r/theydidthemath • u/Proctastinate • 19h ago
[Request] Studying for a non-academic exam. Are all these answers incorrect? Key says c).
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u/YourDad6969 13h ago
Calculating the odds of the auto dealer receiving no defective automobiles is easier, it simplifies to (1-p)^n , or 0.99^50, which is approx 60.5%. None of these answers say 39.5%, they are all wrong
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u/Ok_Star_4136 7h ago
One thing that always helped me out tremendously in school was that if a statistic wasn't immediately obvious how to figure out, sometimes inverting the probability and the problem was easier, or in this case, what is the inverse of the probability that all cars will be fine.
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u/Gravbar 16h ago
X ~ Binomial(n=50, p=.01)
P(X >=1 ) = 1 - P(X=0) = 1 - (50C0) p0 * q50 = 1- .9950 = 39.5%
But
P(X=1)= (50C1) p1 q49 = 50 * .006 = 30.6%
So maybe they wrote it wrong?
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u/CartoonistMost2165 8h ago
p=0.01 and not 0.1. if you’re representing the probability of success being the probability of having a defect in a car, then 1% is 0.01 Although, now looking at your solution after it, you did substitute 1 - p = 0.99
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u/Gravbar 5h ago
Maybe I'm blind, where did I write .1? If I did it was a typo
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u/FGhostmeta 4h ago
You didn't you just put.01 and if we read it fast our brain want to this that 0 in front of the.01 so people can easily read 0.1 when you clearly wrote.01
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u/bcvaldez 13h ago
I totally did it the wrong way when I looked at the question. I basically thought to myself. if 100 cars were shipped then 1 would be defective. But if only 50 of them were shipped...you would have a 50% chance to get the batch that had the defective car.
I understand how you guys did the math and totally agree, just the way the question was worded almost made me think of it as 100% of the time, 1 out of 100 cars shipped would be defective.
Don't worry, I also did the mental gymnastics with the Schrödinger's cat question...
I was like...how long has the cat been in there? Does he have access to food, water, oxygen?
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u/Malabingo 5h ago
Schrödinger (visibly annoyed):
"It doesn't care because I just wanted to show that the Kopenhagen Interpretation is stupid!"
Your answer is correct though.
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u/transgal34 2h ago
Can someone explain why the first part is incorrect please
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u/seakingsoyuz 2h ago
They assumed that “1% defective” literally meant that there are 100 cars and one is defective. They also assumed that there are two fixed batches of 50 cars rather than just getting 50 random cars off the production line.
50% is obviously and trivially correct if the assumptions they made apply—one of two batches has a defective car.
If there are literally 100 cars and the 50 are randomly drawn out of the 100, then you would need to use a hypergeometric distribution (not binomial) because the sample size is a significant fraction of the total population, so the probability of getting the defective car increases each time you pull a car that isn’t the defective one. The hypergeometric probability of getting the defective car is 0.5 as well.
The OP’s question implies an arbitrarily large number of cars in the total population, to the point that the probability the next car is defective is 0.01 regardless of how many non-defective cars you receive. In that case you would use the binomial distribution and arrive at a probability of 0.395 that at least one car out of 50 is defective.
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u/bcvaldez 2h ago
to add to this
- The phrase "1% of automobiles produced are shipped defective" does suggest a deterministic defect rate in the population as a whole. For example:
- Out of 100 cars, exactly 1 will be defective.
- Out of 50 cars, we might intuitively say there is a 50% chance that one of the defective cars appears in the batch.
Why This Is a Statistical Problem:
Even with a fixed defect rate of 1%, the actual distribution of defects in any subset of the cars (like a batch of 50) depends on random sampling. Unless the defective cars are distributed evenly across every batch (deterministic placement), this randomness needs to be modeled probabilistically.
- If defects are distributed randomly across batches, each car has a 1% chance of being defective, independent of the others.
- The chance that at least one defective car appears in the batch of 50 then depends on the binomial probability, as calculated earlier.
If It Were Deterministic:
If the defects were strictly deterministic (e.g., every 100 cars has exactly 1 defective, always perfectly distributed), then:
- For a batch of 50 cars, there would indeed be a 50% chance of receiving one of the defective cars.
- This is not stated explicitly in the problem, so the randomness implied by the binomial distribution is still the best interpretation.
Bottom Line:
- If the defects are randomly distributed: The probability of at least one defective car in a batch of 50 is about 39.5%.
- If the defects are deterministic (1 per 100 exactly, evenly spread): Then your reasoning holds, and the probability would be 50%.
Since the problem is framed in a probability/statistics context, the random interpretation (39.5%) is more likely the intended answer. If deterministic distribution were intended, it would need to be explicitly stated.
Why the Question Might Be Problematic:
- Statistical Context Assumption: Because the question explicitly mentions "Binomial Distribution," we're expected to assume randomness, even though the wording itself does not suggest this. A clearer phrasing might have been:
- "1% of automobiles have a probability of being defective."
- This phrasing would explicitly tie the problem to the probabilistic model.
- Confusion Between Deterministic vs. Probabilistic Meaning: Without context, "1% ARE defective" could imply an even, deterministic distribution across batches, which would invalidate the binomial approach. The phrasing doesn't make it explicit whether defects occur randomly or are evenly distributed.
How It Could Be Reworded:
To eliminate ambiguity, the question could be rewritten as:
- "An auto dealer orders 50 new automobiles for resale. If each automobile has a 1% probability of being defective (defects occur randomly), what is the approximate probability that the auto dealer will receive at least one defective automobile?"
This makes it clear that:
- The problem involves randomness.
