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It's been a while /sci/. I don't have a clever one, but here you go:
>You've just visited the doctor and have been told that you have microscopic hematuria. As it turns out, microscopic hematuria occurs in 10% of all people, but in 100% of people with kidney cancer, which occurs in .0002% of all people. What are the chances, given what your doctor has told you, that you have kidney cancer?
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>>7706316
>riddle
>bayesian statistics
this is at least one of the more cleverly disguised homework threads, OP, i'll give you that.
you're still a faggot, though
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>>7706316
2 in 10,000 people have kidney cancer.
1000 in 10,000 people have microscopic hematuria, including you.
You have a 1 in 500 chance of having kidney cancer.
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>>7706338
It does seem like that now that I think about it because the riddle is too simple. Here's a harder one for you, faggot:
>Joe is a dude with many ducks. Joe went to the market to sell said many ducks, but can't count pass 100. He does, however, know this:
>If you divide the number of ducks by 1, there will be 1 duck left.
>If you divide the number of ducks by 2, there will be 1 duck left.
>If you divide the number of ducks by 3, there will be 1 duck left.
>If you divide the number of ducks by 4, there will be 1 duck left.
>If you divide the number of ducks by 5, there will be 1 duck left.
>If you divide the number of ducks by 6, there will be 1 duck left.
>If you divide the number of ducks by 7, there will be 1 duck left.
>If you divide the number of ducks by 8, there will be 1 duck left.
>If you divide the number of ducks by 9, there will be 1 duck left.
>If you divide the number of ducks by 10, there will be 1 duck left.
>If you divide the number of ducks by 11, there will be no duck left.
>How many ducks did Joe have?
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Hint is middle school Math, I think? And don't tell me again that this is a homework-guised riddle because I said middle school.
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>>7706379
>>If you divide the number of ducks by 1, there will be 1 duck left.
Wait, whut?
d/1=x, remainder 1?
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A: You have microscopic hematuria
B: You have kidney cancer
P(B|A) = P(A AND B) / P(A) = P(A|B) P(B) / P(A) = 100% * .002% / 10% = 0.02%
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>>7706379
>if you divide by 1 you will have 1 remainder
you fucked up already man
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>>7706418
Fuck, I messed up. Delete "If you divide the number of ducks by 1, there will be 1 duck left." from the riddle. My bad anon.
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>>7706423
>.002
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>>7706435
no im gonna call you a faggot, instead. faggot.
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>>7706379
121. that is retard tier easy, man
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>>7706379
At first I read "Joe is a man with many dicks".
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>>7706449
>121. that is retard tier easy, man
Something's retard tier.
121/7 = 17, remainder 2
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>>7706379
Not enough information. He could have 11 (you dont specify that he has more than 100, just that he cant count past 100) or any multiple of 11 using primes larger than 11.
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>>7706379
A solution is 1771561. Use multiples and the hints from the modulus results.
Joe has a fuck ton of ducks.
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>>7706379
25,201 is the smallest number possible
What do I win?
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>>7706470
Finally someone got it. There are more answers though in case someone notice the pattern of the riddle. Infinite amount of answers even.
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>>7706461
>He could have 11
11 divided by 3 is 3, remainder 2
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>>7706472
Ducks.
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>>7706483
Thanks!
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>>7706472
-2519 is also a solution and smaller.
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>>7706379
>cleverly disguised chinese remainder theorem problem
stop this
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>>7706638
Thread posts: 25
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Thread images: 4