19. The probability of the second (black is first pick) IS 8/19. But in each case, the probability of getting an ummatched pair is the same, 0.2573. So the total probability of an unmatched pair over all scenarios (there being only two) is (11/19 x 0.2573) + (8/19 x 0.2573) = 19/19 x 0.2573, or plain 0.2573.
 
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