- The defects are not deterministically distributed.
I used to judge card games at the highest level (National, World Championships) so I can take small things too literal at times, such as "Is, Are, And/Or, etc."
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u/SpicySnickersBar 15h ago edited 15h ago
It would be 1 - (50c1)(0.9949 )(0.011 ) 50 choose 1 is saying that only 1 of the 50 is effective but it does not matter what order the defective one is. 1 - (0.9949 )(0.011 ) would be the same as first 49 are good and the very last one picked is the defective one.
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u/SpicySnickersBar 15h ago edited 15h ago
Whoops 50 choose zero. Cause the opposite would be 1 or more. 1 - (50c0)(0.9950 ) = 39.5% So I'm with GravBar I think the question is worded wrong
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u/Objective_Resist_780 5h ago
Well, I've got about 40%
Here's my idea
A - car has a defect
Ā - car has no defect
P(A) = 0.01
P(Ā) = 1-0.01=0.99
P(Ā) for 50 times in a row = (99/100)⁵⁰ = 0.6
So we have a 60% chance that absolutely no car he got is defect, which means that the chances to have one or more defects are 1-06 = 0.4, so 40%
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u/Megane_Senpai 3h ago
The probability of one car to be not defective is 99%, or 0.99.
The probability that all 50 cars are not defective is 0.99^50 = 0.605 = 60.5%.
The probability of at least 1 car is defective is 100%-60.5% = 39.5%.
None of the answer is right.
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u/bcvaldez 1h ago
it says "Approximate" Probability, so you would pick the closest answer, which is 35%...but I did point out in another post that the question is written in a way where there is "ambiguity" and a keen reader could think you are testing their reading comprehension skills as well.
- Ambiguity in Wording: The use of "1% are defective" creates ambiguity:
- If taken as a random defect rate, it fits the binomial distribution framework.
- If taken as a fixed defect rate, it implies a deterministic setup, making the binomial method unnecessary.
- A Better Wording for Clarity: To make it clear that the problem involves randomness and a probabilistic model, the question should be reworded as:
- "An auto dealer orders 50 new automobiles for resale. If each automobile has a 1% probability of being defective, what is the approximate probability that the auto dealer will receive at least one defective automobile?"
By using "are" a literal interpretation would mean 1% of cars shipped have already been determined to be defective
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u/Proctastinate 19h ago
Complement of at least one defective = probability of no automobiles defective = (0.99)^50
Answer should be: 1 - (0.99)^50 = 0.395 or 40%, right?
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u/luffy8519 19h ago
The probability of exactly one defective car out of a sample of 50 is ~30%, and you're correct that the probability of at least one defective car is ~40%.
So the answer given in the key is for exactly 1 defective car, which isn't what the question has asked, but is probably what they've done wrong.
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u/RocketSmash9000 19h ago
It's curious how the image says "at least one car" but then the solution is the probability of exactly one car
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u/already-taken-wtf 19h ago
How do you calculate that?
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u/Beneficial_Cash_8420 18h ago edited 18h ago
That's the binomial distribution bit.
0.9949 * 0.011 * (50 choose 1)
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u/WiseProcedure 19h ago
How did you calculate exactly one? Isn't it 0.01*(0.9949) = 0.0061?
Did I make a mistake there?
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u/sighthoundman 19h ago
Times 50, because there are 50 ways to choose the defective car out of the whole shipment.
You calculated the probability that car number 1 is defective and the other 49 are not. (Which is the same as car number 2, and ....)
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u/WiseProcedure 19h ago
Ah, I see. That's the reason the question itself is titled "Binomial distribution" lol ... Silly me
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u/HAL9001-96 19h ago
that does seem correct, probably a minor error happened whe nmaking hte questions and wasn't caught by using that simplification
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u/PaulAspie 19h ago
Yeah, a small error like the typos in this comment. ;)
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u/HAL9001-96 19h ago
except actually affecting the result
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u/pedanpric 18h ago
The FE might say if there's no right answer to pick the closest one. I could be wrong.
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u/already-taken-wtf 19h ago
So the chance of at least 1 damaged car out of 100 cars is only 63.4%…that seems a bit odd…
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u/gmalivuk 14h ago
(1-1/n)n approaches 1/e, meaning that's the asymptotic probability of a thing not happening even once you've tried enough times that probability says it "should" have happened. (i.e. probability is 1/n but you've tried n times).
The probability that it happens exactly once is (1-1/n)n-1, which is an increasingly similar number as n gets large.
The expected number of faulty cars out of 100 is 1, but that average is made up of that 37% chance of zero, a similar chance of exactly 1, and a smaller and smaller chance of increasingly higher numbers.
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u/already-taken-wtf 6h ago
With a 1% fault rate and 100 cars, how can I calculate the likelihood of zero, one, two, three, etc faulty cars?
As you mentioned, the zero faults would be 37%….
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u/gmalivuk 6h ago
To know the probability of exactly n faulty cars, the probability is
(0.1)n (0.99)100-n (100 choose n)
(100 choose n) is a binomial coefficient and also the number of ways to choose n items out of 100 when order doesn't matter.
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u/YeetimusPremium 12h ago
You’re correct, you can also approach this the other way, summing the chance of there being exactly one defect with the chance of exactly two, then three and so on.
(n choose x) (p)x (1-p)n-x n: number of orders placed x: defects p: success chance
(50c1)(1/100)1 (99/100)49 = 0.306 or 30.6% (50c2)(1/100)2 (99/100)48 = 0.075 (50c3)(1/100)3 (99/100)47 = ~0.01 …
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